Module 4 : Algebraic Processes

Module 4 : Algebraic Processes
Module 4
Junior Secondary
Mathematics
Algebraic Processes
Science, Technology and Mathematics Modules
for Upper Primary and Junior Secondary School Teachers
of Science, Technology and Mathematics by Distance
in the Southern African Development Community (SADC)
Developed by
The Southern African Development Community (SADC)
Ministries of Education in:
• Botswana
• Malawi
• Mozambique
• Namibia
• South Africa
• Tanzania
• Zambia
• Zimbabwe
In partnership with The Commonwealth of Learning
COPYRIGHT STATEMENT
The Commonwealth of Learning, October 2001
All rights reserved. No part of this publication may be reproduced, stored in a retrieval
system, or transmitted in any form, or by any means, electronic or mechanical, including
photocopying, recording, or otherwise, without the permission in writing of the publishers.
The views expressed in this document do not necessarily reflect the opinions or policies of
The Commonwealth of Learning or SADC Ministries of Education.
The module authors have attempted to ensure that all copyright clearances have been
obtained. Copyright clearances have been the responsibility of each country using the
modules. Any omissions should be brought to their attention.
Published jointly by The Commonwealth of Learning and the SADC Ministries of
Education.
Residents of the eight countries listed above may obtain modules from their respective
Ministries of Education. The Commonwealth of Learning will consider requests for
modules from residents of other countries.
ISBN 1-895369-65-7
SCIENCE, TECHNOLOGY AND MATHEMATICS MODULES
This module is one of a series prepared under the auspices of the participating Southern
African Development Community (SADC) and The Commonwealth of Learning as part of
the Training of Upper Primary and Junior Secondary Science, Technology and
Mathematics Teachers in Africa by Distance. These modules enable teachers to enhance
their professional skills through distance and open learning. Many individuals and groups
have been involved in writing and producing these modules. We trust that they will benefit
not only the teachers who use them, but also, ultimately, their students and the communities
and nations in which they live.
The twenty-eight Science, Technology and Mathematics modules are as follows:
Upper Primary Science
Module 1: My Built Environment
Module 2: Materials in my
Environment
Module 3: My Health
Module 4: My Natural Environment
Junior Secondary Science
Module 1: Energy and Energy
Transfer
Module 2: Energy Use in Electronic
Communication
Module 3: Living Organisms’
Environment and
Resources
Module 4: Scientific Processes
Upper Primary Technology
Module 1: Teaching Technology in
the Primary School
Module 2: Making Things Move
Module 3: Structures
Module 4: Materials
Module 5: Processing
Junior Secondary Technology
Module 1: Introduction to Teaching
Technology
Module 2: Systems and Controls
Module 3: Tools and Materials
Module 4: Structures
Upper Primary Mathematics
Module 1: Number and Numeration
Module 2: Fractions
Module 3: Measures
Module 4: Social Arithmetic
Module 5: Geometry
Junior Secondary Mathematics
Module 1: Number Systems
Module 2: Number Operations
Module 3: Shapes and Sizes
Module 4: Algebraic Processes
Module 5: Solving Equations
Module 6: Data Handling
ii
A MESSAGE FROM THE COMMONWEALTH OF LEARNING
The Commonwealth of Learning is grateful for the generous contribution of the
participating Ministries of Education. The Permanent Secretaries for Education
played an important role in facilitating the implementation of the 1998-2000
project work plan by releasing officers to take part in workshops and meetings and
by funding some aspects of in-country and regional workshops. The Commonwealth of
Learning is also grateful for the support that it received from the British Council (Botswana
and Zambia offices), the Open University (UK), Northern College (Scotland), CfBT
Education Services (UK), the Commonwealth Secretariat (London), the South Africa
College for Teacher Education (South Africa), the Netherlands Government (Zimbabwe
office), the British Department for International Development (DFID) (Zimbabwe office)
and Grant MacEwan College (Canada).
The Commonwealth of Learning would like to acknowledge the excellent technical advice
and management of the project provided by the strategic contact persons, the broad
curriculum team leaders, the writing team leaders, the workshop development team leaders
and the regional monitoring team members. The materials development would not have
been possible without the commitment and dedication of all the course writers, the incountry reviewers and the secretaries who provided the support services for the in-country
and regional workshops.
Finally, The Commonwealth of Learning is grateful for the instructional design and review
carried out by teams and individual consultants as follows:
•
Grant MacEwan College (Alberta, Canada):
General Education Courses
•
Open Learning Agency (British Columbia, Canada):
Science, Technology and Mathematics
•
Technology for Allcc. (Durban, South Africa):
Upper Primary Technology
•
Hands-on Management Services (British Columbia, Canada):
Junior Secondary Technology
Dato’ Professor Gajaraj Dhanarajan
President and Chief Executive Officer
ACKNOWLEDGEMENTS
The Mathematics Modules for Upper Primary and Junior Secondary Teachers in the
Southern Africa Development Community (SADC) were written and reviewed by teams
from the participating SADC Ministries of Education with the assistance of The
Commonwealth of Learning.
iii
CONTACTS FOR THE PROGRAMME
The Commonwealth of Learning
1285 West Broadway, Suite 600
Vancouver, BC V6H 3X8
Canada
National Ministry of Education
Private Bag X603
Pretoria 0001
South Africa
Ministry of Education
Private Bag 005
Gaborone
Botswana
Ministry of Education and Culture
P.O. Box 9121
Dar es Salaam
Tanzania
Ministry of Education
Private Bag 328
Capital City
Lilongwe 3
Malawi
Ministry of Education
P.O. Box 50093
Lusaka
Zambia
Ministério da Eduação
Avenida 24 de Julho No 167, 8
Caixa Postal 34
Maputo
Mozambique
Ministry of Education, Sport and
Culture
P.O. Box CY 121
Causeway
Harare
Zimbabwe
Ministry of Basic Education,
Sports and Culture
Private Bag 13186
Windhoek
Namibia
iv
COURSE WRITERS FOR JUNIOR SECONDARY MATHEMATICS
Ms. Sesutho Koketso Kesianye:
Writing Team Leader
Head of Mathematics Department
Tonota College of Education
Botswana
Mr. Jan Durwaarder:
Lecturer (Mathematics)
Tonota College of Education
Botswana
Mr. Kutengwa Thomas Sichinga:
Teacher (Mathematics)
Moshupa Secondary School
Botswana
FACILITATORS/RESOURCE PERSONS
Mr. Bosele Radipotsane:
Principal Education Officer (Mathematics)
Ministry of Education
Botswana
Ms. Felicity M Leburu-Sianga:
Chief Education Officer
Ministry of Education
Botswana
PROJECT MANAGEMENT & DESIGN
Ms. Kgomotso Motlotle:
Education Specialist, Teacher Training
The Commonwealth of Learning (COL)
Vancouver, BC, Canada
Mr. David Rogers:
Post-production Editor
Open Learning Agency
Victoria, BC, Canada
Ms. Sandy Reber:
Graphics & desktop publishing
Reber Creative
Victoria, BC, Canada
v
TEACHING JUNIOR SECONDARY MATHEMATICS
Introduction
Welcome to Algebraic Processes, Module 4 of Teaching Junior Secondary Mathematics!
This series of six modules is designed to help you to strengthen your knowledge of
mathematics topics and to acquire more instructional strategies for teaching mathematics in
the classroom.
The guiding principles of these modules are to help make the connection between
theoretical maths and the use of the maths; to apply instructional theory to practice in the
classroom situation; and to support you, as you in turn help your students to apply
mathematics theory to practical classroom work.
Programme Goals
This programme is designed to help you to:
•
strengthen your understanding of mathematics topics
•
expand the range of instructional strategies that you can use in the mathematics
classroom
Programme Objectives
By the time you have completed this programme, you should be able to:
•
develop and present lessons on the nature of the mathematics process, with an
emphasis on where each type of mathematics is used outside of the classroom
•
guide students as they work in teams on practical projects in mathematics, and help
them to work effectively as a member of a group
•
use questioning and explanation strategies to help students learn new concepts and
to support students in their problem solving activities
•
guide students in the use of investigative strategies on particular projects, and thus to
show them how mathematical tools are used
•
guide students as they prepare their portfolios about their project activities
vi
How to work on this programme
As is indicated in the programme goals and objectives, the programme provides for you to
participate actively in each module by applying instructional strategies when exploring
mathematics with your students and by reflecting on that experience. In other words, you
“put on your student uniform” for the time you work on this course.
Working as a student
If you completed Module 1…did you in fact complete it? That is, did you actually do the
various Assignments by yourself or with your students? Did you write down your answers,
then compare them with the answers at the back of the module?
It is possible to simply read these modules and gain some insight from doing so. But you
gain far more, and your teaching practice has a much better chance of improving, if you
consider these modules as a course of study like the courses you studied in school. That
means engaging in the material—solving the sample problems, preparing lesson plans when
asked to and trying them with your students, and so on.
To be a better teacher, first be a better student!
Working on your own
You may be the only teacher of mathematics topics in your school, or you may choose
to work on your own so you can accommodate this programme within your schedule.
Module 1 included some strategies for that situation, such as:
1. Establish a regular schedule for working on the module.
2. Choose a study space where you can work quietly without interruption.
3. Identify someone whose interests are relevant to mathematics (for example, a science
teacher in your school) with whom you can discuss the module and some of your ideas
about teaching mathematics. Even the most independent learner benefits from good
dialogue with others: it helps us to formulate our ideas—or as one learner commented,
“How do I know what I’m thinking until I hear what I have to say?”
It is hoped that you have your schedule established, and have also conversed with a
colleague about this course on a few occasions already. As you work through Module 4,
please continue!
Resources available to you
Although these modules can be completed without referring to additional resource
materials, your experience and that of your students can be enriched if you use other
resources as well. There is a list of resource materials for each module provided at the
end of the module.
vii
ICONS
Throughout each module, you will find the following icons or symbols that alert you to a
change in activity within the module.
Read the following explanations to discover what each icon prompts you to do.
Introduction
Rationale or overview for this part of the course.
Learning Objectives
What you should be able to do after completing
this module or unit.
Text or Reading Material
Course content for you to study.
Important—Take Note!
Something to study carefully.
Self-Marking Exercise
An exercise to demonstrate your own grasp of
the content.
Individual Activity
An exercise or project for you to try by yourself
and demonstrate your own grasp of the content.
Classroom Activity
An exercise or project for you to do with or
assign to your students.
Reflection
A question or project for yourself—for deeper
understanding of this concept, or of your use of it
when teaching.
Summary
Unit or Module
Assignment
Exercise to assess your understanding of all the
unit or module topics.
Suggested Answers to
Activities
Time
Suggested hours to allow for completing a unit or
any learning task.
Glossary
Definitions of terms used in this module.
viii
CONTENTS
Module 4: Algebraic processes
Module 4 – Overview ...................................................................................................... 2
Unit 1: Learning and teaching algebra .......................................................................... 4
Section A1 What is algebra? ........................................................................................ 5
Section A2 Historical development of algebra.............................................................. 5
Section A3 Aims of learning algebra............................................................................ 7
Section A4 Pupils’ learning difficulties with algebra ................................................... 15
Answers to self mark exercises Unit 1....................................................... 19
Unit 2: Expansion of expressions................................................................................... 20
Section A1 Collecting like terms................................................................................. 21
Section A2 Expansion of the product of a monomial and a binomial ........................... 25
Section A3 Expansion of the product of two binomials ............................................... 31
Section B Models for expansions .............................................................................. 37
Section B1 Multiplication table algorithm for expansion............................................. 37
Section B2 Expansion using concrete manipulatives (algebra tiles)............................. 39
Section B3 The FOIL algorithm.................................................................................. 44
Section B4 Illustrating the identity (a + b)(a – b) = a2 – b2 with paper models ........... 45
Section B5 The product (a + b)(a – b) from generalised arithmetical pattern............... 46
Answers to self mark exercises Unit 2....................................................... 49
Unit 3: Factorisation of expressions .............................................................................. 56
Section A Factorising................................................................................................ 58
Section A1 Factorising expressions with one common factor ...................................... 58
Section A2 Outline of a worksheet for pupils on factorisation ..................................... 61
Section B Factorising quadratic expressions.............................................................. 65
Section B1 Factorising a2 – b2 using a pattern approach............................................. 65
Section B2 Factorising x2 + ax + b.............................................................................. 68
Section C Fallacies ................................................................................................... 77
Answers to self mark exercises Unit 3....................................................... 79
Unit 4: Binomial theorem and factor theorem.............................................................. 85
Section A1 Product of polynomials. ............................................................................ 86
Section A2 Binomial expansion .................................................................................. 87
Section B1 Investigating Pascal’s triangle................................................................... 98
Section B2 Pascal’s pyramid....................................................................................... 99
Section C1 Dividing polynomials ..............................................................................102
Section C2 Remainder theorem..................................................................................104
Section C3 Factor theorem.........................................................................................106
Answers to self mark exercises Unit 4......................................................109
References
................................................................................................................115
Glossary
................................................................................................................116
Appendices
Appendix 1 Biographical data on Mohammed ibn-Musa al-Khwarizmi .....................118
Appendix 2 Biographical data on Diophantus ............................................................123
Appendix 3 Biographical data on Viète......................................................................129
Module 4
1
Algebraic processes
Module 4
Algebraic processes
Introduction to the module
This module deals with algebra. Algebra deals with generalised numbers
(variables) and studies the relationships among these changing quantities.
Algebra can be a useful tool to describe and model real life situations. It
requires a certain level of abstract thinking and can be introduced to learners at
the time they start to move from the concrete thinking level to a more abstract
level of thinking (14 – 16 years). If algebra is introduced too early or at a time
when the pupil does not really need it to answer the questions he/she is
working on, it might become simply manipulation of letters without meaning.
Aim of the module
The module aims at:
(a) justifying the inclusion of algebra in the mathematics curriculum
(b) practising pupil-centred teaching methods
(c) raising your appreciation for use of games and investigations in pupils’
learning of mathematics
(d) reflection on your present practice in the teaching of algebra
(e) making you aware of a variety of learning aids that can be used in the
teaching of algebra
(f) extending your knowledge on expansion and factorisation
Structure of the module
This module first looks into the purpose of algebra at the lower secondary
level (Unit 1) before moving to the topics traditionally covered: Expansion
(Unit 2) and factorisation (Unit 3). The emphasis is on looking at activities
for the classroom to make algebra more meaningful and understandable for
pupils. A pupil centred approach is encouraged throughout the module by
looking at activities that can be set to pupils to involve them making sense of
abstract concepts.
The final unit (Unit 4) covers content to extend your knowledge in expansion
(binomial expansion) and factorisation of polynomials (factor theorem).
Unit 4 content is normally not part of Junior Secondary mathematics
teaching, and is presented without the emphasis of classroom activities or
practical meaning.
The module does not yet make use of the current available technology:
symbolic-manipulation algebraic calculators. This new technology is presently
changing dramatically the study of algebra in the western world: from
symbolic paper-and-pencil manipulation towards conceptual understanding,
symbol sense and mathematical modelling of real life problems.
Module 4
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Algebraic processes
Objectives of the module
When you have completed this module you should be able to create with
confidence, as your own understanding and knowledge has been enhanced, a
learning environment for your pupils in which they can:
(i) acquire, with understanding, the knowledge of expansion and
factorisation of algebraic expressions
(ii) use paper models to illustrate identities
(iii) use algebraic tiles to model expansion and factorisation
(iv) investigate algebraic fallacies
Prerequisite knowledge
This module assumes you have covered Module 1, especially the unit on
sequences and finding the general rule for the nth term in a sequence.
Module 4
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Algebraic processes
Unit 1: Learning and teaching algebra
Introduction to Unit 1
The traditional, formal approach to teaching algebra is to look at algebra as a
purely mathematical discipline with no emphasis on relating algebra to day
to day situations. Pupils have ‘done algebra’ without really seeing a need for
it, resulting in numerous common errors. The current trend is to place
algebra, and mathematics in general, in a context. The emphasis is on doing
mathematics, recognising connections and modelling real life situations.
Algebra is a powerful tool in the hands of the learners provided they
understand the uses and the limitations of the tool.
Purpose of Unit 1
In this unit you will look at what algebra is and the stages in the historical
development of algebra. You also learn about pupils’ main problems in the
learning of algebra. This is important, because being aware of pupils’
problems might guide you to look for ways to present the content to pupils in
a way so as to avoid the common errors. You will also reflect on the reasons
for including algebra as a topic in the learning of mathematics at the lower
secondary level.
Objectives
When you have completed this unit you should be able to:
•
explain what algebra is
•
state the three main stages in the historical development of algebra
•
justify the inclusion of algebra in the mathematics curriculum
•
list and exemplify pupils’ main problems in the learning of algebra
•
justify the use of manipulatives and models in the learning of algebra
Time
To study this unit will take you about 12 hours. Trying out and evaluating
the activities with your pupils in the class will be spread over the weeks you
have planned to cover the topic.
Module 4: Unit 1
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Learning and teaching algebra
Unit 1: Learning and teaching algebra
Section A1: What is algebra?
The word algebra is from Arabic origin, it is part of the title of a book
written by Mohammed ibn-Musa al-Khwarizmi (approximately 830 AD).
The book’s title was ‘Hisab al-jabr w’al-muqabalah’. Al-jabr means
something like ‘completion’ or ‘restoration’ and refers to the transposition of
subtracted terms to the other side of an equation. Muquabalah refers to
‘reduction’ or ‘balancing,’ that is cancelling out like terms on opposite sides
of the equal sign in equations.
Al-Khwarizmi’s work is based on Hindu work (Brahmagupta, approximately
600) and works from ancient Mesopotamian origin.
Algebra deals with generalised numbers (variable quantities) and studies the
relationships (functions) among these changing quantities. It is sometimes
referred to as ‘generalised arithmetic,’ as rules and patterns in arithmetic can
lead to more general algebraic expressions. Algebra is the study of general
expressions and general results. In arithmetic one might need to compute
(12 + 45)2; in algebra one explores (a + b)2. Algebra allows one to express
in a short form relationships that might otherwise need long sentences, as for
example the Pythagorean theorem.
Section A2: Historical development of algebra
The symbol system of algebra as we know it today has developed over the
ages. Three main stages can be distinguished:
I. The time before Diophantus (ca 250): the rhetoric stage.
Initially ordinary language was used to describe the solution to problems. No
symbols or special designed signs were used. Everything was described in
ordinary language.
II. The time from Diophantus (ca 250) to Viète (ca 1600): the abbreviation
or syncopated stage.
Diophantus was the first known person to start using symbolism (abbreviated
words) and is frequently referred to as the father of algebra. (History does
not mention a mother!) Diophantus is still remembered today for the
“Diophantine Equations”: any equation in (usually) several unknowns,
whose solutions are required to be integers, for example to find all pairs of
integers (x, y) that satisfy
(i) 2x – 3y = 4 or x2 + y2 = 25.
In this period mathematicians started to use abbreviations for unknown (but
specific) quantities. The main concern during this period was the solving of
equations, i.e., finding the value of the unknown quantity represented by a
letter(s).
For example in n + 5 = 13, with n representing a specific unknown, solving
the equations means finding the value of that unknown quantity.
Module 4: Unit 1
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Learning and teaching algebra
III. From Viète (1540 - 1603, also known as Vieta) to today: the abstract
stage.
As from the time of Viète, letters were used not only to represent unknown
quantities (as in the previous period) but letters were also used to represent
given quantities (used as parameters). This made it possible, among others,
to give general solutions to problems and to generalise patterns.
For example the solution to an equation can be x = 3t + 4, the nth term of a
sequence can be stated as un = 2n – 3 for n a natural number, the general
form of the equation of a line can be stated as ax + by + c = 0, here a, b, and
c are parameters (variables to distinguish special cases of the general
expression).
Unit 1, Assignment 1
1. Write a quarter page biographical data on each of the following
mathematicians:
(i) Mohammed ibn-Musa al-Khwarizmi
(ii) Diophantus
(iii) Viète
Information can be extracted from books on the History of Mathematics.
If you do not have access to such books you can find some material taken
from books in appendices 1, 2 and 3.
2. Using examples illustrate the three major stages in the historical
development of algebra.
3. Justify why you would include ‘History of Mathematics’ in the
teaching/learning of mathematics at lower secondary level.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 1
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Learning and teaching algebra
Section A3: Aims of learning algebra
The topics appearing in our syllabi are often taken for granted. We do not
question them, we just try to make the best of it by trying to find activities
for our students to cover the topic. With algebra, more than with other topics
in the syllabus, the questions “Why should pupils at the lower secondary
level learn about algebra? What is the use to them of being able to expand,
factorise, substitute, solve equations?” are to be faced. The pupils are asking
us as teachers. What is our response?
1. What do you consider the main reasons to include algebra in the
learning/teaching of mathematics?
2. Read the rational and general aims of the mathematics syllabus you use
in your school. Are the reasons you mentioned in 1 in line with the
rational and general aims stated in the syllabus?
3. List the content topics you consider core knowledge for pupils aged 12 15 and justify each topic. (Why should a pupil have knowledge about the
topic you listed?). In which form would you cover the topics you listed?
4. Compare the list you made in 3 with the topics the syllabus wants you to
cover with the pupils in each form level. Comment on differences (you
might feel that certain topics mentioned in the syllabus should be left out
or covered at another level or you might feel that a topic you consider
core knowledge is not included, etc).
5. Go through the algebra topics in the mathematics book you use in your
school in the classes 1, 2 and 3. How practical are the topics? How useful
to pupils in the world of work?
6. A pupil questions why he/she should learn expansion and factorisation of
algebraic expressions. What answer would you give the pupil?
When reading the following section refer to the notes you made in your
above ‘reflection’. See whether your own thinking is confirmed or
challenged by the following. If the statements below do not agree with your
own reflections critically look at both.
The mathematics syllabus for the forms 1 - 3 of the Junior Secondary School
in Botswana emphasises that the learning of mathematics should be
“geared towards developing qualities and skills needed for the world of
work”
How practical and useful is school algebra in the world of work and day to
day situations?
Most traditional mathematics books present algebra as nothing but
manipulation with ‘letters’. The algebra presented is not realistic, practical
and applicable. If the pupils cannot relate to A + B, and (a + b)(c + d), what
practical situation relates to such expressions? Where do they meet a(c + d),
apart from in a maths book? Little to no explicit use of algebra is required in
the majority of jobs.
Module 4: Unit 1
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Learning and teaching algebra
Formulas, using sometimes single letter variables, but more often expressed
in words, are widely used by technicians, craftsmen, clerical workers and
nurses.
For example:
(height of adult in cm) = (mass in kg) + 100;
(amount to pay in P) = (fixed rate in P) + (number of units used × unit price
in P);
Child’s dose of medicine =
Age × Adult dose
Age + 12
All that is required is the substitution of numbers in these formulae and
the appropriate use of a calculator in the computation. It is not necessary to
transform a formula (change of subject). Any form which is likely to be
required will be readily available or can be looked up in a manual.
In the world of work and day to day situations it is not necessary to expand
expressions (remove brackets), to collect like terms, to factorise (insert
brackets) or to solve equations. Yet all these do appear in most if not all
syllabi at this level.
It is good to be aware of this: you are to cover content that will be of no use
to more than half of the pupils. It is extremely hard to justify—at this
level—the value of using letters to represent variables, because pupils do not
see any need for it. This is partly due to the fact that the examples used DO
NOT need any use of letters.
Consider a question such as: solve for x the equation x + 2 = 5. The use of a
variable does not make much sense, any primary school child ‘knows’ what
to add to 2 to obtain 5, so why such a complicated way of writing that down?
Similarly in examples on substitution: find the value of a + b if a = 2 and
b = 3; what use do variables have in such cases? You are facing the task to
motivate pupils to engage in an activity which has no practical use to them
and which is hard to relate to every day work. It is no surprise that, to many
pupils, manipulation of letters (which are ill understood to be representing
variables or specific unknowns) becomes an isolated practice required to
pass an exam.
What are meaningful activities at junior secondary level in algebra?
It might be useful to express an abstract idea or relationship in an algebraic
form when this provides a concise way of writing down a structure which is
already recognised and is difficult to express neatly in words. However
imposition of an algebraic format or technique which is not seen to be
necessary and not related to any familiar situation cannot serve any useful
purpose.
Let’s look at examples.
1. In Cash Build customers can buy square tiles ( 50 cm by 50 cm) to make
paths. The manager noted that often a customer will come and say: “I
want to make a path two tiles wide around a rectangular region of 3 m by
4 m. How many tiles do I need?” Another customer will come with the
Module 4: Unit 1
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Learning and teaching algebra
same question but to surround a 3 m by 5 m, rectangular region. A third
customer will come with again different dimensions of the region to be
surrounded and might want the path to be three tiles wide.
2. On the bank of a river are 8 adults, 2 children and 1 very small rowboat.
The owner of the boat charges P0.25 for each crossing and the boat has
to end up at the same bank from which it started. The owner does not
charge for crossings in which he is involved himself. The group wants to
cross the river making as few crossings as possible. The boat can hold
2 children, or 1 child or 1 adult. All people in the group can row. A
‘crossing’ is moving from one bank to the opposite in either direction.
What has the owner to charge for the crossing of the 8 adults and 2
children?
Work the next exercise to find out how these problems lead to algebra. You
might want to use a problem solving approach: start with simple cases,
tabulate your results and look for a pattern. A systematic approach is needed.
Self mark exercise 1
1. The manager asks you to solve the problem in general. How many square
tiles (50 cm × 50 cm) are needed to surround a p m by q m rectangular
region with a path of s tiles wide? What relationship between p, q and s
would you present to the manager to find the number of tiles needed?
2. The owner of the boat has always counted the number of crossings, but
now he has asked you to give him a formula in which he can substitute
the number of adults and children in a party to find the amount he is to
charge for a party of a adults and c children. What formula would you
give to the owner of the boat?
Check your answers at the end of this unit.
Module 4: Unit 1
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Learning and teaching algebra
What is meaningful for 12 - 15 year old pupils to learn in algebra?
1. Pupils are to become users of algebra, not designers.
Only those aspects of algebra which will be of use to the pupils in their
future job, their day to day life or further study (mainly directed towards
NON science studies) should be included. Pupils should learn to use
algebra as they might meet it in those future situations. Seldom, if ever,
will they have to develop new formulas, tables, graphs or algebraic
techniques. What is required is to use a given formula, to read a table, to
interpret a graph etc.
2. Pupils have to learn to interpret tables, graphs, relationships and NOT to
MANIPULATE formulas. If manipulation is required it follows from the
interpretation, in order to support the interpretation. In the lower
secondary school the emphasis should be on interpretation of
relationships, graphs and tables as these appear in newspapers,
magazines and books.
If a pupil describes the general term of a pattern as 3n(n – 1) + 1 while
another pupil in the group comes up with the relation n3 – (n – 1)3 the
need arises to check whether or not the expressions express the very
same thing.
3. Pupils are to study general techniques instead of case specific
techniques. It is to be preferred that pupils learn a general approach
which is applicable in many situations than a narrow technique which is
applicable to one situation only.
For example, there is no need to restrict oneself to linear relationships,
for many relationships are nonlinear. Interpretation of aspects of a graph
can be practised on various types of graphs not necessarily linear.
The technique of solving equations by trial and improvement method
works for all equations, while techniques for solving linear equations
(cover up method, reverse flow method, ‘balance’ method) or quadratic
equations (formula, completing the square) are case specific. These
special techniques apply to special types of equations only.
Techniques are to be introduced only when there is a clear need for it.
For example when introducing solving of linear equations, using a
balance model, and starting with expressions such as x + 7 = 10 or
3x + 1 = 22 does not make much sense as the pupil can ‘see’ the
answer immediately. Insisting that pupils write down all the steps
x + 7 = 10
x + 7 – 7 = 10 – 7
x=7
hardly makes sense and pupils don’t see the need of it.
Module 4: Unit 1
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Learning and teaching algebra
There is no motivation nor justification for learning a new technique
unless the pupil meets with a problem which is ‘difficult’ or impossible
to solve without a new technique. It is only after it is seen that the
techniques available are insufficient to answer the question that the need
to look for and introduce other techniques arises.
This is a general principle in the learning of mathematics (and other
subjects): before introducing new concepts and techniques a need for it is
to be created. The pupils must feel the need to extend their knowledge
base with new ideas because a question they like to answer cannot be
answered with the knowledge they have. The manager of Cash Build and
the rowboat owner (in self mark exercise 1) felt a need for algebraic
expressions so they could quickly find the results (number of tiles, price
to charge) they wanted.
4. Pupils are to learn to use algebra in order to describe in a precise way
relationships between variables, for example, the relationship between
length and mass, height and arm span, number of days after planting and
the height of the maize plant, general patterns in arithmetic etc.
When an algebra course starts with the use of letters as mathematical
objects and proceeds to operate on these objects, it remains obscure to
pupils what these objects represent and how they can be related (if at all)
to realistic situations. Links between arithmetic and algebra are absent.
The approach since the seventies is to construct algebra from generalised
arithmetic (starting with what pupils do know and are familiar with). The
first step is using word variables before the move to letter variables is
made. Pupils devise and explore situations, recognise and make
generalisations from patterns and use algebraic notations. The
‘situations’ are not to be realistic in the sense that they do occur in a
work situation. Pupils, in general, are fascinated by patterns of tiles,
matches, circles, dots, cubes, etc. and interested in finding expressions
for the general case. The number of patterns one can design is nearly
limitless.
At all times abstract manipulation of expressions without a context is to
be avoided as this will lead only to “fruit salad or cattle post algebra”—
the adding of apples and bananas or cows and goats, using letters to label
objects instead of using them as variables.
The following exercise illustrates how algebraic expressions follow
‘naturally’ from an arithmetic pattern of numbers.
Module 4: Unit 1
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Learning and teaching algebra
Self mark exercise 2
Objective: To generalise from a numerical pattern to an algebraic (word)
formula.
1. Here is a pattern of tiles. How many white and black tiles do you need to
build the nth pattern? If you have N black tiles, how many white tiles do
you need?
Tabulate your results and generalise:
Pattern
1st
2nd
Number of black tiles
1
2
Number of white tiles
5
..
3rd
3
..
4th
4
..
...
nth
N
..
2. Find, in each of the following cases, a generalised expression for the
relation between the number of white squares and the number of black
squares.
Tabulate your results from simpler cases and look for a pattern.
2a. Pattern
1st
2nd
3rd
4th ...
Number of white tiles
0
1
...
...
...
Number of black tiles
4
...
...
..
...
nth
w
b
2b. Pattern
Number of white tiles
Number of black tiles
nth
w
b
1st
2
2
2nd
3
...
3rd
..
...
4th
..
...
...
...
...
Self mark exercise 2 continued on next page
Module 4: Unit 1
12
Learning and teaching algebra
Self mark exercise 2 continued
3. A tile pattern is such that (number of white tiles) = 2 × (number of black
tiles) – 1. Draw the pattern.
4. When making a cable for a suspension bridge many strands are
assembled into a hexagonal formation and then ‘compacted’ together.
The diagram illustrates a “size” 1, 2 and 5 cable made up of respectively
1, 7 and 61 strands.
A bridge needs a size 10 cable: How many strands are needed?
The manager of the cable factory wants to know how many strands she
needs to make a cable of any size (let call it size N).
Tabulate results and look for a pattern as in questions 1 and 2.
Check your answers at the end of this unit.
Unit 1, Practice activity
1. a) Design a worksheet for pupils to investigate the number of tiles
needed to make a path around a region. Indicate objective(s) of the
activity. To which form would you set the activity?
b) Try out your worksheet and write an evaluative report.
Some questions you might want to answer could be: What were the
strengths and weaknesses of the activity? What needs improvement?
How was the reaction of the pupils? What did you learn as a teacher
from the lesson? Could all pupils participate? Were your objective(s)
attained? Was the timing correct? What did you find out about
pupils’ investigative abilities? What further activities are you
planning to strengthen pupils’ understanding. Were you satisfied with
the outcome of the activity?
2. a) Design a worksheet for pupils to investigate the crossing the river
problem. Indicate objective(s) of the activity. To which form would
you set the activity?
b) Try out your worksheet and write an evaluative report.
Unit 1, Practice activity continued on next page
Module 4: Unit 1
13
Learning and teaching algebra
Unit 1, Practice activity continued
3. In the text we mentioned four aspects of meaningful learning of algebra
for pupils in the junior secondary school age range. What they learn
should be
(i) applicable to situations they are likely to meet in the world of work
(ii) interpretation of situations expressed in algebraic format (graphs,
word formulas)
(iii) general techniques rather than case specific techniques
(iv) using algebra to describe a pattern or relationship they are interested
in
Describe clearly, illustrating with examples, the implications for the
teaching of algebra of the above i.e. how does it affect the teaching of
algebra? Is that reflected in the book you are using in the class? If not,
what changes are needed?
4. It is emphasised in the text that pupils will be motivated to learn new
concepts and skills when they feel a need for it.
What problems/questions would you set to pupils to create a need for
(i) fractions
(ii) negative numbers
(iv) solving linear equations
(iii) variables
(v) quadratic formula
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 1
14
Learning and teaching algebra
Section A4 : Pupils’ learning difficulties with algebra
Algebra is an ill understood part of mathematics. Although pupils might
manage to manipulate with letters this does not imply understanding. The
numerous errors made by pupils in algebra point to the fact that ill
understood ‘rules’ are applied to situations in which they are not applicable.
1. Make a list of the common errors you have seen in pupils’ work in
algebra.
Try to categorise the errors. That is, place errors you feel have the same
root cause in the same category.
2. Write down what you understand by variable.
3. Express in algebraic format: there are 4 times as many pupils in the
school as there are teachers, using the variables p for the number of
pupils and t for the number of teachers.
Now read the following text and compare with your own reflection on the
issue.
Research (Hart, 1981) mentions the following as the main difficulties of
pupils in the learning of algebra.
I. Concept of variable.
Research agrees that the main difficulty in algebra is the concept of
variable. The idea that a letter is used to represent ANY NUMBER taken
from a certain set (usually N, Z, Q or ℜ—the notation used is from module
1) is hard for pupils to understand. The distinction between a letter as an
unknown (specific) quantity (to be found by solving the equation:
3x2 + 2x – 8 = 0) and a letter as a parameter (the gradient m and the yintercept n in y = mx + n are parameters) is also hard for many pupils.
(a) Letters for objects
Letters in algebraic expressions are frequently viewed in a different way by
pupils. They might see the letter to represent an object. If in a square the
measure of the length of the side is indicated with n, pupils see this as
denoting the side rather than the measure of the side. This error of labelling
is extremely common. For example pupils will write: “Let the side be x”,
“Let the pencil be p”, “Let the exercise book be b” etc.—giving labels to
objects, instead of introducing a variable.
This misconception is reinforced by several teachers who say, in order to
explain 2a + 3a, 2 apples plus three apples (so called “fruit salad” algebra).
But letters are NOT shorthand for objects! 4y does not mean that you have
four y’s (which is not the same as four times the number y). Using a letter as
an object, which amounts to reducing the letter’s meaning from the abstract
variable to something far more concrete and ‘real’, allows many pupils to
arrive at correct answers. However it will break down at some point,
particularly in cases in which it is essential to distinguish between the objects
themselves and their number.
Module 4: Unit 1
15
Learning and teaching algebra
The classic example is to express in algebraic form the statement: There are
8 times as many pupils in the school as teachers. The frequently given
answer is the error response 8p = t (letters as objects), instead of the correct
relationship p = 8t.
Expressed in word variables: (number of pupils) = 8 × (number of teachers).
Here is another example: if the number of days is d, and the number of
weeks is w, what is the relation between d and w? Common error: w = 7d
(there are seven days in a week), instead of 7w = d or expressed using word
variables: 7 × (number of weeks) = (number of days).
The idea of seeing letters as labels (truncated words) rather than as a variable
might stem from the use of letters l and b in the relation for the area enclosed
by a rectangle. l is seen as truncated “length” and b as truncated “breadth”,
but l and b are representing the measurements i.e. number of length units and
NOT the object (the sides).
(b) Different letters represent different numbers
The following questions are from research into pupils’ understanding of
algebra.
Questions:
1. Is the following statement a + b = a + c true?
A always
B never
C sometimes, that is when .....
2. Is the following statement a + b = a + 2b true?
A always
B never
C sometimes, that is when .....
3. Find a relation between x, y and z if given that
x + 5 = b and y + z + 5 = b
The general response of 11-13 years old to the first and second item is
NEVER. They reason in question 1: as b and c are different letters they
could not be the same number. To respond correctly to this item demands the
concept of both b and c running through the whole set of possible numbers.
In the second question it is forgotten that zero is also a possible value the
variables can take. In the third, they miss the fact that in a + a the variable
MUST represent the same number, while in x + y, the variables x and y
MIGHT be the same or MIGHT be different.
II. Conjoining
Initially many pupils have difficulties in understanding that for example
n + 4 means add 4 to n, and at the same time is representing the result of that
addition (four more than n). Pupils are used to expressions such as: 7 + 8,
which is equal in value to 15, so the expression (7 + 8) and its value (15) are
represented in different ways. Pupils are used to deterministic answers, a
practice frequently reinforced by teachers encouraging an answer oriented
culture in the classroom.
‘Has everybody the correct answer?’ ‘What is the answer?’ are common
questions in a classroom. The emphasis is to be on the process: how did you
work that?
Module 4: Unit 1
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Learning and teaching algebra
The difficulty to accept n + 4 as the result of the addition of n and 4 leads to
conjoining: giving ‘one’ answer:
n + 4 = 4n, a + a + a + b = 3ab or aaab, 3a + 4b = 7ab.
Expressions such as n + 4, a + b are not recognised as legitimate ‘answers’ (a
single term answer), the (in this example) + sign being seen as indicating an
operation which still needs to be completed.
Pupils have to learn the difference between an operation schema such as for
example, 15 – 7 (subtract 7 from 15) and 3a + b (add b to 3a) and the result
of such an operation schema (respectively 8 and 3a + b meaning b more than
3a).
Inability to accept lack of ‘closure’ is also demonstrated in the responses to
the question:
Find a relation between x, y and z if given that x + 5 = b and y + z + 5 = b
1
with an incorrect answer y = z = x; pupil is not able to accept y + z as
2
an entity.
III. Difference in convention of notation used in algebra and arithmetic
(a) If digits are conjoined in arithmetic it represents an implicit addition.
43 = 40 + 3;
1
1
2 =2+
2
2
In algebra ab ≠ a + b, but means a × b, i.e., an implicit multiplication.
(b) In arithmetic a letter refers to a unit 2 m = 2 metres = 200 cm;
in algebra 2m = 2 × m
This leads to errors such as: If y = 3 than 4y = ...; answers given included
7 and 43
IV. Problems with the grammatical structure of the problem
Language problems lead to translating an English sentence wrongly into an
algebraic equivalent, for example in: ‘There are ten times more pupils in the
school than teachers’, giving the answer 10p = t.
V. Common errors in algebra
When algebra is not related to relational understanding and to realistic
models it becomes a ‘meaningless’ manipulation of letters. “Rules” that
apply in one situation are erroneously applied to other situations, i.e. rules
are over generalised. The following examples you will recognise and must
have come across yourself. You will be able to extend the list!
Module 4: Unit 1
17
Learning and teaching algebra
Correct
Error form
(x × y)2 = x2 × y2
( x + y)2 = x2 + y2 or
x+y
x y
= +
z
z z
x
x x
= +
y+z
y z
ab
b
=
ad
d
ab + c b + c
=
ad
d
x 2 + y2 = x + y
Unit 1, Assignment 3
1. Describe four difficulties pupils have in the learning of algebra. Illustrate
each with several examples you met when covering algebra with your
classes.
2. Study each of the following pupils’ statements. What misconception is
reflected in the pupil’s answer? Describe in detail the four remedial steps
(diagnosing the error, creating conflict, resolving the conflict,
consolidation of the correct concept) you would take to assist the pupil to
overcome the misconception.
a. You can’t add 2a and 3b together because it’s like 2 apples and 3
bananas.
b. 2x + 1 = 7 and 2y + 1 = 7 are different equations because they have
different letters.
c. If you add 3 onto 4a you get 7a.
d. You can’t do p + q = 10 because there isn’t an answer.
e. (i) In this school there are three times as many girls than boys, so if
b stands for boys and g for girls then in this school it is true that
b = 3g.
(ii) -x is a negative number.
Present your assignment to your supervisor or study group for discussion.
Summary
Awareness of the “pitfalls of algebra,” which were introduced in this unit, is
pivotal to your teaching. In the ensuing units, study how the authors seek to
make the traditional manipulations relevant to this age group—and thus less
prone to common errors—by creating a need, or by using citations from life
to make use of algebra realistic.
Module 4: Unit 1
18
Learning and teaching algebra
Unit 1: Answers to self mark exercises
Self mark exercise 1
1. If the number of square tiles is N then N = 8(s2 + ps + qs).
2. To be paid P0.25 × f, where f = 4a + 2c, c > 2
Self mark exercise 2
1. Tabulate cases and generalise
Pattern
1st 2nd
Number of black tiles 1
2
Number of white tiles 5
7
3rd
3
9
4th
4
...
11
nth
N
2N + 3
2a.
Pattern
Number of white tiles
Number of black tiles
1st
0
4
2nd
1
8
3rd
4
12
4th
9
...
40
nth
w
4w + 4
Pattern
Number of white tiles
Number of black tiles
1st
2
2
2nd
3
6
3rd
4
12
4th
5
...
20
nth
w
w(w – 1)
2
1
3
7
4
19
2b.
3. For example:
4. size 10 needs 271 strands
Size of cable
Strands
Module 4: Unit 1
1
19
...
37
N
3N(N – 1) + 1
Learning and teaching algebra
Unit 2: Expansion of expressions
Introduction to Unit 2
Expanding algebraic expression is a traditional topic in algebra. With
symbolic-manipulation algebraic calculators available the paper-and-pencil
approach will become less and less important in the years to come. The
processes involved will get more attention than the actual algorithm. A
similar change has taken place in arithmetic: paper-and-pencil long
multiplication and division have lost much of their importance. What has
become more important is to know when to multiply, when to divide and to
use approximations to verify calculator displays. This unit, by providing
concrete models pupils can relate to, moves away from the abstract approach.
Purpose of Unit 2
In this unit you will learn about models that can be used in the teaching of
expansion of expressions of the format a(b + c), (ax + by)(cx +dy),
(ax + by)2. If pupils at this age do not relate abstract ideas to concrete
models, algebra becomes pure manipulation of ‘letters’ without real
relational understanding of what is happening and what it can represent.
Objectives
When you have completed this unit you should be able to:
•
•
•
•
•
•
•
•
•
•
•
use the area model in expansion of expressions
use a multiplication table structure to multiply two polynomials
use the FOIL acronym in expanding the product of two binomials
use algebra tiles in the expansion of (ax + b)(cy +d) where a, b, c and d
are integers
illustrate geometrically the identities
(a + b)(a – b) = a2 – b2, (a + b)2 = a2 + 2ab + b2 and
(a – b)2 = a2 – 2ab + b2 using paper models.
use generalised arithmetical patterns to obtain algebraic identities
justify the use of manipulatives and models in the learning of algebra
implement with confidence the models for expansion (area model,
multiplication table, FOIL, algebra tiles) in the teaching of algebra
set investigative activities in which pupils generalise arithmetical
patterns and express the generalization algebraically
use paper models in the classroom to illustrate algebraic identities
evaluate the effectiveness of using models in the teaching of expansions
as compared to a more abstract method
Time
To study this unit will take you about 12 hours. Trying out and evaluating
the activities with your pupils in the class will be spread over the weeks you
have planned to cover the topic.
Module 4: Unit 2
20
Expansion of expressions
Section A1: Collecting like terms
One of the ‘first steps’ in algebra is simplifying expressions. How did you
learn about it? What were you told by your teacher? What are you doing
with your own pupils?
1. Write down an outline for a lesson in which you want pupils to learn the
following:
collecting of like terms
Use your outline and compare when working through the following
section.
2. Is your outline different from the way you were taught collecting of like
terms when at secondary school? Explain and justify similarities and
differences.
In setting activities to cover the concept of collecting like terms, you have to
keep in mind the basic ideas mentioned in the previous section:
(i) you must avoid pupils seeing ‘letters’ as labels of objects. That letters
are representing variables should be clear in the model.
(ii) you must use concrete models to which pupils can relate.
What is clearly to be avoided is ‘fruit salad and cattle post’ algebra. Do not
explain a + a as one apple and another apple makes 2 apples so a + a + 2a,
or 3c + 2g + c + 4g as 3 cows and one cow make 4 cows, two goats and four
goats makes 6 goats, so 3c + 2g + c + 4g = 4c + 6g. Although this gives
‘correct’ answers, the conceptual idea (letters representing objects) is invalid
and is exactly what you like to avoid taking root into pupils’ minds.
Here are two suggestions.
I. Perimeter model for collecting like terms
You have different rods, some of the same length and others of different
lengths. You did not measure them, but you know that some rods have a
length of p cm, others of q cm, etc. You have also some rods which you
know are 1 cm in length. The rods are used to make polygonal shapes and
you find their perimeters in terms of p, q, etc.
Here are three examples illustrated.
Example 1
What is the
perimeter of this
polygon? The
lengths of the sides
are in cm.
Module 4: Unit 2
21
Expansion of expressions
Expected is that pupils will write p + p + p + q + p + q = 4p + 2q. Perimeter
is (4p + 2q) cm
In words: the perimeter is the sum of 4 rods of length p cm and 2 rods of
length q cm.
Example 2
Find the perimeter of this
polygon. The length of the sides
are in cm.
Example 3
Find the perimeter of this
polygon. The length of the sides
are in cm.
Pupils should also relate given expressions to the perimeter model.
Questions such as: represent each of the following expressions using a
perimeter model (i) 2a + b (ii) 2a + 3 should be set.
Self mark exercise 1
1. Write down the working you expect a pupil to give in example 2 and
example 3.
2. Using the perimeter model represent 2a + b and 2a + 3.
Check your answers at the end of this unit.
Module 4: Unit 2
22
Expansion of expressions
II. Bags with coins model
You have bags containing unknown numbers of coins and some single coins
(not in the bags). How many coins are there altogether in each of the
following cases?
You expect pupils to answer respectively (i) 2n (ii) 3n + 2 (iii) n + m + 3
coins.
Unit 2, Practice activity
1. Use the above models for collecting like terms in your classroom. You
might have to develop some worksheets for the pupils to work on.
Write an evaluative report on the activity. Comment on the two models
(perimeter/coins in bags). Is one more appealing to pupils than the other?
2. Develop another concrete model to model collecting of like terms and try
it out with your pupils.
3. Do the models always work? How would you model 2a – 2? 3x – 2y?
4. Consolidation of concepts is always needed. On the following page you
find a game like situation for pupils to consolidate the collecting of like
terms. Develop it fully and try it out with your class and write an
evaluative report. What is the disadvantage of the activity?
5. Look at the outline of the lesson for collecting like terms you wrote at the
beginning of this section. Comment on differences (if any) and
similarities with the activities suggested above.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 2
23
Expansion of expressions
Consolidation of collecting like terms: a pupils’ activity
Use a square grid placing the designated number of red (r) and yellow (y)
counters in each cell. For example:
Moving from Start to End from one cell to the other—to the right (R) or
up (U)—only a number of possible questions can be looked at:
(i) given a route UURRRURRRUR what number of red and yellow counters
do you collect following that route?
(ii) what is the route to be followed if you want the number of red counters
to be as small as possible?
what is the route to be followed if you want the number of yellow
counters to be as small as possible?
(iii) what is the route to be followed if you want the number of red counters
to be as large as possible?
what is the route to be followed if you want the number of yellow
counters to be as large as possible?
(iv) which route will give the least TOTAL of counters?
which route will give the maximum number of counters?
Extension:
The up/right movement can be changed to up /down/right/left (diagonal).
Module 4: Unit 2
24
Expansion of expressions
Section A2: Expansion of the product of a monomial
and a binomial
In this section you are going to look at the expansion of a(b + c), a(b – c)
and similar expressions.
1. Write down an outline for a lesson in which you want pupils to learn the
following:
expansion of algebraic products such as p(x + y), a(b – c)
Use your outline and compare when working through the following
sections.
2. Is your outline different from the way you were taught expansion when
at secondary school?
Explain and justify similarities and differences.
3. Expansion is frequently referred to as “removing brackets”. What
objections could you bring against the use of this expression?
The area model for expansion
The area model can be used to model expansions. The expansion concept is
developed from arithmetic, moving to use of one variable, next to more
variables.
Step 1
Pupils were asked to calculate the area of a school garden. The garden is
partly used for the growing of vegetables and partly for the growing of fruit
trees.
Pupil 1 wrote: 4 × (6 + 3) = 4 × 9 = 36 m2
She explained: I multiplied the total length of the garden with the width.
Pupil 2 wrote: 4 × 6 + 4 × 3 = 24 + 12 = 36 m2
He explained: I calculated the area of the vegetable garden and the fruit tree
garden and added the two together.
They agreed that 4 × (6 + 3) = 4 × 6 + 4 × 3 (this is called the distributive
law: multiplication is distributive over addition).
Module 4: Unit 2
25
Expansion of expressions
The pupils are set an exercise:
(i) to consolidate the two ways of finding the area of a rectangular garden
consisting of two parts (a vegetable and a fruit tree part)
(ii) to draw ‘gardens’ to illustrate a given calculation such as 5 × (8 + 4)
Examples of pupil exercises.
1. Calculate the area of each of the following gardens using the method of
the first pupil (multiplying the total length of the garden with the width)
and also using the method of the second pupil (adding the areas of the
two parts).
V Vegetable garden
F Fruit trees
2. Draw vegetable/fruit trees gardens to illustrate these calculations
a. 5 × (9 + 8) = 5 × 9 + 5 × 8
b. 8 × (7 + 5) = 8 × 7 + 8 × 5
c. 6 × (6 + 4) = 6 × 6 + 6 × 4
Self mark exercise 2
1. Write down the working you expect a pupil to give to question 1 and 2.
2. Why do we have tilted sketches of the gardens and not representations
like this?
Check your answers at the end of this unit.
Module 4: Unit 2
26
Expansion of expressions
Step 2
The school is to extend the length of the vegetable garden but does not know
exactly by how much. The plan is shown in the diagram.
The two pupils use their same method to find the area of the whole garden.
Pupil 1 uses the total length times the width: 4 × (x + 3) = 4(x + 3) m2.
Pupil 2 adds the area of the vegetable garden to the area of the flower
garden:
4 × x + 4 × 3 = (4x + 12) m2
They agree that 4(x + 3) = 4x + 12.
Self mark exercise 3
1. Write an exercise, covering the two types of questions as in step 1, to
consolidate the second step.
Check your answers at the end of this unit.
Step 3
A school has a plot to be used for planting vegetables and flowers. They do
know the width of the plot but are not yet sure how to subdivide the plot into
vegetable part and flower part. They used this plan:
Pupils use the two methods to find an expression for the area of the plot.
Using the total length times the width: 4 × (x + y) = 4(x + y) m2.
Using addition of the area of the vegetable garden and the area of the flower
garden:
4 × x + 4 × y = (4x + 4y) m2
They agree that 4(x + y) = 4x + 4y.
Module 4: Unit 2
27
Expansion of expressions
Examples of questions for the consolidation exercise for the pupils.
1. Calculate the area of each of the following gardens in two ways.
V Vegetable garden
F Fruit trees
2. Draw vegetable/flower gardens to illustrate these calculations
a) 5 × (x + 2y) = 5 × x + 10 × y
or
5(x + 2y) = 5x + 10y
b) 8 × (2a + b) = 16 × a + 8 × b
or
8(2a + b) = 16a + 8b
c) 6 × (6x + 4y) = 36 × x + 24 × y or
6(6x + 4y) = 36x + 24y
Self mark exercise 4
1. Write down the working you expect a pupil to give to question 1 and 2.
2. Write out the next step, Step 4, to cover the expansion of expressions
involving three
variables such as
a(x + y), a(2b + c), 4(2x + y + 3), 2a(3b + c + 5), 2a(a + 3b + 6)
Use the last format only if you want to include powers.
Make it into a pupil’s worksheet.
Check your answers at the end of this unit.
Module 4: Unit 2
28
Expansion of expressions
Step 5A
The school is expanding and need more space for buildings. The area
available for use as garden is to be reduced.
This is the plan:
To obtain the area of the remaining part the two methods can be applied
again:
(i) multiplying the length of the remaining garden section with the width:
5 × (12 – 3) = 5 × 9 = 45 m2
(ii) subtracting from the original area of the garden the part that is removed:
5 × 12 – 5 × 3 = 60 – 15 = 45 m2
Step 5B
In another school it is not yet known what part is going to be cut. Their plan
looks like this:
The two ways of calculating the remaining area of the garden looks now as
follows
(i) multiplying the length of the remaining garden section with the width:
5 × (12 – x) = 5(12 – x) m2
(ii) subtracting from the original area of the garden the part that is removed:
5 × 12 – 5 × x = (60 – 5x) m2
So that it is concluded that 5(12 – x) = 60 – 5x
Module 4: Unit 2
29
Expansion of expressions
Self mark exercise 5
1. Set an exercise for the pupils to consolidate the above steps 5A & 5B.
Remember to cover the two types of questions (expressing given area in
two ways, illustrating pairs of equal areas) as in the previous steps.
2. Write out the next steps, step 5C (using two variables) and step 5D
(using three variables or more), to cover the expansion of expressions
such as 5 (x – y), a(2b – 4), 4x(2y – 3) in step 5C and expressions such as
a(3b – c – 5), 2a(20 – 3b – 2c), a(4b – c – d) in step 5D. Make it into a
pupil’s worksheet.
3. Can the area model be used to cover and illustrate all possible expansions
of a monomial and a polynomial? Justify and illustrate your answer.
Check your answers at the end of this unit.
Multiplication table algorithm for expansion
The area model is a concrete model. A step towards a more abstract
approach is using the multiplication table. This is close to the area model, the
difference being that two of the sides of the rectangle are deleted. Below are
examples in a multiplication table algorithm format.
×
6
×
3
3
4
24
12
4(6 + 3) = 24 + 12
4
4x
12
4(x + 3) = 4x + 12
×
×
a
b
x
y
5
5a
5b
5(a + b) = 5a + 5b
a
ax
ay
a(x + y) = ax + ay
×
12
-3
×
18
-x
5
60
-15
10
4x
12
5(12 – 3) = 60 – 15
10(18 – x) = 180 – 10x
×
a
-b
×
x
-y
5
5a
-5
a
ax
-ay
5(a – b) = 5a – 5b
Module 4: Unit 2
x
a(x – y) = ax – ay
30
Expansion of expressions
The multiplication table algorithm can be generalised to expressions such as
-a(b + c), -a(b – c) and -a(-b – c) which cannot be interpreted/illustrated with
an area model.
×
b
c
×
b
-c
×
-b
-c
-a
-ab
-ac
-a
-ab
ac
-a
ab
ac
-a(b + c) = -ab – c
-a(b – c) = -ab + c
-a(-b – c) = ab + c
Unit 2, Practice activity
1
Use the above area model for expansion in your classroom. You should
have developed worksheets for the pupils to work through the various
steps.
Write an evaluative report on the activity.
2. Compare the area model with the multiplication algorithm pointing out
advantages and disadvantages of each model. Which would you use in
the classroom? Justify.
3
Look at the outline of the lesson for expansion you wrote at the
beginning of this section. Comment on differences (if any) and
similarities with the activities suggested above.
Present your assignment to your supervisor or study group for discussion.
Section A3:
Expansion of the product of two binomials
In this section you will extend the work of the previous section (modelling
the product of a monomial and a polynomial) to modelling the product of
two binomials. The same steps in developing the concepts, using the area
model, can be used.
Step 1: Numerical values used only.
A rectangular garden plot is divided into four rectangular parts. The area of
the whole garden can be found in two ways.
Method 1: Multiplying the total length of the plot with the total width.
Method 2: Adding the areas of the four parts that make up the total garden.
Module 4: Unit 2
31
Expansion of expressions
Area of the garden is
(4 + 8)(6 + 2) = 12 × 8 = 96 m2
or
24 + 8 + 48 + 16 = 96 m2
Hence (4 + 8)(6 + 2) = 24 + 8 + 48 + 16
Step 2: One variable is introduced.
This will cover expressions such as (6 + 3)(a + 4).
The first method will give (6 + 3)(a + 4) = 9(a + 4)
The second method gives 6a + 24 + 3a + 12. This simplifies to 9a + 36
Hence (6 + 3)(a + 4) = 9(a + 4) = 9a + 36
Step 3: Two variables are used to cover expressions such as (a + 3)(b + 6).
The first method, multiplying total length with total width of the plot, gives
the expression
(a + 3)(b + 6).
The second method, adding the areas of the four parts of the garden, will
give
ab + 6a + 3b + 18.
Hence (a + 3)(b + 6) = ab + 6a + 3b + 18.
You might want to include cases with a = b. For example:
The school farm is to be divided into four sections. The plan is shown. Next
to it is written the two methods that can be used to find the area.
(x + 2) (x + 3)
or
x2 + 2x + 3x + 6 = x2 + 5x + 6
Pupils will agree that (x + 2)(x + 3) = x2 + 5x + 6
Module 4: Unit 2
32
Expansion of expressions
Step 4: Extend to the general format (a + b)(c + d).
In cases the numbers involved are positive (a + b)(c + d) can be modelled as
the area enclosed by a rectangle with sides (a + b) and (c + d).
The area enclosed by the big rectangle equals the sum of the areas enclosed
by the four smaller rectangles. Illustrating that
(a + b)(c + d) = ac + ad + bc + bd.
In the consolidation exercise for each step you are to cover two types of
questions
(i) Given a plot divided into four parts, find two expressions for the total
area of the plot. For example a step 3 question could look as illustrated.
(ii) Given an expression, pupils are to illustrate the expression using an area
model and use the model to expand the given expression.
Given expression could be for example:
at step 3 level
a (a + 3) (b + 4)
b (b + 5)(c + 1)
c (c + 2)(c + 2)
d (2x + 1)(3y + 4)
at step 4 level
e (x + y)(x + y)
Module 4: Unit 2
f (2x + 3y)(3x + y)
33
Expansion of expressions
Self mark exercise 6
1. Illustrate steps 1 to 4 with area diagrams.
2. Set an exercise, one for each step, for the pupils to consolidate the steps 1
to 4. Remember to cover the two types of questions (expressing given
area in two ways, illustrating pairs of equal areas) and to use context (the
school garden for example).
Check your answers at the end of this unit.
In the first four steps all variables involved are (assumed) to be positive real
numbers. The area model will work well as long as the variables involved
are all positive. In the examples positive whole numbers were used only.
However, for higher achievers, including decimals and fractions could be
considered.
Expanding (2.1x + 3.7)(4.8y + 3.6) can well be modelled with the area
model. The reason to keep the computational work simple is to allow pupils
to concentrate on the concept of expansion (use of distributive law) and not
to divert their attention by including ‘heavy’ computational work.
Step 5: moving to expressions of the format (a – b) (p + q), (a – b)(p – q).
Remember that you modelled in the previous section, expressions such as
a(p – q), by cutting part of the school garden.
Let’s look at the same idea again. The school reducing the garden by 3 m
drew the following plan:
before “cutting”
after cutting (a – 3)(b + 5)
The first method, multiplying length and width of the remaining part of the
garden, will give (a – 3)(b + 5).
The second method, subtracting from the original area of (ab + 5a) m2, the
two shaded parts with areas respectively of 3b and 15, will give the
expression ab + 5a – 3b – 15.
The conclusion being that (a – 3)(b + 5) = ab + 5a – 3b – 15.
This works well. Now what happens if the school garden is reduced in size
by cutting in both directions?
Module 4: Unit 2
34
Expansion of expressions
The plan looks like this:
before “cutting” ab
after cutting (a – 3)(b – 5)
The area remaining of the garden after the cutting is, using the first method,
(a – 3)(b – 5).
Using the second method the area must also be equal to the original area of
ab m2 minus the areas of the three shaded parts with areas of respectively 15,
3(b – 5) = 3b – 15 and 5(a – 3) = 5a – 15.
This leads to the conclusion that
(a – 3)(b – 5) = ab – (15 + 3b – 15 + 5a – 15) = ab – (3b + 5a – 15).
But here we get stuck with our area model as the model does not allow us to
interpret -(3b + 5a – 15) as -3b – 5a + 15.
Is there a possibility to get around this problem?
Consider the following diagrams.
The first diagram illustrates the initial situation: a garden with area of ab m2.
The second diagram illustrates the ‘cutting’ of an area of 3 m by b m
i.e. 3b m2.
Module 4: Unit 2
35
Expansion of expressions
In the next diagram 15 m2 is added in order to make it possible to remove a
strip with area 5a m2.
What you have done now is to remove areas of 3b and 5a and added an area
of 15. The remaining area is therefore ab – 3b – 5a + 15.
As the two methods give the same area it must be true that
(a – 3)(b – 5) = ab – 5a – 3b + 15.
Self mark exercise 7
1. Set an exercise for the pupils to consolidate step 5.
Remember to cover the two types of questions (expressing given area in
two ways, illustrating pairs of equal areas) and to use context (the school
garden for example).
2. Have all products of two binomials now been covered? Justify your
answer.
3. Set an exercise for higher achievers to extend expansion, using the area
model, to expressions such as (a + b)(p + q + r), (a – b) (p + q + r),
(a – b)(p + q – r), (a + b + c)(p + q + r), etc.
Check your answers at the end of this unit.
Module 4: Unit 2
36
Expansion of expressions
Section B: Models for expansions
The area model discussed above is not the only possible model you can use
in the classroom. In this section some more models are considered.
Section B1:
Multiplication table algorithm for expansion
The area model fails when negative numbers are involved, as negative
lengths of sides of rectangular gardens cannot occur. The area model, being
an extremely useful model for expansion, needs some abstraction to allow
expansions of expressions such as (-a – b)(p – q).
Presenting the calculations in a multiplication grid is not so big a step as it
remains close to the area model.
Some examples of expansions using the multiplication grid are below.
Example 1: (p + 3)(p + 2)
The factors to be multiplied are placed as row and column headings. To
expand
(p + 3)(p + 2) the grid looks as shown.
×
p
2
p
p2
2p
3
3p
6
Each variable or number at the top is multiplied by each variable or number
at the left, and the results are added : p× p + p ×2 + 3× p + 3×2 = p2 + 5p + 6
Example 2: Expansion of (a + b)(c + d) and (a – b)(-c + d)
×
c
d
×
-c
d
a
ac
ad
a
-ac
ad
b
bc
bd
-b
bc
-bd
(a + b)(c + d)
= ac + ad + bc + bd
Module 4: Unit 2
(a – b)(c – d)
= -ac + ad + bc – bd
37
Expansion of expressions
Example 3: Expansion of
1. (a – 2b)( 2a + b)
2. (b – 3)(2 – a)
3. (1 – 2a)(2a + 1)
×
2a
b
×
2
-a
×
2a
1
a
2a2
ab
b
2b
-ab
1
2a
1
-2b
-4ab
-2b2
-3
-6
-3a
-2a
-4a2
-2a
(a – 2b)(2a + b)
(b – 3)(2 – a)
(1 – 2a)(2a + 1)
= 2a2 + ab – 4ab – 2b2
= 2b – ab – 6 – 3a
= 2a – 1 – 4a2 – 2a
= 2a2 – 3ab – 2b2
= -3a – ab + 2b – 6
= -4a2 – 1
Self mark exercise 8
1. Use an area model and a multiplication table to expand
a) (2a + 3)(b + 4)
d) (d – 4)(5 – d)
b) (b + 1)(c – 3)
c) (2c – 3)(3d – 1)
e) (5e – 11)(5f + 11) f) (5 – 3f)(2 – 5f)
2. Use a multiplication table to work (2a + 3)(a2 – 2a + 3).
Copy and complete:
×
a2
2a
2a3
-2a
3
6a
3
9
(2a + 3)(a2 – 2a + 3) = 2a3 . . ..
3. Copy and complete the multiplication table for (a + b)2 = (a + b)(a + b)
×
a
b
a
b
(a + b)2 =
Self mark exercise 8 continues on next page
Module 4: Unit 2
38
Expansion of expressions
Self mark exercise 8 continued
4. Copy and complete the multiplication table for (a – b)2 = (a – b)(a – b).
×
a
-b
a
-b
(a – b)2 =
5. Copy and complete the multiplication table for (a + b)(a – b).
×
a
-b
a
b
(a + b)(a – b) =
6. Apply the relationships you obtained in question 3, 4, and 5 to expand:
a) (p + q)2
b) (p – q)2
c) (2p + 3q)2
d) (2p – 3q)2
e) (5p + 6)2
f) (-p + 2)2
g) (-4p – 3q)2
h) (4p + 3q)2
i) Compare your answer to g and h.
Will it always be true that (ap + bq)2 = (-ap – bq)2? Justify your answer.
1
1
j) ( p + q)2
2
3
k) (3k + 2)(3k – 2)
1
1
1
1
l) ( a + b)( a – b)
3
4
3
4
Check your answers at the end of this unit.
Section B2: Expansion using concrete manipulatives
(algebra tiles)
Some pupils understand better when they have concrete objects to
manipulate. The objects represent the concepts.
For expansion of expressions of the format
(a + (1st number))(b + (2nd number))
so for example (a + 2)(b + 3), (a – 2)(b + 3), etc.
You need a set of algebra tiles to represent a, b, 1, ab, -a, -b, -ab and -1. You
can cut them out, after photocopying them, preferably onto stiffer paper or
card stock.
Module 4: Unit 2
39
Expansion of expressions
Algebra tiles
Module 4: Unit 2
40
Expansion of expressions
Module 4: Unit 2
41
Expansion of expressions
The tiles can be used on an overhead projector for use with the whole class
and at the same time pupils can be given individual sets to model expansions
with the algebra tiles.
Here is an example of the expansion of (a – 2)(b – 3).
Place tiles representing a and -1, -1 along one side to represent the factor
(a – 2).
Along the other side you place tiles b and -1, -1, -1 to represent the factor
(b – 3).
The region is now filled with the appropriate tiles, keeping in mind the rules
for multiplication of directed numbers.
The result gives the following tiles:
ab
3 × -a = -3a
2 × -b = -2b
6×1 =6
Adding gives the result of the expansion, thus
(a – 2)(b – 3) = ab – 3a – 2b + 6.
Self mark exercise 9
1. Use your algebra tiles to expand
a. (2a – 1)(b + 4)
b. (-a + 2)(2b – 1)
c. (-a – 3)(-2b + 3)
2. Compare using algebra tiles with the area model. What are the
similarities? What are the differences?
Check your answers at the end of this unit.
Module 4: Unit 2
42
Expansion of expressions
Remember that the area model was abstracted into a multiplication table
model. The use of algebra tiles also allows a gradual move to a more abstract
representation/algorithm for expansion.
Here are the steps you could consider:
Step 1
Allowing pupils to use the concrete representations as long as they want.
A move can be made to a more abstract representation by only placing the
factors and writing the tiles ‘values’. Some pupils might do this without
being prompted!
The diagram illustrates this first step
Step 2
No longer using any concrete manipulative but representing the tiles by their
‘values’ in a multiplication table:
Module 4: Unit 2
×
b
-3
a
ab
-3a
-2
-2b
6
43
Expansion of expressions
Section B3: The FOIL algorithm
The algorithm traditionally used for the expansion of the product of two
binomials is the FOIL algorithm. FOIL is an acronym for
F
First
O
Outer
I
Inner
L
Last
It looks like this:
This algorithm tells pupils HOW to do it, but the WHY is not clear. This is
one of the characteristics of an algorithm: it tells you how you can obtain the
required result, but does not explain why the suggested method works.
Many pupils can tell you: “to divide two fractions you multiply the first
fraction with the reciprocal of the other,” and in algebraic notation they are
a c a d
saying
÷ = × . Yet few pupils can explain why this algorithm
b d b c
works.
Unit 2, Practice activity
1. Use the area model and the use of algebra tiles for expansions of the
product of two binomials in your classroom. You will have to develop
worksheets for the pupils to work through the various steps.
Compare the two methods.
Write an evaluative report on the activity. Pay attention to pupils’
reaction to use of area model and algebra tiles. What are their
preferences?
2. Compare the area model, and the use of algebra tiles with the FOIL
algorithm, pointing out advantages and disadvantages of each approach.
Which would you use in the classroom? Justify.
3. Explain why the FOIL algorithm works.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 2
44
Expansion of expressions
Section B4: Illustrating the identity
(a + b)(a – b) = a2 – b2 with paper models
One way to move away from the abstract nature of algebra is to represent
identities in a more concrete way. For several pupils this helps their
understanding. Concrete presentation allows pupils to link the abstract idea
to something they can see/touch.
The identity (a + b)(a – b) = a2 – b2.
The diagram below illustrates how pupils should use paste and cut to
illustrate the identity (a + b)(a – b) = a2 – b2.
Starting from a large square (side 10 cm for example) mark another square
(side 3 cm for example) and also the region Z. Cut out X, Y and Z and
rearrange as in second diagram.
The diagram on the left illustrates that the area of the regions Y + Z is equal
to the area enclosed by the larger square (with side of length a) minus the
area enclosed by the smaller square (with side b). Hence the area of the
region Y + Z is numerically equal to a2 – b2. The diagram on the right is a
restructuring of the first diagram. The area of the region Y + Z is now that of
a rectangular region with sides of length (a + b) and (a – b) respectively.
Hence the area measure is numerically equal to (a + b)(a – b). This
illustrates that (a + b)(a – b) = a2 – b2.
Another way to illustrate the same identity is illustrated in the following
diagram. Again it is a cut and paste illustration which can be done by pupils
themselves.
From a square (side of length a) a square with a side of length b is removed.
The remaining part is cut out (see second diagram), and the two parts I and II
are placed to form a rectangle.
Module 4: Unit 2
45
Expansion of expressions
Comparing the area enclosed by the first diagram (illustrating a2 – b2) with
the area enclosed by the last diagram (illustrating a rectangle with sides of
length a + b and a – b) leads to an illustration of the identity
(a + b)(a – b) = a2 – b2.
For classroom use making a set for the overhead projector might be a useful
teaching aid.
Section B5: The product (a + b)(a – b) from generalised
arithmetical pattern
Arithmetic pattern when generalised can inductively lead to algebraic
identities. Work through the following instructions:
Study the following pattern:
8 × 8 = 64
12 × 12 = 144
104 × 104 = 10816
9 × 7 = 63
13 × 11 = 143
105 × 103 = 10815
10 × 6 = 60
14 × 10 = 140
106 × 102 = 10812
...
Add four more rows to the pattern.
Look at the first row.
8 × 8 = 64
12 × 12 = 144
104 × 104 = 10816
What is the general pattern in that row?
Did you recognise the squares?
The pattern in the first row could be written as:
(number) × (number) = n × n = n2
Compare the numbers used in the second row with the numbers in the first.
8 × 8 = 64
12 × 12 = 144
104 × 104 = 10816
n × n = n2
↓ ↓
↓
↓
↓ ↓
↓
9 × 7 = 63
↓
↓
13 × 11 = 143
↓
↓
↓
105 × 103 = 10815
The first 8 ⇒ 9 (add 1) The second 8 ⇒ 7 (subtract 1)
and 64 ⇒ 63 (subtract 1)
Check that the same is true for the next expressions in the row.
What has to go with the first n?
What with the second?
What has to go with the n2?
Did you find (n + 1), (n – 1) and n2 – 1?
Module 4: Unit 2
46
Expansion of expressions
The pattern in the second row, when generalised gives
(number + 1) × (number – 1) = (number) × (number) – 1
or in algebraic notation: (n + 1) (n – 1) = n2 – 1
Check that the third row, when compared with the first, will give you the
generalisation:
(n + 2) (n – 2) = n2 – 4.
Write down the generalization for the fourth, fifth and pth row.
Unit 2, Practice activity
1. Use the paper demonstration of the identity (a + b)(a – b) = a2 – b2 in
your classroom. Write an evaluative report.
2. Try out the pattern method for pupils to generalise to the identity
(a + b)(a – b) = a2 – b2. You will have to adapt the above outline to the
pupils’ level. In your evaluation compare with the other models you tried
out for expansion of the product of two binomials.
3. Develop an worksheet and try it out in the classroom, which when
generalised will lead to pupils to discover the identities:
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a – b)2 = a2 – 2ab + b2
4. Make the following tiles
Use the four pieces to make a square. What is the length of the side of
the square? What identity have you illustrated?
5. Expand
(a + 1)
=a+1
(a + 1)2 = (a + 1)(a + 1) = ...
(a + 1)3 = (a +1)(a + 1)(a + 1) = ...
(a + 1)4 = (a + 1)(a + 1)(a +1)(a + 1) = ...
(a + 1)5 = (a + 1)(a + 1)(a + 1)(a + 1)(a + 1) = ...
....
Practice activity continued on next page
Module 4: Unit 2
47
Expansion of expressions
Practice activity continued
Describe any pattern you can see.
Find out about Pascal’s Triangle and how it is related to the expansion.
6. Investigate how you could extend the suggested models, algorithms and
concrete representations to the expansion of the product of two
polynomials. For example to expand (ax + by + cz)(px + qy + rz). Pay
attention to ‘short comings’ of some models.
7. How would you encourage able pupils to further investigate expansion?
Write an outline for a worksheet on the topic pupils could use to guide
their investigation.
Present your assignment to your supervisor or study group for discussion.
Summary
The message of this unit has been clear: all algebraic expansions should be
taught from examples of real objects that “behave” that way.
Module 4: Unit 2
48
Expansion of expressions
Unit 2: Answers to self mark exercises
Self mark exercise 1
1. p + p = 2p + q = q = 4p + 2q
x + x + 1 + x + x + 1 = 4x + 2
2. For example
Self mark exercise 2
1. a. 6 × (8 + 3) = 6 × 11 = 66 m2
6 × 8 + 6 × 3 = 48 + 18 = 66 m2
b. 7 × (10 + 8) = 7 × 18 = 126 m2
7 × 10 + 7 × 8 = 70 + 56 = 126 m2
c. 5 × (5+ 11) = 5 × 16 = 80 m2
5 × 5 + 5 × 11 = 25 + 55 = 80 m2
d. 25 × (9 + 3) = 25 × 12 = 300 m2
25 × 9 + 25 × 3 = 225 + 75 = 300 m2
2. If pupils ONLY meet representations of rectangles in ‘standard’ position
(sides parallel to edges of the page) they might erroneously think that
representations of rectangles in tilted positions are NOT rectangles. A
multiple format of representation is required to build up the correct
concept.
Module 4: Unit 2
49
Expansion of expressions
Self mark exercise 3
A few examples are:
1.
V Vegetable garden
F Fruit trees
2. Draw vegetable/fruit trees gardens to illustrate these calculations:
a) 5 × (a + 6) = 5 × a + 5 × 6
b) 7 × (b + 3) = 7 × b + 7 × 3
c) 9 × (6 + c) = 9 × 6 + 9 × c
Self mark exercise 4
1. a) 6(p + q) = 6p + 6q
b) 7(2a + b) = 14a + 7b
c) 5(3x + 4y) = 15x + 20y
d) 25(3a + b) = 75a + 25b
Or equivalent expressions.
Module 4: Unit 2
50
Expansion of expressions
2.
V Vegetable garden
F Fruit trees
2. Outline of Step 4 to be extended into pupils’ worksheet.
A school will get a plot to be used for planting vegetables and flowers.
They do not know the width of the plot and are not yet sure how to
subdivide the plot into vegetable part and flower part. They used this
plan:
Pupils use the two methods to find an expression for the area of the plot.
Using the total length times the width: a × (x + y) = a(x + y) m2.
Using addition of the area of the vegetable garden and the area of the
flower garden:
a × x + a × y = (ax + ay) m2
They agree that a(x + y) = ax + ay.
Module 4: Unit 2
51
Expansion of expressions
Examples of questions for the consolidation exercise for the pupils.
1. Calculate the area of each of the following gardens in two ways.
V Vegetable garden
F Fruit trees
2. Draw vegetable/flower gardens to illustrate these calculations:
a) p × (x + 2y) = p × x + p × y
or
p(x + 2y) = px + py
b) c × (2a + b) = c × 2a + c × b
or
c(2a + b) = 2ac + bc
c) k × (6x + 4y) = k × 6x + k × 4y or
k(6x + 4y) = 6kx + 4ky
Self mark exercise 5
1&2. Similar to previous consolidation exercises.
3. The model will fail in cases such as -2(4a – 3) , -3b(5x – 2y) as negative
‘length’ (-2 and -3b respectively, also note that the variable b is assumed
to be positive, as otherwise -3b could represent a positive number)
cannot be modelled with area. The model also assumes that the variables
take positive values i.e. in modelling a(b + c) with an area model a, b,
and c are assumed positive.
Self mark exercise 6
Set examples similar to those illustrated in the text. You can find a
similar approach in Maths in Action Pupils book 1 and 2 with numerous
consolidation exercises.
Module 4: Unit 2
52
Expansion of expressions
Self mark exercise 7
1 & 3. Set examples similar to those in the text.
2. No! Cases such as (-p – q)(a + b) cannot be represented by an area
model.
Self mark exercise 8
1. For example f (5 – 3f)(2 – 5f)
Start with 5 by 2 rectangle: area 10.
Indicate the strips of width 3f and 5f to be removed.
First remove a strip of width 3f and area 6f (second diagram).
The third diagram is the amount remaining.
To allow a strip of 5f by 5 to be removed first add 5f by 3f rectangle
(third diagram).
Now the strip with area 25f can be removed.
Left is (5 – 3f )(2 – 5f ).
This must be equal to the original 10 minus 6f plus 15f 2 minus 25f
10 – 6f + 15f 2 – 25f = 10 – 31f + 15f 2
Hence (5 – 3f )(2 – 5f ) = 10 – 31f + 15f 2
×
2
-5f
5
10
-25f
-3f
-6f
15f 2
Hence (5 – 3f )(2 – 5f ) = 10 – 31f + 15f 2
Module 4: Unit 2
53
Expansion of expressions
2.
×
a2
-2a
3
2a
2a3
-4a2
6a
3
3a2
-6a
9
(2a + 3)(a2 – 2a + 3) = 2a3 – 4a2 + 6a + 3a2 – 6a + 9 = 2a3 – a2 + 9
3.
×
a
b
a
a2
ab
b
ab
b2
(a + b)2 = a2 + ab + ab + b2 = a2 + 2ab + b2
4.
×
a
-b
a
a2
-ab
-b
-ab
b2
(a – b)2 = a2 – ab – ab + b2 = a2 – 2ab + b2
5.
×
a
-b
a
a2
-ab
b
ab
-b2
(a + b)(a – b) = a2 – ab + ab + b2 = a2 – b2
6. a) p2 + 2pq + q2
b) p2 – 2pq + q2
c) 4p2 + 12pq + 9q2
d) 4p2 – 12pq + 9q2
e) 25p2 + 60p + 36
f) p2 – 4p + 4
g) 16p2 + 24pq + 9q2
h) 16p2 + 24pq + 9q2
i) (ap + bq)2 = (-ap – bq)2 = a2p2 + 2abpq + b2q2
1
1
1
1
1 2
j) p2 + pq + q2
k) 9k2 – 4
l) a2 –
b
4
3
9
9
16
Module 4: Unit 2
54
Expansion of expressions
Self mark exercise 9
1. Straightforward placing of the tiles.
2. Both make use of ‘areas’, but ‘negative’ areas occur with the tile model.
The tiles allow us to model all the expansions of expressions of the form
(pa + q)(sb + t) where p, q, s and t are integers. The area model did not
allow negative values for p and s.
Module 4: Unit 2
55
Expansion of expressions
Unit 3: Factorisation of expressions
Introduction to Unit 3
Like expansion (Unit 2), factorisation is not a goal in itself but a tool to be
used in, for example, the solution of equations. As the inverse process of
expansion of expressions, the methods covered in Unit 2 can be worked
backwards to factorise a given expression. Emphasis is on relational
understanding by using models.
Purpose of Unit 3
In this unit you will look at the inverse process of expansion of the algebraic
expression you studied in Unit 2. Expansion refers to writing a product of
factors as the sum of terms. The inverse process is writing the sum of terms
as a product. This is called factorisation. You will focus mainly on how the
topic can be presented to pupils in your class using concrete models and
manipulatives. Your own content knowledge is reviewed, but not extended
to, for example, the factor theorem and more general factorisation of
polynomials. You might want to look at these topics in any mathematics
book for advanced level (form six secondary school) or additional
mathematics (forms 4 and 5 secondary school).
In Unit 1 it was mentioned that algebra will rarely be useful for students
once they are in the workforce. The examples that introduced expansions in
Unit 2—tiles around a fountain, garden areas with vegetables and fruit
trees—suggested situations where a worker might actually find expansions
useful. Realistic uses for factorising are by contrast so rare, however, that
even that kind of suggestion is not feasible. That is why this unit treats “unexpanding” or checking one’s work after expansion, as the practical entry
point for teaching factorising.
North American textbooks tend to use the shorter term ‘factoring’ for
factorising.
Objectives
When you have completed this unit you should be able to:
Module 4: Unit 3
•
identify the highest common factor in pairs of expressions
•
set activities to pupils to factorise expressions that have a common factor
using an area model or a multiplication table structure
•
set activities to pupils to discover how to factorise quadratics of the form
x2 + px + q
•
set activities for pupils to discover the factorisation of the difference
between two squares
•
use algebraic tiles to factorise quadratics
•
use algebra tiles in the classroom with pupils in factorisation
•
detect the error(s) in algebraic fallacies
56
Factorisation of expressions
•
use fallacies in the teaching of algebra
•
justify the use of fallacies in the teaching of mathematics
•
use small group discussion as a learning technique
Time
To study this unit will take you about 12 hours. Trying out and evaluating
the activities with your pupils in the class will be spread over the weeks you
have planned to cover the topic.
Module 4: Unit 3
57
Factorisation of expressions
Section A: Factorising
Section A1: Factorising expressions with one common
factor
In this section you are going to look at writing expressions such as ab + ac,
ab – ac, x3 + 2x2 – 4x and similar expressions, as a product of two factors.
This is the reverse process of expansion.
1. Write down an outline for a lesson in which you want pupils to learn the
following:
factorising expressions such as ab + ac, ab – ac, x3 + 2x2 – 4x
Use your outline and compare when working through the following
sections.
Is your outline different from the way you were taught expansion when
at secondary school?
Explain and justify similarities and differences.
The inverse process of expansion is factorisation. This is writing an
expression as a product of factors. As factorisation is the reverse process, the
models, algorithms and manipulatives used in expansion should be used in
‘reverse’ order. It also implies that a factorisation can always be checked by
expanding the answer. Expanding the answer has to give the original
expression. At all times pupils should check answers (not only in this
context).
Consider the following example:
Expansion:
3(2a + 3) = 6a + 9
Bringing back brackets: 6a + 9 = 3(2a + 3),
3 is the highest common factor of 6a and 9.
Represented in the area model:
Expansion
Factorisation
Expansion: given the sides of the rectangle, find the expression for the area
of the rectangle.
Factorisation: given the area enclosed by a rectangle, find the length of the
sides.
Module 4: Unit 3
58
Factorisation of expressions
Self mark exercise 1
1. If the area of a rectangle is 24 cm2, what are the length and width of the
rectangle?
2. If the area of a rectangle is (6a + 9) square units, what are the length and
width of the rectangle?
Check your answers at the end of this unit.
Knowing the length of the sides of a rectangle gives one and only one
possible answer for the area enclosed by that rectangle.
But knowing the area enclosed by a rectangle you can find infinitely many
possible measures for the sides.
How many possibilities did you write down for the rectangle enclosing an
area of 24 cm2? Did you stop because your list is infinitely long?
You might have started with rectangles with dimensions 1 cm × 24 cm,
1
1
2 cm × 12 cm, ...etc. Did you include cm × 48 cm, cm × 72 cm,
2
3
2 cm × 12 2 cm, etc., etc.? In other words there are an infinite number of
rectangles enclosing 24 cm2.
Is the situation any different if the area enclosed is (6a + 9) square units?
The answer is NO. Again an infinite number of rectangles will enclose an
area of (6a + 9) square units. To list but a few:
1 × (6a + 9), 2 × (3a + 4.5), 3 × (2a + 3), 1.5 × (4a + 6), 6(a + 1.5),
-3(-2a – 3), 100(0.06a + 0.09), a × (6 + 9 ), ....
a
24 and 6a + 9 can be expressed as a product of two factors in infinitely many
ways. This would make the process of factorisation not very meaningful.
However when the word factorisation is used two things are assumed—and
pupils should be made aware of these—
(i) factorisation is over a set of numbers. This set over which the
factorisation is to be carried out is, at secondary school level, assumed to
be the (positive) integers.
(ii) factorisation always assumes that the highest common factor is used.
Module 4: Unit 3
59
Factorisation of expressions
Self mark exercise 2
1. Look at the suggested factorisations of 6a + 9.
a) Which have to be rejected because the factorisation is over the
integers?
b) Which have to be rejected because you are looking for the highest
common factor?
c) Does 6a + 9 have a unique factorisation over the integers?
2. Find the highest common factor of each pair of expressions.
a) 4x and 12
c) 15x and 60x2
b) 24a and 9
d) 4x2 and 26x3
3. Find the highest common factor of each pair of expressions.
a) 4x2 and 20x
c) 6x4 and 30x2
b) 24a3 and 9a2
d) 12x2 and 16x3
4. Find the highest common factor of each pair of expressions.
a) 20x2y and 12xy2
c) 5xy3 and 10x2y2
b) 24a2b2 and 9ab3
d) 3x2y4z and 27x3y3z2
Check your answers at the end of this unit.
Let’s return to the area model to represent the factorisation of 6a + 9.
Factorisation means: You are to find the dimensions of the rectangle given
its area enclosed.
As the width is to be the highest common factor, the width is 3. (step 2).
Dividing the width into the area:
6a + 9
= 2a + 3 gives the dimension of the
3
length (step 3).
Factorisation
Step 1: write down
the area.
Factorisation
Step 2: Width will be
equal to the highest
common factor.
Factorisation
Step 3: Divide
width into each
of the terms of
the area to find the
length.
Notice that this approach requires that the students be comfortable when
handling proper fractions. Some review may be necessary.
Module 4: Unit 3
60
Factorisation of expressions
To assist pupils in factorisation you go through various stages, starting at
numerical level and gradually including variables and factorisation of
increasing difficulty level.
Here are some examples of questions you could set to pupils.
Section A2: Outline of a worksheet for pupils on
factorisation
Objective: Pupils should be able to use the area model to factor out a
common factor.
It is assumed that pupils have covered an exercise on finding the highest
common factor of pairs of expressions.
Worksheet (outline)
Worked example:
Find the length of the missing side of each rectangle.
Now write down the factorisation of the expression.
The expected working is: the length of the rectangle is
24 ÷ 2 = 12 cm and
12 a + 9
= 4a + 3 respectively.
3
Factorisation: 24 = 2 × 12
Module 4: Unit 3
12a + 9 = 3(4a + 3)
61
Factorisation of expressions
Question 1
Find the missing side of each rectangle. Write down the expression you
factorised.
Question 2
Factorise using the area model. Check your answers using expansion.
a) 3a + 6
b) 12a + 6b c) 4x – 6y
d) 5x + 10y
f) 6a2 – 2b2
g) 8y – 12x
i) 2x + 4y – 6
h) 24a – 38b
e) 4a2 + 12b2
Question 3
Factorise using the area model then check by expanding your answer.
a) 3a2 + 6a
d) 5x3 + 10x2
g) 8y3m – 12y2
b) 12b + 6b2
e) 4b2 + 12b
h) 4 – 4abc3
c) 4x2 – 6x
f) 6a2 – 2a
i) x3 + x2 – x
Question 4
Factorise using the area model then check by expanding your answer.
a) 3ab2 + 6ab
d) 5x3y2 + 10x2y
g) 8xy3 – 12x2y3
Module 4: Unit 3
b) 12a3b + 6a2b
e) 4a2b2 + 12b2
h) 4ab – 4abc3
62
c) 4xy2 – 6y2
f) 6a2b – 2ab2
i) 2x2y2 – 6x2y + 8xy2
Factorisation of expressions
Self mark exercise 3
1. Work through the above pupils’ worksheet on factorisation.
2. Can any expression, having a common factor, be factorised using the
area model? Justify your answer.
Check your answers at the end of this unit.
The area model in factorisation has obviously the same weakness as it had in
expansions: negative numbers as factors do not fit in the model. A (slight)
step towards abstraction is using the multiplication table structure. In
expansion the entries in the first row and column were given, then the body
of the table had to be filled in. The expansion of a(b – 3) appeared in
multiplication table form as illustrated.
×
b
-3
a
In the case of factorisation: the entries in the cell of the table are known, but
the ‘headings’ are to be found. The factorisation of 2a2 + 6a is illustrated.
A pupils’ exercise could include examples similar to the following
Question 1: Complete the multiplication tables and the factorisations:
a)
b)
×
a
..
×
..
..
×
..
..
..
8a
24
3
12b
-9
..
t2
-8t
8a + 24 = . .(a + . .)
Module 4: Unit 3
c)
12b – 9 = 3(. . – . .)
63
t2 – 8t = t(. . – . .)
Factorisation of expressions
Question 2: Use a multiplication table to factorise:
a) 4a + 12
d) -a2 – 4a
b) 5p – 30q
e) -3ab – 6ab2 + 9a2
c) -3d + 9e
Self mark exercise 4
1. Work through the pupils’ sample questions for an exercise presented
above.
Check your answers at the end of this unit.
Unit 3, Practice activity
1. Use the area model and the multiplication table to factorise expressions
with a common factor in your classroom. You will have to develop
worksheets for the pupils to work through the various steps and levels of
difficulty.
Write an evaluative report on the activity.
2. Develop and try out a worksheet to extend factorisation of expressions
having a common term. Justify the structure of your worksheet and the
type of questions included.
Write an evaluative report.
3. Compare the activities suggested in this section with your own lesson
outline you wrote at the beginning of this section. Comment on
difference and similarities. Would you change your original lesson
outline after having gone through this section? Justify your answer.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 3
64
Factorisation of expressions
Section B: Factorising quadratic expressions
In this section you are going to look at various ways in which factorising of
quadratic expressions can be presented to pupils. Quadratic expressions are,
in general, of the form
ax2 + bx + c where a, b and c are real numbers and a ≠ 0.
Section B1:
Factorising a2 – b2 using a pattern approach
The following illustrate a possible discovery approach for the factorisation of
the difference of two squares. The objective of the activity is that pupils will
discover that a2 – b2 will factorise as (a + b)(a – b). Pupils sit in groups of
four, work individually through the worksheet, but can discuss ideas at any
moment with each other. If each pupil in the group has completed the
worksheet they are to compare, discuss and come up with a conjecture as a
group. The first part considers factorisation with a and b being positive
integers. The last question g extends to non integer values of a and b.
Question h is a challenge for the pupils and can only be tackled after they
have met with square roots.
This is only a first step in the factorisation of the difference of squares:
restricted to numerical cases, although leading to a generalised form
a2 – b2 = (a + b)(a – b).
The next step is to apply the relation to algebraic expressions.
Pupils are to factorise expressions such as
p2 – q2
4p2 – 9q2
16x2 – 36y2
a2b2 – 9c2
-25a2 + 64c2
And some more challenging expressions such as
2.25a2 – 6.25c2
Module 4: Unit 3
1 2 1 2
a – b
4
9
65
1
1
–
4 a 2 9d 2
Factorisation of expressions
Outline of a worksheet on factorisation of a2 – b2. A group/class
discussion.
a) Complete the table:
Difference of two
squares
Calculation
Factors
Pattern
42 – 12
16 – 1= 15
3×5
(4 – 1)(4 + 1)
52 – 22
25 – 4 = 21
3×7
(5 – 2)(5 + 2)
62 – 12
62 – 32
72 – 42
82 – 32
92 – 22
102 – 32
b) Can you find a pattern that connects the first column to the last?
c) Check your pattern for
122 – 82
452 – 212
Add these to the table.
d) Check three other differences of two squares and write your results in the
table.
e) How can you factorise a2 – b2?
f) Check your factorisation by working backwards i.e. expanding your
expression to see if you get a2 – b2.
g) Does your expression work for decimals? fractions? Check for example
3 2
1 2
(ii)   −  
 4
 2
(i) (3.2)2 – (1.3)2
h) Can you factorise 3 – 2?
Module 4: Unit 3
66
Factorisation of expressions
Self mark exercise 5
1. Work through the pupil’s activity described above to factorise a2 – b2
2. A pupil said that you can factorise the difference of two squares in more
than one way. For example 16 – 25 = 42 – 52 = (4 – 5)(4 + 5)
or also 16 – 25 = (-4)2 – 52 = (-4 – 5)(-4 + 5)
or also 16 – 25 = 42 – (-52) = (4 – -5)(4 + -5).
For algebraic forms such as a2 – 4b2 the pupil gave as factorisation
(-a + 2b)(-a – 2b) and (a + 2b)(a – 2b).
How would you respond to this pupil?
3. Factorise completely.
a) 72 – a2
d) 1 – 2.25d2
1
2
g) ( )2 – ( )2
3
3
2
j) 5b – 45
b) b2 – (0.5)2
e) 982 – 22
c) 9c2 – 16
f) 542 – 462
h) (9.8)2 – (0.2)2
i) 2a2 – 32
4. Factorise completely.
1
4
16 y 2
2
a) (x – 2 )
b) ( 2 –
)
25
x
9x
d) (a + b)2 – ( a – b)2 e) (x + 5)2 – 9
g) a4 – b4
h) x2 – 3
c) x4 – 1
f) (3x – 1)2 – (4x + 5)2
i) 2x2 – 5
Check your answers at the end of this unit.
Unit 3, Practice activity
1. Try out the activities in the pupil’s outline worksheet to discover the
factorisation of the difference of two squares from an arithmetic pattern.
Give sufficient consolidation with numbers before moving to algebraic
expressions.
Write an evaluative report.
2. Develop and try out a worksheet to assist your pupils in the learning of
the factorisation of the difference of two algebraic squares. Justify the
structure of your worksheet and the type of questions included. The
question 3 and 4 in the Self mark exercise 5 might give you some ideas
on questions that could be included. You might have to differentiate by
content to allow both low and high achievers to meet challenges.
Write an evaluative report.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 3
67
Factorisation of expressions
Section B2: Factorising x2 + ax + b
In this section you are going to look at ways to help pupils to factorise
trinomials of the form x2 + ax + b. Two methods are discussed
(i) generalisation from patterns (ii) using algebra tiles
I. Generalizing a pattern to discover the factorisation of x2 + ax + b
As factorisation is the reverse of expansion, pupils are expanding binomials
(using a multiplication algorithm) and comparing the expanded form with
the original factor product. Reading backwards—from expanded form to
factorised form—gives the factorisation of the trinomial. The coefficient
of x2 is kept 1 and is later to be extended.
Pupils sit in groups of four and work individually, but can consult each other
at any time. The completed worksheets need comparing and discussion in the
group. Some challenges are included at the end of the worksheet to convey
to pupils: you cannot always factorise over the integers.
The objective is to have pupils discover that to factorise x2 + ax + b they are
to find two numbers p and q, with sum a and product b. If such numbers can
be found the expression factorises as (x + p)(x + q). The multiplication table
is used to find the values of p and q.
The factorisation is restricted to factorisation over the integers. The value for
p and q to look for are integers. This also indicates the limitation of the
method, that very, very few trinomials can be factorised. To impress this on
pupils in the challenge activities, one can include some trinomials which do
not factorise over the integers or not factorise at all. (Although strictly
speaking ALL trinomials can be factorised over C, the set of complex
numbers).
Outline of a worksheet on factorisation of x2 + ax + b. A group/class
discussion.
a) Expand the expression in the first column using area model or
multiplication table model.
×
a
4
a
a2
4a
3
3a
12
7a
Module 4: Unit 3
68
Factorisation of expressions
Place the result below and complete the table.
Factorized form
(a + p)(a + q)
p
q
Expanded form
a2 + sa + t
s
t
(a + 4)(a + 3)
4
3
a2 + 7a + 12
7
12
(a + 1)(a + 2)
1
2
(a + 2)(a + 4)
(a – 2)(a + 5)
(a – 3)(a + 4)
(a – 2)(a – 5)
b) Can you find a pattern that connects the values of s and t to the values of
p and q?
Write it down in words and in formula form s = . . . and t = . . .
c) Check your relationship for three more expansions. Add your results to
the table.
d) Does your relationship work for decimals, e.g. (a – 2.1)(a + 3.7)?
1
3
fractions, e.g.,  a −   a +  ? Check and place in the table.

2
4
Module 4: Unit 3
69
Factorisation of expressions
e) Rewrite your table in the form:
Expanded form
a2 + sa + t
s
t
Factorised form
(a + p)(a + q)
p
q
a2 + 7a + 12
7
12
(a + 4)(a + 3)
4
3
(a + 1)(a + 2)
(a + 2)(a + 4)
(a – 2)(a + 5)
(a – 3)(a + 4)
(a – 2)(a – 5)
If you know the expanded form, for example a2 + 7a + 12, how can you
find the factorised form (a + p)(a + q)? In other words how can you find
the value of p and the value of q?
Make a conjecture.
f) Check whether your conjecture works for a2 + 14a + 33 by first
completing the multiplication table.
×
a
..
a
a2
..
..
..
33
Write down the factorisation of a2 + 14a + 33 = (a + .. )(a + ...)
g) Do the same for:
(i) a2 + 7a + 10 (ii) a2 + 11a + 10 (iii) a2 – a – 6 (iv) a2 + a – 12
h) Does your method also work for the difference of two squares?
Try b2 – 25.
Module 4: Unit 3
×
b
..
b
b2
..
..
..
25
70
Factorisation of expressions
i) Does your method always work?
Try to factorise c2 – 2c – 6, or c2 – 4c + 3.75 or x2 + 4.
Discuss when your method works and when it does not work.
Take some more practice with these
1. Complete the multiplication tables and write down the factorisation of
the expression.
a) a2 + 8a + 15
×
a
..
a
a2
..
..
..
15
a2 + 8a + 15 = (a .......)(a .......)
b) b2 – 12b + 35
×
b
..
b
b2
..
..
..
35
b2 – 12b + 35 = (b .......)(b .......)
c) c2 – 6c + 5
×
b
..
b
b2
..
..
..
35
c2 – 6c + 5 = (c .......)(c .......)
d) d2 – 6d – 7
×
d
..
d
d2
..
..
..
-7
d2 – 6d – 7 = (
)(
)
e) e2 – e – 56
×
e
..
e
e2
..
..
..
-56
e2 – e – 56 = (
Module 4: Unit 3
)(
)
71
Factorisation of expressions
f) f 2 + 21f – 72
×
f
..
f
f2
..
..
..
-72
f 2 + 21f – 72 = (
2.
)(
)
Factorise
a)
d)
g)
x2 + 11x + 24
z2 – 4z – 21
c2 – 13c – 30
b)
e)
h)
y2 + 14y + 40
a2 – a – 30
d2 + 13d – 30
c)
f)
x2 + 14x + 45
b2 – 13b + 30
Self mark exercise 6
1. Work through the pupils’ worksheet outlined above to factorise
x2 + ax + b.
Check your answers at the end of this unit.
II. Using algebra tiles in factorisation of trinomials ax2 + bx + c
Algebra tiles can be used as a concrete representation of factorisation. A
good supply of tiles representing the unit, x and x2 are needed.
To factorise for example x2 + 5x + 6, one algebra tile representing x2, 5 tiles
representing x and 6 unit tiles are to be placed such that they form a
rectangle (see next page). The dimensions of the rectangle will give the
factors. Cut out tiles from pages 40 and 41.
Module 4: Unit 3
72
Factorisation of expressions
Given tiles
The given tiles are to be placed such that they form a rectangle. The length
and width will be the required factors.
This is illustrated in the following diagram:
The sides of the rectangle measure (x + 3) and (x + 2).
This illustrates the factorisation of x2 + 5x + 6 = (x + 2)(x + 3).
Self mark exercise 7
1. Use your algebra tiles to factorise the following trinomials
a) 2x2 + 3x + 1
b) 3x2 + 4x + 1
c) 4x2 + 8x + 3
Check your answers at the end of this unit.
Module 4: Unit 3
73
Factorisation of expressions
The restrictions of the above model are:
(i) positive integral coefficients are needed and
(ii) relatively small coefficient are needed (10x2 + 17x + 3 would need 30
pieces!)
The first restriction can be overcome by using different coloured pieces to
represent positive (white) and negative (shaded) terms.
‘Negative’ regions are to be placed on top of a ‘positive’ region cancelling
out the region. If need be, equal positive regions and negative regions can be
added to the diagram.
Use your tiles to practice with the following cases:
(i) The diagram illustrates how to factorise x2 – 2x + 1 and x2 + x – 2.
You are to use one x2 tile, two (-x) tiles and one 1 tile to form a rectangle.
This can be done as illustrated.
The unshaded area in the last diagram is (x – 1)2
(ii) The next diagram illustrates how to factorise x2 + x – 2.
You are to use one x2 tile, one x tile and the two (-1) tiles to form a
rectangle. This can be done as illustrated, with adding a (x) tile and a (-x)
tile.
Module 4: Unit 3
74
Factorisation of expressions
x2 + x – 2 = (x + 2)(x – 1)
Self mark exercise 8
Use your algebra tiles to factorise the following trinomials:
a) x2 + 2x – 3
b) x2 – 2x – 3
c) 2x2 + 3x – 2
Check your answers at the end of this unit.
The working with algebra tiles to factorise might be unfamiliar to you. When
at school you might have been introduced to factorisation by a different
method. Therefore your first impression might be that this approach is
‘difficult’. Remember that any new method introduced to pupils is ‘difficult’
to them and that the method they learn to work with becomes ‘easy’ as
compared to another method they might be introduced to later. Research has
indicated that pupils introduced to factorisation by using concrete
manipulatives (tiles) prefer the concrete approach over a more abstract
sum/product method.
The concrete working with tiles can be gradually replaced by a more abstract
approach if pupils are ready for it. However the tiles should be available at
all times so pupils can use them if they want. A more permanent set—wood,
plastic, plasticized cardboard—is a valuable teaching/learning aid. The set
can also be used on the overhead projector. The more abstract method was
described in the previous section: the sum/product algorithm.
Module 4: Unit 3
75
Factorisation of expressions
Unit 3, Practice activity
1. Try out in your class the activity for pupils to discover the factorisation
of x2 + ax + b and the consolidation exercise.
Write an evaluative report.
2. Try out the use of the algebra tiles in your class in the factorisation of
trinomials.
Write an evaluative report.
Compare the use of the concrete manipulatives with the more abstract
‘sum/product’ method (use different classes!).
Which method has your preference? Justify.
Which method has pupils’ preference?
3. Develop and try out a worksheet to assist pupils in the learning of the
factorisation of ax2 + bx + c. Justify the structure of your worksheet and
the type of questions included.
Write an evaluative report.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 3
76
Factorisation of expressions
Section C: Fallacies
Pupils do at times succeed without real understanding of the concepts: they
have operational understanding i.e. they know how to apply certain
algorithms without knowing why they work. Going through fallacies is a
way to make pupils reflect on each and every step taken in an algebraic
argument. It helps to sharpen their relational understanding.
Here are some examples of “algebraic proofs”. The question is: What is
wrong in the following ‘proofs’ as the outcome is clearly false?
(i) ALL people are the same age.
Proof: If my age is x years and your age is y years then our average age is
1
m = ( x + y) .
2
so now
2m = x + y
2m(x – y) = (x + y)(x – y)
2mx – 2my = x2 – y2
y2 – 2my = x2 – 2mx
y2 – 2my + m2 = x2 – 2mx + m2
(y – m)2 = (x – m)2
y–m=x–m
y=x
So we are the same age.
(ii) Negative numbers are the same as positive numbers i.e. -a = +a because:
(-a)2 = (+a)2 Taking the square root of both sides
-a = +a
(iii) 1 = 2
Proof:
Let a = b
a2 = ab (multiply both sides by a)
a2 – b2 = ab – b2 (subtract b2 from both sides)
(a – b)(a + b) = b (a – b) (factorise)
a + b = b (divide both sides by a – b)
Then
But as
(iv)
a = b: a + a = a
2a = a (divide by a)
2=1
1
m = 25 cm
4
Taking the square root at both sides gives
1
m = 5 cm
2
Module 4: Unit 3
77
Factorisation of expressions
Self mark exercise 9
Clearly identify the error in each of the above ‘proofs’.
Check your answers at the end of this unit.
Unit 3, Practice activity
1. There are numerous algebraic fallacies. Most of them are based on the
two erroneous ideas illustrated above (i) ignoring that the square root of a
number or expression is to be a non negative value (ii) division by 0 is
undefined, and (iii) a few based on operation on the numbers only and
ignoring units.
Find or construct at least three algebraic fallacies.
2. Present fallacies to your pupils. In groups they are to discuss what is
wrong.
Write an evaluative report.
Present your assignment to your supervisor or study group for discussion.
Summary
This unit has continued to build algebraic concepts through examples and
working models (tiles), and has added the use of fallacies to your teaching
arsenal. It achieved realism—meaning applicability to the present or future
lives of most students—by avoiding any suggestion of using factorising in
the workplace.
Module 4: Unit 3
78
Factorisation of expressions
Unit 3: Answers to self mark exercises
Self mark exercise 1
1. Rectangles with dimensions 1 cm × 24 cm, 2 cm × 12 cm,
1
1
cm × 48 cm,
cm × 72 cm, 2 cm × 12 2 cm, etc. etc.
2
3
In other words there are an infinite number of rectangles enclosing
24 cm2.
2. To list but a few:
1 × (6a + 9), 2 × (3a + 4.5), 3 × (2a + 3), 1.5 × (4a + 6),
9
6(a + 1.5), -3(-2a – 3), 100(0.06a + 0.09), a × (6 + ), ....
a
Self mark exercise 2
1. a) 2 × (3a + 4.5), 1.5 × (4a + 6), 6(a + 1.5), 100(0.06a + 0.09),
9
a × (6 + )
a
b) 1 × (6a + 9)
c) NO there remain two possibilities, namely 3(2a + 3) and -3(-2a – 3),
both meeting the set conditions: factorised over the integers and hcf
is outside the brackets.
2. a) 4
b) 3
c) 15x
d) 2x2
3. a) 4x
b) 3a2
c) 6x2
d) 4x2
4. a) 4xy
b) 3ab2
c) 5xy2
d) 3x2y3z
Self mark exercise 3
N.B. Area diagrams have not been included here.
1. Question 1
a) 4p + 12 = 4(p + 3)
b) 5q + 10 = 5(q + 2) c) 6p + 9 = 3(2p + 3)
d) 8a + 16 = 8(a + 2)
e) c2 + 3c = c(c + 3) f) 9pq – 6p = 3p(3q – 2)
Question 2
Module 4: Unit 3
a) 3(a + 2)
b) 6(2a + b)
c) 2(2x – 3y)
d) 5(x + 2y)
e) 4(a2 + 3b2)
f) 2a(3a – 1)
g) 4(2y – 3x)
h) 2(12a – 19b)
i) 2(x + 2y – 3)
79
Factorisation of expressions
Question 3
a) (a + 2)
b) 6b(2 + b)
c) 2x(2x – 3)
d) 5x2(x + 2)
e) 4b(b + 3)
f) 2a(3a – 1)
g) 4y2(2y – 3)
h) 4(1 – abc3)
i) x(x2 + x – 1)
a) 3ab(b + 2)
b) 6a2b(2a + 1)
c) 2y2(2x – 3)
d) 5x2y(xy + 2)
e) 4b2(a2 + 3)
f) 2ab(3a – b)
g) 4xy3(2 – 3x)
h) 4ab(1 – c3)
i) 2xy(xy – 3x + 4y)
Question 4
2. Negative numbers as factors do not fit in the model.
Self mark exercise 4
Tables have not been included here.
1. a) 8a + 24 = 8(a + 3)
2. a) 4(a + 3)
c) t2 – 8t = t(t – 8)
b) 12b – 9 = 3(4b – 3)
c) -3(d – 3e) = 3(-d + 3e)
2b) 5(p – 6q)
d) -a(a + 4) = a(-a – 4)
e) 3a(-b – 2b2 + 3a) = -3a(b + 2b2 – 3a)
Self mark exercise 5
1. h) 3 – 2 = ( 3 − 2 )( 3 + 2 )
2. The factorisations are correct. Pupil should be allowed to use any of the
forms but should be made aware that the expressions are equivalent (if
he/she is not aware of that). To set task to “show that the expressions are
equivalent” would be a useful exercise in algebra.
3. a) (7 + a)(7 – a)
b) (b + 0.5)(b – 0.5)
c) (3c + 4)(3c – 4)
d) (1 + 1.5d)(1 – 1.5d)
e) (98 + 2)(98 – 2)
f) (54 + 46)(54 – 46)
2 1 2 1
g)  +   − 
 3 3  3 3
h) (9.8 + 0.2)(9.8 – 0.2)
i) 2(a2 – 16) = 2(a +4)(a – 4)
1
1
4. a)  x +   x − 

x
x
j) 5(b + 3)(b – 3)
2 4y 2 4y
b  +  − 
 3x 5   3x 5 
c) (x2 + 1)(x + 1)(x – 1)
d) (a + b + a – b)(a + b – a + b) [= 2a . 2b = 4ab]
e) (x + 5 + 3)(x + 5 – 3) = (x + )(x + 2)
f) ([3x – 1] + [4x + 5]) ([3x – 1] – [4x + 4]) = (7x + 4)(-x – 5)
g) (a2 + b2)(a + b)(a – b) h) (x + √3)(x – √3) i) (x√2 + √5)(x√2 – √5)
Module 4: Unit 3
80
Factorisation of expressions
Self mark exercise 6
1. a) See e
b) s = P + q, t = p × q
e)
Expanded form
a2 + sa + t
s
t
Factorized form
(a + p)(a + q)
p
q
a2 + 7a + 12
7
12
(a + 4)(a + 3)
4
3
a2 + 3a + 2
3
2
(a + 1)(a + 2)
1
2
a2 + 6a + 8
6
8
(a + 2)(a + 4)
2
4
a2 + 3a – 10
3
-10
(a – 2)(a + 5)
-2
5
a2 + a – 12
1
-12
(a – 3)(a + 4)
-3
4
a2 – 7a + 10
-7
10
(a – 2)(a – 5)
-2
-5
a2 + 1.6a + 7.77
1.6
7.77
(a – 2.1)(a + 3.7)
-2.1
3.7
1
4
-3
8
(a –
-1
2
3
4
a2 +
1
3
a–
4
8
1
3
)(a + )?
2
4
f) a2 + 14a + 33 = (a + 11)(a + 3)
g) (i) (a + 2)(a + 5)
(ii) (a + 1)(a + 10)
(iii) (a + 2)(a – 3)
(iv) (a + 4)(a – 3)
i) The quadratics do not factor over the integers.
More practice
1. a) (a + 3)(a + 5)
b) (b – 7)(b – 5)
c) (c – 1)(c – 5)
d) (d – 7)(d + 1)
e) (e – 8)(e + 7)
f) (f – 3)(f –+ 24)
b) (y + 4)(y + 10)
c) (x + 4)(x + 9)
d) (z – 7)(z + 3)
e) (a – 6)(a + 5)
f) (b – 3)(b – 10)
g) (c – 15)(c + 2)
h) (d + 15)(d – 2)
2. a) (x + 3)(x + 8)
Module 4: Unit 3
81
Factorisation of expressions
Self mark exercise 7
For example 1c:
You need to form a rectangle with 4 tiles of x2, 8 tiles of x and 3 unit tiles.
The diagram illustrates how the tiles are to be placed.
Hence 4x2 + 8x + 3 = (2x + 3)(2x + 1).
Module 4: Unit 3
82
Factorisation of expressions
Self mark exercise 8
Self mark exercise 9
(i) and (ii) both move from a quadratic form to a linear by taking a square
root. However by convention p is standing for the non negative number
which when squared gives p.
In (I) up to the line (y – m)2 = (x – m)2 the working is correct.
If m is the average of x and y then either x < m < y or y < m < x.
In the first case x – m < 0, in the second case y – m < 0
As the square root is to be a non-negative number
(y – m)2 = (x – m)2 is to be followed by either
y – m = m – x (if x < m < y)
or by
m – y = x – m (if y < m < x)
Both statements return to the (correct) starting line x + y = 2m.
Module 4: Unit 3
83
Factorisation of expressions
In (ii) if a > 0: (−a)2 = a (not -a) and
square roots are to be non negative.
a2 = a (not -a), as by convention the
If a < 0: (−a)2 = -a (not a) and a2 = -a (not a), as by convention the
square roots are to be non negative.
(iii) “Hiddenly” the equation is divided by 0 (in line three, dividing by a – b,
but as a = b you are dividing by 0, which is undefined).
(iv) is based on performing the operation (root extraction) only on the
numbers and ignoring the units involved.
Module 4: Unit 3
84
Factorisation of expressions
Unit 4: Binomial theorem and factor theorem
Introduction to Unit 4
Expanding and factorisation belong to the major manipulative techniques in
algebra. The nature of algebra is to generalise. Expanding the product of
(ax +b)(cx + d) nearly automatically suggests to look at products of the form
(ax + b)(cx + d)(ex + f ), (Ax2 + Bx + C)(ax2 + bx + c), (ax + by)3, etc. or in
general the product of m polynomials. Similarly for factorisation: after
having looked at trinomials one wonders “What about expressions with four
terms, five terms, .. n terms?”
Purpose of Unit 4
In this unit you will look at extension work on expansion, the product of
polynomials and binomial theorem, and extension work on factorisation,
remainder and factor theorem. The intention is to extend and/or strengthen
your content knowledge. The material covered in this unit is above the junior
secondary level, although the high achiever could investigate some aspects,
for example Pascal’s triangle, covered in this unit. There is no attempt to
relate these theorems to everyday life. Scientific calculators have supplanted
all the practical uses of these theorems and their derived techniques.
Objectives
When you have completed this unit you should be able to:
• define a polynomial in one variable
• give examples and non-examples of polynomials in one variable
• state the degree of a polynomial
• multiply two polynomials by each other
• apply the binomial theorem for integer and fractional indices
• use Pascal’s triangle in expansion of (ax + by)n, where n is a positive
integer less than 10
• define factorial n
n
• state the meaning of  r 
• write the binomial expansion for positive integer values of the exponent
n
using factorial notation and  r  notation.
• recognise Pascal’s triangle pattern in various situations
• divide a polynomial of degree n by a polynomial of degree m where (n > m)
• use the remainder theorem to find the remainder of a polynomial when
divided by a linear factor ax + b
• use the factor theorem to find a factor of a polynomial and factorise the
polynomial completely
Time
To study this unit will take you about 10 hours.
Module 4: Unit 4
85
Binomial theorem and factor theorem
Unit 4: Binomial theorem and factor theorem
Section A1: Product of polynomials
An expression of the form axn is called a monomial in one variable x, where
a is a real number and n is a non negative integer.
axnym is a monomial in two variables x and y, where a is a real number and
n and m are non negative integers.
The sum of monomials is called a polynomial. For one variable the general
form is
a0 + a1x + a2x2 + ... + anxn, coefficients are real numbers.
If an ≠ 0, the one variable polynomial in x is said to be of degree n i.e. the
highest index of x occurring with a non zero coefficient.
Two polynomials A(x) and B(x) can be multiplied by applying the ‘extended’
distributive law. The lay-out can take different formats.
Example:
(i)
(x2 – 3x + 1)(2x3 + x – 3)
= x2(2x3 + x – 3) – 3x(2x3 + x – 3) + 1(2x3 + x – 3)
= 2x5 + x3 – 3x2 – 6x4 – 3x2 + 9x + 2x3 + x – 3
= 2x5 – 6x4 +3x3 – 6x2 + 10x – 3.
(ii) The working can also be represented in a multiplication table:
×
2x3
x
-3
x2
2x5
x3
-3x2
-3x
-6x4
-3x2
9x
1
2x3
x
-3
Taking like terms together gives: 2x5 – 6x4 +3x3 – 6x2 + 10x – 3.
As in the previous working, like terms have to be ‘found’—they are not
neatly standing below each other to allow convenient taking together of the
like terms.
Module 4: Unit 4
86
Binomial theorem and factor theorem
(iii) Lay out as with ‘long multiplication’ of numbers. Leaving ‘space’ for
‘missing’ powers, the multiplication allows one immediately to order the
powers such that like terms are placed in the same column, making addition
easier.
2x3
+x –3
x2
2x3
-6x4
2x5
– 3x + 1
+x –3
-3x2 + 9x
+ x3 – 3x2
2x5 – 6x4 +3x3 – 6x2 + 10x – 3
Self mark exercise 1
1. State the degree and the coefficient of x3 for each of these polynomials:
a) 2x3 – 2x + 17
b) 8 – 2x2 + 5x3
c) 4x4 – 2x2 – 9x
d) 3
e) 3x8 – 4x6 – x3 + 1
2. Expand and write the answer in descending powers of x.
a) (2x2 – 3x + 3)(3x2 + x – 2)
b) (3 – x)( 2 – x2 + 4x3)
c) (2x6 – 3x2 – 2)(3x6 + 4x2 – 1)
d) (2x – 3)(4x + 5)(x2 – 2x + 3)
3. a) A polynomial Pn of degree n is multiplied by a polynomial Pm of
degree m. What is the degree of the product polynomial?
b. A polynomial Pn of degree n is squared. What is the degree of the
resulting polynomial?
c. A polynomial Pn of degree n is raised to the power t. What is the degree of
the resulting polynomial?
Check your answers at the end of this unit.
Section A2: Binomial expansion
Check the expansion of powers of (x + 1) listed below by multiplying the
previous result by (x + 1) to obtain the next line.
(x + 1)1 = x + 1
(x + 1)2 = x2 + 2x + 1
(x + 1)3 = (x + 1)(x + 1)2 = x3 + 3x2 + 3x + 1
(x + 1)4 = (x + 1)(x + 1)3 = x4 + 4x3 + 6x2 + 4x + 1
(x + 1)5 = (x + 1)(x + 1)4 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1
Module 4: Unit 4
87
Binomial theorem and factor theorem
The triangle formed with the coefficients of the expansion of (x + 1)n is
known as Pascal’s triangle as it was intensively studied by the French
mathematician Blaise Pascal (1623 - 1662). The triangle had already been
described much earlier by Chinese mathematicians (1303 Chu Shih-chieh)
and by Omar Khayyam (ca 1100, Persian poet and mathematician). The first
rows, following from the above results, are as follows
Self mark exercise 2
1. Expand
a) (x + 1)6
b) (x + 1)7
c) Using your results of a and b write the next two rows in Pascal’s
triangle.
2. a) Study the triangle and see whether you can find a pattern to construct
the next row from the previous one.
b) Using the pattern write down two more rows in Pascal’s triangle.
Check your answers at the end of this unit.
Module 4: Unit 4
88
Binomial theorem and factor theorem
To find an expression for the expansion of (x + 1)n study the following
pattern:
(x + 1)1 = x + 1
(x + 1)2 = x2 + 2x + 1
3
3×2
(x + 1)3 = x3 + 3x2 + 3x + 1= x3 + x2 +
x+1
1
1× 2
4
4×3 2 4×3×2
x +
x+1
(x + 1)4 = x4 + 4x3 + 6x2 + 4x + 1= x4 + x3 +
1
1× 2
1× 2 × 3
(x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1=
5
5× 4 3 5× 4×3 2 5× 4×3×2
x +
x +
x+1
x5 + x4 +
1
1× 2
1× 2 × 3
1× 2 × 3× 4
(x + 1)6 = x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1=
x6 +
6 5 6×5 4 6×5×4 3
x +
x +
x +
1× 2
1
1× 2 × 3
6×5× 4×3 2 6×5× 4×3×2
x +
x+1
1× 2 × 3× 4
1× 2 × 3× 4 × 5
(x + 1)7 = x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1
(x + 1)8 = x8 + 8x7 + 28x6 + 56x5 + 70x4 + 56x3 + 28x2 + 8x + 1
Self mark exercise 3
1. In a similar way write down the coefficients in the expansion of (x + 1)7
and (x + 1)8.
2. Write down the expansion of (x + 1)n by generalising the pattern in (1).
Check your answers at the end of this unit.
The result inductively obtained in the self mark exercise 3, question 2 is
known as the binomial expansion for positive integer index or binomial
theorem.
n × (n - 1) 2 n × (n - 1) × (n - 2) 3
(1+ x)n = 1 + nx +
x +
x +
1× 2
1× 2 × 3
n × (n - 1) × (n - 2) × (n - 3) 4
x + ... +
1× 2 × 3× 4
... +
Module 4: Unit 4
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × .... × (n -{r - 1}) r
x + ... + ... xn
1 × 2 × 3 × 4 × 5 × .... × r
89
Binomial theorem and factor theorem
The coefficients in the expansion can be obtained from Pascal’s triangle for
lower values of n. For larger values of n the above formula is more
economical to use.
Note the following:
(i) The number of terms in the expansion is (n + 1).
(ii) The expansion can be written down in ascending or descending powers
of x due to the symmetry in the coefficients.
(iii) the general term in the expansion xr has as coefficient
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × .... × (n -{r - 1})
1 × 2 × 3 × 4 × 5 × .... × r
(iv) No proof of the binomial theorem for positive integer index has been
given. To prove the theorem for positive integer index proof by induction
is generally used. This is not discussed in this unit.
(v) The expansion is symmetric as you can see in the structure of Pascal’s
triangle: you can write down the expansion in ascending or in descending
powers of x using the same row in the triangle. The coefficients of xr and
xn-r are the same.
(vi) The product 1 × 2 × 3 × 4 × 5 × ..... × n is written in mathematics as n!
and called ‘n factorial’. There should be a factorial key on your
calculator. For example: 5! = 1 × 2 × 3 × 4 × 5 = 120.
0! is taken as 1.
This allows a shorter notation for the binomial coefficient of the general
term:
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × ... × (n -{r - 1})
=
1 × 2 × 3 × 4 × 5 × ... × r
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × ... × (n -{r - 1}) × (n - r ) × (n - r - 1) × ... × 2 × 1
=
1 × 2 × 3 × 4 × 5 × ... × r × (n - r ) × (n - r - 1) × ... × 2 × 1
n!
n
=  r 
r! ×(n - r )!
Using the factorial notation the binomial theorem can be written as
n
n
n
n
(1+ x)n = 1 + 1  x +  2 x 2 +  3 x 3 + ... +  r  x r + ... + x n
Module 4: Unit 4
90
Binomial theorem and factor theorem
(vii) The expansion of (1+ x)n can be used to obtain the expansion of the
more general expression
(p + q)n as follows:
[p(1 +

q n
q
q
n × (n - 1)  q  2
)] = pn(1 + )n = pn  1 + n × +
  +

p
p
p
1 × 2  p
n × (n - 1) × (n - 2) q 3
n × (n - 1) × (n - 2) × (n - 3)  q  4
( ) +
  + ... +
 p
p
1× 2 × 3
1× 2 × 3× 4
 q 
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × .... × (n -{r - 1})  q  r
  + ... + ...   n  =
 p
 p 
1 × 2 × 3 × 4 × 5 × .... × r
pn + n × pn-1 q+
n × (n - 1) n-2 2 n × (n - 1) × (n - 2) n-3 3
p q +
p q +
1× 2
1× 2 × 3
n × (n - 1) × (n - 2) × (n - 3) n-4 4
p
q + .....+
1× 2 × 3× 4
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × ... × (n -{r - 1}) n-r r
p q + ... + ... qn
1 × 2 × 3 × 4 × 5 × ... × r
In factorial notation this becomes:
(p+ q)n =
 n
 n
 n
 n
pn +   pn-1q +   pn-2 q 2 +   pn-3 q3 + ... +   pr q n-r + ... + q n
 1
 2
 3
 r
Note that in this expansion:
The number of terms in the expansion is (n + 1).
The general term is pn-r qr
The sum of the indices of p and q in each term is always n.
Due to the symmetry you can interchange p and q in the expression and by
doing so the expansion is either in ascending powers of p (and descending
powers of q) or in descending powers of p (and ascending powers of q).
Example: (3x – 4y)3 expanded in ascending powers of x (and hence
descending powers of y) gives:
(-4y)3 + 3 × (3x) × (-4y)2 + 3 × (3x)2 × (-4y) + (3x)3 =
-64y3 + 144xy2 – 108x2y + 27x3
or in descending powers of x (and ascending powers of y):
(3x)3 + 3 × (3x)2 × (-4y) + 3 × (3x) × (-4y)2 + (-4y)3 =
27x3 – 108x2y + 144xy2 – 64y3
Module 4: Unit 4
91
Binomial theorem and factor theorem
Self mark exercise 4
Using the binomial expansion expand in descending powers of x.
1
1. (1 + 2x)4
2. (1 – 2x)3
3. (1 + x)5
2
1
1
4. (2x – 3y)4
5. (x2 + y2)3
6. ( x + y)3
3
2
7. Use the binomial expansion to obtain the value of (71.02)3 to 2 decimal
places accurately given that 713 = 357 911. Hint: expand (71 + 0.03)3.
8. Expand (1 + x – x2)5 in ascending powers of x up to and including the
term in x3 by writing the expression as {1 +( x – x2)}5.
9. a) Find the coefficient of x4 in the expansion of (3x – 2)11.
b) Find the coefficient of x3 in the expansion of (2 – 5x)9.
10. Find the term independent of x in the expansion of (3x +
2 12
) .
x
11. Use the expansion of (2 + x)7 to obtain correct to 3 decimal places the
value of:
a) (2.08)7
b) (1.99)7
12. Find the value of the following binomial coefficients:
8
a)  3
8
12
b)  5 c)  4 
12
25
d)  8  e) 13 
25
f) 12 
 n 
n
g) What does this suggest for r  and n - r  ?


 
h) Prove your conjecture made in g.
Check your answers at the end of this unit.
The binomial expansion for (1 + x)n holds in general for all real values of n.
For the proof you need Maclaurin’s theorem which is covered in calculus
courses. Here we will assume that the expansion can be applied and
investigate whether it makes sense.
p
First we look at positive rational index. Remember that x
and q are positive integers.
q
= x p where p
q
1
Check that for (1 + x ) 3 = 3 1 + x you will get the following expansion
(Note that the factorial notation has no meaning in this case and the original
expression for the expansion is to be used).
Module 4: Unit 4
92
Binomial theorem and factor theorem
1 1
1 1
1
× ( − 1)( − 2)
× ( − 1)
1
3
1+ x + 3 3
x2 + 3 3
x 3 + ... =
3
1× 2
1× 2 × 3
1
1
5
1 + x − x 2 + x 3 − ...
3
9
81
Work out the next term in the expansion to confirm that the next term is
10 4
−
x
243
The series, unlike in the case for positive integer index, will never end as the
coefficient of the powers of x will never become 0.
Does the infine series give reasonable results?
Let’s check this for some values of x.
Checking for x = 7
x = 7 gives
1
(1 + 7) 3
= 3 1+ 7 = 3 8 = 2
The expansion, for x = 7, gives
1+
14
196
1
1
5
10
7
4
× 7 − × 72 + × 73 −
× 74... = 1 + − 5 + 21 − 98
+ ...
243
3
9
81
243
3
9
81
Tabulating the sum obtained against the number of terms used in the
expansion we get:
Number of terms
Sum
1
1
2
3.33
-1.11
3
4
20.06
5
78.74
This in no way comes closer to the expected value of 2. The more terms used
in the expansion the further away from the value 2 one gets. The series is
divergent for x = 7. This is illustrated in the graph below:
Checking x = 2
Module 4: Unit 4
93
Binomial theorem and factor theorem
Expected outcome of the series is
1
(1 + 2) 3
= 3 1 + 2 = 3 3 ≈ 1.442
The expansion, for x = 2, gives
1+
1
1
5
10
2 4 40 160
× 2 − × 2 2 + × 23 −
× 2 4..... = 1 + − +
−
+ .....
3
9
81
243
3 9 81 243
Tabulating the sum obtained against the number of terms used in the
expansion we get:
Number of terms
1
2
3
4
5
Sum
1
1.67
1.22
1.72
1.06
As in the previous case the values of the sum do not come closer to the
expected value of 1.442. To get closer to the answer the terms in the
expansion should become smaller and smaller (in absolute value). The graph
below illustrates that the values do not get closer to 1.442. The series is
divergent for x = 2
Checking x = 1
Expected outcome of the series is
1
(1 + 1) 3
The expansion, for x = 1, gives 1 +
= 3 1 + 1 = 3 2 ≈ 1.260
1 1 5 10
− + −
.....
3 9 81 243
Tabulating the sum obtained against the number of terms used in the
expansion we get:
Number of terms
1
2
3
4
5
Module 4: Unit 4
Sum
1
1.33
1.22
1.28
1.24
94
Binomial theorem and factor theorem
Presented in a graph
The values are alternatively a little above or below the true sum, but getting
closer each time. The series converges to the value 3 2 ≈ 1.260 .
Self mark exercise 5
1
(1 + x ) 3
1 1
5
10 4
x + ...
= 3 1 + x = 1 + x - x2 + x3 −
3 9
81
243
to check whether the series converges or diverges if
Use the expansion of
1. x =
−7
8
2. x = 0.1
3. x = -1
Check your answers at the end of this unit.
1
(1 + x ) 3
The result of investigating the expansion of
= 3 1 + x is that the
series diverges for values of x > 1 or x < -1 and is convergent for -1< x < 1.
This is generally true for the expansion of
(1 + x)n , where n is a positive rational number. The series is convergent for
-1< x < 1 and divergent for values of x outside this range.
The binomial expansion has lost its utility, even for scientists and engineers,
as calculators are easily available now. In the days when no calculators were
available, the binomial expansion allowed the computation of surds to any
degree of accuracy required. Tables of square roots, cube roots etc. were
constructed using the binomial expansion.
For example, using the expansion
1
(1 + x ) 2
= 1+ x = 1+
and taking x =
Module 4: Unit 4
1 1 2 1 3
5 4
x- x + x −
x + ...
2 8
16
128
2
2
11 1
the left hand side gives us 1 + =
=
11
9
9 3
9
95
Binomial theorem and factor theorem
The right hand side (the first terms of the series) gives
1+
1 2 1  2 2 1  2 3
5  2 4
× - ×
+ ×
−
×
+ ... =
2 9 8  9
16  9 
128  9 
1 + 0.11111 − 0.00617 + 0.00068 − 0.000095 + ....
Hence
1
11 = 1.10553 to 5 decimal places and 11 = 3.3166 to 4 decimal
3
places.
Remember that in order to obtain a final result to 4 decimal places
accurately, in the working you have to use (at least) one more decimal place
and round to 4 decimal places in the last step.
The expression (p + q)n cannot be expanded using the expansion you
obtained when n is a positive integer, if n is a (positive) rational number.
More general: the expansion for non positive integer values of n applies to
the form (1 + t)n only i.e. the expression in the brackets is to start with 1. For
1
example, to expand the expression ( 4 + x ) 2 it is to be restructured into the
format
1
1
1
x
x
(1 + t)n by writing ( 4 + x ) 2 = [ 4(1 + )] 2 = 2(1 + ) 2 before applying the
4
4
binomial theorem.
Self mark exercise 6
1. Expand 5 1 + x in ascending power of x up to and including the term in
x3. Use the expansion and take x = 0.02 to find to 4 decimal places
accurately 5 1.02
2. Expand 5 x + 1 in ascending power of x up to and including the term in
1
x3. Use the expansion and take x =
to find to 4 decimal places
100
accurately 105
3. Expand 3 1 - 4 x in ascending power of x up to and including the term in
1
x3. Use the expansion and take x =
to find to 4 decimal places
100
accurately 3 15
4. Expand 1 − 3 x in ascending power of x up to and including the term in
1
x3. Use the expansion and take x =
to find to 4 decimal places
30
accurately 10
Self mark exercise 6 continued on following page
Module 4: Unit 4
96
Binomial theorem and factor theorem
Self mark exercise 6 continued
3
5. Expand (1 – 2x) 2 in ascending power of x up to and including the term
in x3. Use the expansion and take x =
1
to find to 4 decimal places
8
accurately 3
2 + x in ascending power of x up to and including the term
6. Expand
in x3.
7. Assume that the expansion of (1 + x)n is valid for negative integers and
rational numbers
Find the first 5 terms in the expansion in ascending powers of x of the
following expressions and investigate the range of values of x for which
the expansion is valid.
a) (1 + x)-1
b) (1 + x)-2
c) (1 + x)
1
-2
8. Expand (1 + x)-3 in ascending powers of x up to and including the term
in x4. Use the expansion and take x = - 0.01to find to 4 decimal places
accurately (0.99)-3.
9. Expand in ascending powers of x up to and including the term in x4:
a) (1 + 3x)-2
b) (1 – 4x)-5
c) (2 – x)-2
10. Expand in ascending powers of x up to and including the term in x3:
1
-2
a) (1− x)
b) (1 + 2 x)
1
-2
Check your answers at the end of this unit.
In summary: If the binomial expansion of (1 + x)n leads to an infinite series,
the series
(i) if n > 0 converges for -1< x < 1
(ii) if -1 < n < 0 converges for -1 < x < 1
(iii) if n < -1 converges for -1 < x < 1
Module 4: Unit 4
97
Binomial theorem and factor theorem
Section B1: Investigating Pascal’s Triangle
Pascal’s triangle has been studied for many years and mathematicians
continue to find interesting patterns and situations related to or leading to
Pascal’s triangle.
Reading along diagonals you might have spotted the counting numbers 1, 2,
3, 4, .. and the triangular numbers 1, 3, 6, 10, 15, .... The following
investigations all relate to Pascal’s triangle.
Problem Solving 1
Investigate the following:
1. The sum of the numbers in the nth row of the Pascal’s triangle.
Number of row
Sum of numbers
1
2
3
4
1
2
4
8
2. Find the sum of all the numbers in Pascal’s triangle contained in
rectangles with as one corner the top number 1. Relate it to one of the
numbers outside the rectangle. For example:
Continues on next page.
Module 4: Unit 4
98
Binomial theorem and factor theorem
3. Starting at the top how many downwards paths can you take to spell
“MATHEMATICS”?
4. Find the sum of the numbers on the perimeter of Pascal triangles with 2,
3, 4, ..... n rows.
5. Lebogang lives in house 1 in a quarter where the houses are placed as
shown in the diagram. There are paths between the houses. Her friend
lives in house number 12. In how many different ways can Lebogang
reach her friend’s house by always moving either right or up in the
diagram? She leaves by the front door of her own house and enters her
friend’s house by the front door (bold line in diagram).
Check your answers at the end of this unit.
Section B2: Pascal’s Pyramid
Pascal’s triangle can be used to determine the coefficients of a binomial
expansion (a + b)n. To obtain the coefficients in the expansion of trinomials
(a + b + c)n Pascal’s triangle can be extended to Pascal’s pyramid. Generate
first some data by using multiplication of polynomials to find the expansion
of (a + b + c)2, (a + b + c)3 and (a + b + c)4.
Module 4: Unit 4
99
Binomial theorem and factor theorem
Self mark exercise 7
1. Obtain the expansion of
a) (a + b + c)2
b) (a + b + c)3
c) (a + b + c)4.
Check your answers at the end of this unit.
The coefficients in the expansion of (a + b + c)n can be represented by
sections of a triangular pyramid.
The first expansion (a + b + c)0 has the single coefficient 1, represented by
the vertex of the pyramid.
The next expansion (a + b + c)1 with coefficients 1, 1, 1 is represented by
the next section of the pyramid with the 1 on the lateral edges of the
pyramid.
The next layer is to represent the coefficients in the expansion of
(a + b + c)2.
The numbers on the outer edges of each layer (the numbers between the
vertices) are the sum of the two numbers directly above it in the previous
layer. The number(s) in the interior of the triangle is the sum of the three
numbers above it in the previous layer.
Module 4: Unit 4
100
Binomial theorem and factor theorem
To assign these coefficients to the correct term in the expansion consider the
pattern in the following diagrams:
Looking, for example, at the third layer: starting from the vertex with a3
moving towards the vertex with b3, the exponents of a decrease by 1 in each
step and the exponents of b increase by 1. This applies similar along the
other sides of the triangle. The coefficient in the middle, 6, was the result of
adding 2 + 2 + 2 from the previous layer. These were the coefficients of ab,
bc, and ca. These have as lowest common multiple abc—the term in the
middle of the next layer.
Self mark exercise 8
1. Using the third layer section represent the coefficients of the fourth layer
in a triangle and assign the coefficient to the appropriate terms.
2. What do you notice about the numbers along the sides of the triangles in
each layer?
Check your answers at the end of this unit.
The observation in question 2 of the Self mark exercise 8 allows for an
alternative way of obtaining the coefficients.
The Pascal triangle ending with the coefficients 1
looks as follows:
4
6
4
1
The lower side is as required in our 4th pyramid layer.
Module 4: Unit 4
101
Binomial theorem and factor theorem
Multiplying the first row by 1, the second by 4, the third by 6, the fourth by 4
and the fifth by 1 to obtain the numbers as required on the sides of the
pyramid triangle generates the required layer of the pyramid.
Self mark exercise 9
1. Starting with the appropriate Pascal’s triangle, obtain the coefficients for
the expansion of (a + b + c)5
2. Obtain the expansion of
a) (a + b + c)5
b) (a + 3b + 2c)3
c) (a + 2b + c)4.
3. Investigate how the coefficients in the expansion of (a + b + c)n can be
expressed using factorial notation.
4. Investigate the ‘multinomial theorem’ i.e. the expansion of
(a1 + a2 + a3 + .... + ar)n
Check your answers at the end of this unit.
Section C1: Dividing polynomials
Polynomials can be divided using long division. For example to divide
6x3 – 13x2 + 4x + 5 by (2x + 1) you could work as follows:
)
3x 2 − 8x + 6
2 x + 1 6 x 3 − 13 x 2 + 4 x + 5
6 x + 3x 2 + 4 x
−
16 x 2 + 4 x
−
16 x 2 − 8 x
12 x − 5
12 x + 6
−
1
Consider only the 2x in the divisor and the first term of the polynomial 6x3.
2x into 6x3 goes 3x2 times.
Multiply (2x + 1) by 3x2 to give 6x3 + 3x2.
Subtract this from 6x3 – 13x2, which will give you -16x2.
2x into -16x2 goes -8x times.
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Binomial theorem and factor theorem
Multiply (2x + 1) by -8x to give -16x2 – 8x.
Bring the + 4x down and subtract -16x2 – 8x from -16x2 + 4x
to give you + 12x.
2x into + 12x goes 6 times. Multiplying (2x + 1) by 6 leads to 12x + 6.
Bring down +5 and subtracting leads to the remainder -1.
Hence (6x3 – 13x2 + 4x + 5) ÷ (2x + 1) gives as quotient (3x2 – 8x + 6) and
as remainder -1.
Compare with 13 ÷ 3 which gives 4 remainder 1.
This can be written as 13 ÷ 3 = 4 +
1
3
or 13 = 3 × 4 + 1, i.e., (dividend) = (divisor) × (quotient) + (remainder).
For the polynomial example you can write
–
6 x 3 – 13 x 2 + 4 x + 5
1
= (3 x 2 – 8 x + 6) +
2x + 1
2x + 1
6x3 – 13x2 + 4x + 5 = (2x + 1)( 3x2 – 8x + 6) + (-1)
Which is of the same format as the numerical example:
(dividend) = (divisor) × (quotient) + (remainder).
1
1
The zero of the divisor is - , i.e., 2x + 1 = 0 for x = - .
2
2
1
Verify that P(- ) = -1. Is this a coincidence, that you get the remainder?
2
That you will investigate further in the next self mark exercise.
Self mark exercise 10
1. Find quotient and remainder when P(x) is divided by D(x) where
a) P(x) = x2 – 3x + 4 and D(x) = x + 4
b) P(x) = 3x3 – 13x2 – 20 and D(x) = x – 5
c) P(x) = 4x3 + 6x2 – 2x – 1 and D(x) = 2x + 1
2. Divide P(x) by D(x) and find the remainder. Calculate also the value of
P(x) for the value of making the divisor zero. Start by entering the results
already obtained.
Continued on next page.
Module 4: Unit 4
103
Binomial theorem and factor theorem
Self mark exercise 10 continued.
Complete the following table:
P(x)
D(x) = ax + b
Remainder
b
P(- )
a
6x3 – 13x2 + 4x + 5
2x + 1
-1
1
P(- ) = -1
2
x2 – 3x + 4
x+4
3x3 – 13x2 – 20
x–5
4x3 + 6x2 – 2x – 1
2x + 1
6x2 – 5x – 3
x–1
x3 – 10
x–2
4x3 – 13x2 + 3x – 3
4x – 1
3. Can you make a conjecture about the remainder when a polynomial P(x)
is divided by (ax + b)?
Check your answers at the end of this unit.
Section C2: Remainder theorem
The conjecture you were to come up with in the last exercise is not so hard to
prove.
Let us look at a simpler case first: dividing a polynomial P(x) by (x – a).
Remember that for all division the identity
(dividend) = (divisor) × (quotient) + (remainder) holds.
If P(x) is divided by (x – a) and the quotient is Q(x) and the remainder R (a
constant, a polynomial of degree 0) you have the identity
P(x) = (x – a) Q(x) + R
Because this is an identity (an equation holding for ALL values of the
variable x) we can take any value for x, for example also the value x = a.
This gives:
P(a) = (a – a) Q(a) + R
P(a) = 0 + R
P(a) = R
In words: The remainder when the polynomial P(x) is divided by (x – a) is
P(a). This is called the remainder theorem.
Module 4: Unit 4
104
Binomial theorem and factor theorem
The theorem can also be written down in case you divided the polynomial
P(x) by the factor (ax + b). In this case you have the identity
P(x) = (ax + b) Q(x) + R
Taking for x the value -
b
a
 − b
 − b
b
P   = (a × – b) Q   + R
a
 a
 a
 − b
P  = 0 + R
 a
 − b
P   = R.
 a
In words: The remainder when the polynomial P(x) is divided by (ax + b) is
 − b
P  .
 a
Self mark exercise 11
1. Find the remainder when
a) x3 – 6x2 + 4 is divided by x+ 1
b) x4 – 8x2 + 3 is divided by x – 1
c) x15 – 1 is divided by x + 1
d) 4x3 – 5x2 + x is divided by x – 2
2. Find the value of k
a) when 2x4 + kx3 + 8 is divided by x + 2 the remainder is 16
b) when kx3 – 4x2 + 6 is divided by x – 3 the remainder is -3
c) when 2x5 + kx3 + x is divided by x + 1 the remainder is -5
3. The remainder when 2x3 + ax2 + bx – 2 is divided by x + 2 is 4. When
divided by x – 2 the remainder is 24. Find the values of a and b.
4. If the polynomial P(x) is divided by (x – a)(x – b) i.e. a quadratic
expression, the remainder will be a linear expression of the form Ax + B.
If the quotient is Q(x) the identity
P(x) = (x – a)(x – b) Q(x) + (Ax + B) will hold.
Find the remainder of Ax + B.
Check your answers at the end of this unit.
Module 4: Unit 4
105
Binomial theorem and factor theorem
Section C3: Factor theorem
If a polynomial P(x) is divided by (x – a) you saw that the identity
P(x) = (x – a) Q(x) + R holds.
In case that the remainder R = 0, x – a is a factor of P(x). This is the factor
theorem:
(x – a) is a factor of the polynomial P(x) if P(a) = 0
or also:
 − b
(ax + b) is a factor of the polynomial P(x) if P   = 0.
 a
The factor theorem is used to find rational factors of polynomials.
Example:
Find the rational factors of P(x) = 2x3 + x2 – 13x + 6.
Try factors of the constant 6: ± 1, ± 2, ± 3, ±6.
P(1) = 2 + 1 – 13 + 6 ≠ 0, so (x – 1) is not a factor.
P(-1) = -2 + 1 + 13 + 6 ≠ 0, so (x + 1) is not a factor.
P(2) = 16 + 4 – 26 + 6 = 0, so (x – 2) is a factor.
P(x) = (x – 2) (2x2 + 5x – 3) by long division.
P(x) = (x – 2) (2x – 1)(x + 3) by factorising the quadratic.
Note that the factorisation now gives an easy way to solve
the equation 2x3 + x2 – 13x + 6 = 0.
2x3 + x2 – 13x + 6 = 0
(x – 2) (2x – 1)(x + 3) = 0
x – 2 = 0 or 2x – 1 = 0 or x + 3 = 0
x = 2 or x =
Module 4: Unit 4
1
or x = -3
2
106
Binomial theorem and factor theorem
Self mark exercise 12
1. Find the value of x if
a) x – 2 is a factor of 3x3 + kx – 20
b) x + 1 is a factor of kx3 + 8x2 + 3x – 2
2. Find the value of p and q if
a) x + 1 and x – 2 are factors of x3 + px2 + qx + 6
b) x + 2 and x – 3 are factors of px3 – x2 – 13x + q
3. Factorise the polynomials:
a) 2x3+ 5x2 + x – 2
b) 3x3 – x2 – 38x – 24
c) x3 – 8
d) x3 + 27
e) x3 – a3
f) x3 + a3
4. a) Factorise 5x2 + 48x + 27. Use tour factorisation to write 54827 as the
product of two prime numbers.
b) What expression would you factorise to express 41309 as the product
of two prime numbers? Factorise the expression and find the prime
numbers.
Check your answers at the end of this unit.
You have come to the end of this module. I hope you found it useful, found
some enjoyment in going through the module, but above all that it has helped
you to improve and renovate your classroom practice.
You should have strengthened your knowledge of pupils’ difficulties in
learning basic algebra and how these difficulties might be avoided by using
concrete materials, precise mathematical language and investigative
methods.
You should have gained confidence in creating a learning environment for
your pupils in which they can
(i) acquire with understanding knowledge on expansion and factorisation of
algebraic expressions
(ii) illustrate with paper models algebraic identities
(iii) investigate arithmetical patterns leading to generalisations expressed in
algebraic form.
(iv) investigate and resolve algebraic fallacies
You may have extended your knowledge in algebra having acquired
knowledge on the binomial theorem, Pascal’s triangle and pyramid,
remainder and factor theorem.
Module 4: Unit 4
107
Binomial theorem and factor theorem
Module 5, Practice activity
1. a) List the knowledge and skills you expect pupils in junior secondary
school to acquire through learning simplification, expansion and
factorisation.
b) Describe how you create opportunities in your class for pupils to
acquire the knowledge and skills you listed in 1a.
2. a) Write down any three digits, for example 653
Reverse the digits (356) and subtract the two numbers (smaller from
the larger 653 – 356 = 297)
Add to this answer (297) the number obtained by reversing the digits
in the answer (792).
Compare with others. Try other numbers. Can you prove the result
using algebra? Try four digit numbers and investigate.
b) Write down a three digit number with the digits different from each
other. Add together all the (6) different two digits numbers you can
form from the three digit number. Divide your answer by the sum of
the digits of the original number.
Compare with others. Try other numbers. Can you prove the result
using algebra?
c) Set the above investigations to your pupils and write an evaluation.
3. a) Make a worksheet for the pupils with number patterns, which when
generalised will lead to the expression:
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a – b)2 = a2 – 2ab + b2
(iii) a – (b – c) = a – b + c
(iv) a – (b + c) = a – b – c
b) Try out the worksheet(s) with your pupils.
c) Write a critical evaluation of the activity.
4. a) Design a game to consolidate factorisation.
b) Try out your game with your pupils.
c) Write an evaluative report.
5. Have a look at the content of unit 4. Could some of the material be
presented to high achievers in form 3? or used in a mathematics club?
Justify your answer. If your answer is positive try out the activity you
feel (some) pupils can do and write an evaluative report.
6. Did this module lead to any changes in the methods you use to assist
pupils in the learning of algebra?
If yes, list the changes and explain why you decided to make a change.
If no, explain why.
Present your assignment to your supervisor or study group for discussion.
Module 4: Unit 4
108
Binomial theorem and factor theorem
Summary
The key to learning algebra at the ages of 11-15 is obvious usefulness
facilitated by working models. The authors hope that your teaching will
supply that key!
Unit 4: Answers to self mark exercises
Self mark exercise 1
Degree
Coefficient of x3
a)
3
2
b)
3
5
c)
4
0
d)
0
0
e)
8
-1
1.
2
a) 6x4 – 7x3 + 2x2 + 9x – 6
b) 4x4 + 13x3 – 3x2 – 2x + 6
c) 6x12 – x8 – 8x6 – 12x4 – 5x2 + 2
d) 8x4 – 18x3 +13x2 + 24x – 45
3
a) (n + m)
b) 2n
c) tn
Self mark exercise 2
1. a) x6 + 6x5 + 15x4 + 20x3 + 15x2 + 6x + 1
b) x7 + 7x6 + 21x5 + 35x4 + 35x3 + 21x2 + 7x + 1
c)
2.
Adding two numbers next to each other in, say row 6, gives the numbers for
row 7. Add a starting and finishing 1 to row 7.
Module 4: Unit 4
109
Binomial theorem and factor theorem
Self mark exercise 3
7
7×6 5 7×6×5 4 7×6×5×4 3
1. a) x7 + x6 +
x +
x +
x +
1
1× 2
1× 2 × 3
1× 2 × 3× 4
7×6×5× 4×3 2 7×6×5× 4×3×2
x +
x+1
1× 2 × 3× 4 × 5
1× 2 × 3× 4 × 5 × 6
8
8×7 6 8×7×6 5 8×7×6×5 4
x +
x +
x +
b) x8 + x7 +
1
1× 2
1× 2 × 3
1× 2 × 3× 4
8×7×6×5×4 3 8×7×6×5×4×3 2
x +
x +
1× 2 × 3× 4 × 5
1× 2 × 3× 4 × 5 × 6
8× 7×6×5× 4×3×2
x+1
1× 2 × 3× 4 × 5 × 6 × 7
n × ( n - 1) 2 n × ( n - 1) × ( n - 2) 3
2. (1+ x)n = 1 + nx +
x +
x +
1× 2
1× 2 × 3
n × ( n - 1) × ( n - 2) × ( n - 3) 4
x + ... + ......+
1× 2 × 3 × 4
n × (n - 1) × (n - 2) × (n - 3) × (n - 4) × .... × (n -{r - 1}) r
x + ... + ... xn
1 × 2 × 3 × 4 × 5 × .... × r
Self mark exercise 4
1. 32x4 + 32x3 + 24x2 + 8x + 1
2. -8x3 + 12x2 – 6x + 1
3.
1 5 5 4 5 3 5 2 5
x + x + x + x + x +1
32
16
4
2
2
4. 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4
5. x6 + 3x4y2 + 3x2y4 + y6
6.
1 5 5 4
5 3 2 5 2 3
5
1 5
x −
x y+
x y −
x y +
xy 4 −
y
32
48
36
54
162
243
7. 358 213.55
8. 1 + 5x + 5x2 – 10x3+ ...
9. a) -3 421 440
b) -672 000
10. 258 660 864
11. a) 168.439
12. ab) 56
b) 123.587
cd) 495
ef) 5 200 300
n
n
g)  r  =  n - r
n!
n!
n!
n
n
and  n − r =
=
h)  r  =
r! ×(n - r )!
(n - r )! {n - (n - r ) }! (n - r )! × r!
Module 4: Unit 4
110
Binomial theorem and factor theorem
Self mark exercise 5
1. convergent
2. convergent
3. convergent
Self mark exercise 6
1
2 2 6 3
1. 1 + x −
x +
x − ...
5
25
125
5
25 2 125 3
x +
x – ...
2. 1 + x –
2
8
16
4
16 2 320 3
3. 1 – x –
x –
x – ...
3
9
81
4. 1 –
9
27 3
3
x – x2 –
x – ...
8
16
2
10.2470
2.4662
3.1623
5. 1 – 3x +
3 2 1 3 3 4
x + x + x + ..
2
2
8
6.
1 2
1 3
1
x–
x +
x + ...)
32
128
4
2 {1 +
1.0040
1.7320
7. a) 1 – x + x2 – x3 + x4 ... convergent for -1 < x < 1
b) 1 – 2x + 3x2 – 4x3+5x5 ... convergent for -1 < x < 1
c) 1 −
1
3
5
35 4
x + x2 − x3 +
x − ... convergent for -1< x < 1
2
8
16
128
8. 1 – 3x + 6x2 – 10x3 + 15x4 – ....
1.0306
9. a) 1 – 6x + 27x2 – 108x3 + 405x4 – ....
b) 1 + 20x + 240x2 + 2240x3 + 17920x4 + ...
c)
1 1
3
1
5 4
+ x + x2 + x3 +
x + ...
4 4
16
8
64
1
1
1
x − x 2 − x 3 − ....
2
8
16
3
5
b) 1 − x + x 2 − x 3 + ....
2
2
10. a) 1 −
Problem Solving 1
1. Sum in nth row 2n – 1.
2. Sum is one less than the number the bottom corner of the rectangle
points to.
Module 4: Unit 4
111
Binomial theorem and factor theorem
3. 252
4. 2n – 1+ 2n – 3
5. 20
Self mark exercise 7
1. a) (a + b + c)2 = a2 + 2ab + 2ac + b2 + 2bc + c2
b) (a + b + c)3 = a3 + 3a2b + 3ab2 + b3 + 3b2c + 3bc2 + c3 + 3c2a +
3ca2 + 6abc
c) (a + b + c)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 + 4b3c + 6b2c2 +
4bc3 + c4 + + 4c3a + 6c2a2 + 4ca3 + 12a2bc + 12ab2c + 12abc2.
Self mark exercise 8
1.
2. The numbers along the sides of the triangular section of the pyramid are
exactly as in Pascal’s triangle. In the above example along the three sides
you find the numbers 1 4 6 4 1 corresponding to the fifth row
in Pascal’s triangle i.e. the coefficients in the expansion of (x + y)4.
Module 4: Unit 4
112
Binomial theorem and factor theorem
Self mark exercise 9
1.
2. a) (a + b + c)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 + 5b4x +
10b3c2 + 10b2c3 + 5bc4 + c5+ 5c4a + 10c3a2 + 10c2a3+ 5ca4 +
20a3bc + 20ab3c + 20abc3 + 30a2b2c + 30ab2c2 + 30a2bc2
b) (a + 3b + 2c)3 = a3 + 9a2b + 27ab2 + 27b3 + 54b2c + 36bc2 + 8c3 +
12c2a + 6ca2 + 36abc
c) (a + 2b + c)4= a4 + 8a3b + 24a2b2 + 32ab3 + 16b4 + 32b3c +
24b2c2 + 8bc3 + c4 + 4c3a + 6c2a2 + 4ca3 + 24a2bc + 48ab2c +
24abc2
3. The coefficient of apbqcr in the expansion of (a + b + c)n where
p + q + r = n is
n!
 n 
. This is also written as 

 p, q, r
p! q! r!
For example the coefficient ab2c2 in the expansion of (a + b + c)5 is
5!
1 . 2 . 3 . 4. .5
 5 
=
= 30

 =
1, 2, 2
1! 2! 2!
1. 2. .1. .2
4. The general term in the expansion will be
n

 a n1 ⋅ a n2 . a n3 . ....... ⋅ a nr
r
2
3
 n1, n2 , n3 , ......, nr  1
Self mark exercise 10
1. a) Quotient (x – 7), remainder 32
b) Quotient (3x2 + 2x + 10) , remainder 30
c) Quotient 2x2 + 2x – 2, remainder 1
Module 4: Unit 4
113
Binomial theorem and factor theorem
 − b
2. Remainder = P  
 a
-1
32
30
1
-2
-18
-3
b
3. The remainder when P(x) is divided by (ax + b) is P(- )
a
Self mark exercise 11
1. a) -3
b) -1
c) -2 d) 14
2. a) 3
b) 1
c) 2
3. a = 4, b = -3
4.
f (a ) – f (b)
af (b) – bf (a )
x +
a – b
a – b
Self mark exercise 12
1. a) -2
b) 3
2. a) p = -4, q = -1
b) p = 2, q = -6
3. a) (x + 1)(x + 2)(2x – 1)
b) (x + 3)(x – 4)(3x + 2)
c) (x + 2) (x2 – 2x + 4)
d) (x – 3) (x2 + 3x + 9)
e) (x + a) (x2 – ax + a2)
f) (x – a)(x2 + ax + a2)
4. a) (5x + 3)(x + 9); 503, 109
b) 4x2 + 13x + 9 = (4x + 9)(x + 1); 409, 101
Module 4: Unit 4
114
Binomial theorem and factor theorem
References
Botswana, Republic of., Three-Year Junior Secondary Syllabus
Mathematics, 1996.
Hart, K. Children’s Understanding of Mathematics 11 - 16, 1981,
ISBN 071 953 772x
Additional References
In preparing the materials included in this module we have borrowed ideas
extensively from other sources and in some cases used activities almost
intact as examples of good practice. As we have been using several of the
ideas, included in this module, in teacher training the original source of the
ideas cannot be traced in some cases. The main sources are listed below.
Boyer, C.B., 1991, A History of Mathematics (2nd Edition),
ISBN 047 154 3977
Eagle M, Exploring Mathematics through history, 1995, ISBN 052 145 6266
NCTM, Algebra in a Technological World, 1995, ISBN 087 353 3267
NCTM, Learning and Teaching Algebra, 1984, ISBN 087 353 2686
NCTM, Patterns and Functions, Addenda Series, 1994, ISBN 087 353 3240
NCTM, Historical Topics for the Mathematics Classroom, 1989,
ISBN 087 353 2813
Open University, Preparing to Teach Equations, PM 753C,
ISBN 033 517 4396
Reimer W, Historical Connections in Mathematics Volume 1, 2 and 3, 1992,
ISBN 188 143 1355, ISBN 188 143 138x, ISBN 188 143 1495
The Penguin Dictionary of Mathematics, 1989, ISBN 014 051 1199
Tauris, I.B., 1991, The Crest of the Peacock. Non European Roots of
Mathematics, ISBN 185 043 2856
UB-INSET, Algebra for Everyone, 1996, University of Botswana
Further reading
The Maths in Action book series are for use by pupils in the classroom. The
books are using a constructivist, activity based approach, including problem
solving, investigations, games and challenges in line with the ideas in this
module.
OUP/Educational book Service, Gaborone, Maths in Action Book 1
ISBN 019 571776 7
OUP/Educational book Service, Gaborone, Maths in Action Book 2
OUP/Educational book Service, Gaborone, Maths in Action Teacher’s File
Book 1
OUP/Educational book Service, Gaborone, Maths in Action Teacher’s File
Book 2
Module 4
115
References
Glossary
Module 4
Algebra
the branch of mathematics that deals with the
general properties of numbers and generalisations
of relationships among numbers
Algebraic expression
any combination of numbers and variables and
algebraic operations
Binomial
a polynomial* consisting of two terms
e.g. 2a + b, 2p3 – 2q2
Common factor
a number and/or expression that divides two or
more given numbers or algebraic expressions
exactly
Degree of polynomial
highest index of the variable x with non zero
coefficient (for polynomials in one variable)
Dividend
a number or polynomial that is divided by another
number or polynomial
Divisor
a number or polynomial that divides another
number or polynomial
Expansion
writing an algebraic expression as the sum of a
number of terms e.g. (x + 3)(x + 4) = x2 + 7x + 12
Factor
a number or polynomial that divides a given
number or polynomial exactly
Factorial n
written n! represent the product of all the integers
1 to n inclusive, i.e., 1 × 2 × 3 × 4 × ... × n. By
convention 0! = 1
Factorising
writing a given sum of terms as a product of
factors, i.e., x2 + 7x + 12 = (x + 3)(x + 4)
Factor theorem
(x – a) is a factor of a polynomial P(x) if P(a) = 0
Fallacy (algebraic)
results obtained through algebra that appear true
but are actually false
HCF
highest common factor: the largest number or
algebraic expression with the highest exponents
that is a common factor of two or more given
numbers or algebraic expressions
Monomial
expression of the form axn or axnym (or with
more variables to an index) where a is a real
number and n and m are non negative integers
Polynomial
sum of monomials. For one variable the general
form is a0 + a1x + a2x2 + ... + anxn
116
Glossary
Module 4
Quotient
the result of dividing one number or polynomial
by another
Remainder
a number or polynomial remaining after one
number or polynomial is divided into another
Remainder theorem
if the polynomial P(x) is divided by (x – a) the
remainder is equal to P(a).
Variable
representation of a changing quantity
117
Glossary
Appendix 1
Muhammad ibn Musa al-Khwarizmi (c780 - c 850)
Arab mathematician from Khiva, now part of the Uzbek Republic of the
USSR. In his Al-jam’ w’al-tafriq ib hisab al-hind (Addition and Subtraction
in Indian Arithmetic), al-Khwarizmi introduced the Indian system of
numerals to the West. He also wrote a treatise on algebra, Hisab al-jabr
w’al-muqabala (Calculation by Restoration and Reduction); from ‘al-jabr’
comes the word ‘algebra’. From al-Khwarizmi’s name was derived the term
‘algorism’ (referring originally to the Hindu-Arabic number system, but later
to computation in a wider sense), from which in turn comes ‘algorithm’.
From: The Penguin Dictionary of Mathematics, 1989, page 14.
====================================================
Al-Khwarizmi
From an intellectual viewpoint, the ninth century A.D. was essentially a
Moslem century. The activity of the scholars of Islam was far superior to that
of any other group. Flourishing under Caliph al-Mamum, the Moslem
mathematician, astronomer, and geographer Mohammed ibn-Musa alKhwarizmi influenced mathematical thought more than did any other
medieval writer on the subject. He is generally known for his work in
algebra and astronomy and for his mathematical tables; but perhaps one of
the greatest achievements was his introduction of the so-called Hindu-Arabic
notational system to the Western world. His writings were the main channel
by which the “new” numerals became known in the West, and through them
he revolutionised the common processes of computation.
Al-Khwarizmi’s best-known book, Hisab al-jabr w’al-muqabalah (“science
of restoration [or reunion] and opposition”), was written to show that “what
is easiest and most useful in arithmetic such as men constantly require in
cases of inheritance, legacies, partition, lawsuits and trade, and in all their
dealings with one another.” This work, containing 79 pages on inheritance
cases, 16 pages of measurement problems, and 70 pages of algebra, begins
with a brief recapitulation of the place-value number system based on ten
and then goes on to teach the use of these positional numerals in
computation.
There is, for example, a systematic treatment of surds. Al-Khwarizmi noted
that, to double a root, one must multiply the square by four; to triple it, by
nine; and to halve it, by one-fourth. He gave examples of the multiplication
and division of surds, concluded the series by saying , “Likewise with other
numbers,” thus indicating that the reader was to deduce a general rule from
the examples given. He used this same technique in mercantile problems,
where many of the examples imply a use of proportionality. “If ten items
cost six [coins], what will four cost?” After seeing how several such
problems are solved, one is to do “likewise with other numbers.”
It should be noted that throughout the treatise numbers are expressed in
words, not symbols, numerals being used only in diagrams and a few
marginal notes. The methods of computation thus taught verbally could be
Module 4
118
Appendices
applied only to the then new place-value system of numeration used by the
Moslems. Thus the promulgation of this work of al-Khwarizmi gave strong
impetus to the scholars of the West to learn what we now call the numerals
of the Hindu-Arabic system and to become skilful in manipulating them.
From: NCTM Historical Topics for the Mathematics Classroom, 1989, page
76 - 77
====================================================
Besides astronomical tables, and treatises on the astrolabe and the sundial,
al-Khwarizmi wrote two books on arithmetic and algebra which played very
important roles in the history of mathematics. One of these survives only in a
unique copy of a Latin translation with the title De numero indorum
(Concerning the Hindu Art of Reckoning), the original Arabic version having
since been lost. In this work, based presumably on an Arabic translation of
Brahmagupta, al-Khwarizmi gave so full an account of the Hindu numerals
that he probably is responsible for the wide-spread but false impression that
our system of numeration is Arabic in origin. Al-Khwarizmi made no claim
to originality in connection with the system, the Hindu source of which he
assumed as a matter of course; but when subsequent Latin translations of his
work appeared in Europe, careless readers began to attribute not only the
book but also the numeration to the author. The new notation came to be
known as that of al-Khwarizmi, or more carelessly, algorismi; ultimately the
scheme of numeration making use of the Hindu numerals came to be called
simply algorism or algorithm, a word that, originally derived from the name
al-Khwarizmi, now means, more generally, any peculiar rule of procedure or
operation—such as Euclidean method for finding the greatest common
divisor.
Through his arithmetic, al-Khwarizmi’s name has become a common
English word; through the title of his most important book, Al-jabr wa’l
muqabalah, he has supplied us with an even more popular term. From this
title has come the word algebra, for it is from this book that Europe later
learned the branch of mathematics bearing this name. Diophantus sometimes
is called “the father of algebra,” but this title more appropriately belongs to
al-Khwarizmi. It is true that in two respects the work of al-Khwarizmi
represented a retrogression from that of Diophantus, First, it is on a far more
elementary level that that found in the Diophantine problems and, second,
the algebra of al-Khwarizmi is thoroughly rhetorical, with none of the
syncopation found in the Greek Arithmetica or in Brahmagupta’s work. Even
numbers were written out in words rather than symbols! It is quite unlikely
that al-Khwarizmi knew of the work of Diophantus, but he must have been
familiar with at least the astronomical and computational portions of
Brahmagupta; yet neither al-Khwarizmi nor other Arabic scholars made use
of syncopation (syncopation) or of negative numbers. Nevertheless, the Aljabr comes closer to the elementary algebra of today than the works of either
Diophantus or Brahmagupta, for the book is not concerned with difficult
problems in indeterminate analysis but with a straightforward and
elementary exposition of the solution of equations, especially of second
degree. The Arabs in general loved a good clear argument from premise to
Module 4
119
Appendices
conclusion, as well as systematic organization—respects in which neither
Diophantus nor the Hindus excelled. The Hindus were strong in association
and analogy, in intuition and an aesthetic and imaginative flair, whereas the
Arabs were more practical-minded and down-to-earth in their approach to
mathematics.
The Al-jabr has come down to us in two versions, Latin and Arabic, but in
the Latin translation, Libber algebra et almucubala, a considerable portion
of the Arabic draft is missing. The Latin, for example, has no preface,
perhaps because the author’s preface in Arabic gave fulsome praise to
Mohammed, the prophet, and to al-Mamum, “the Commander of the
Faithful.” Al-Khwarizmi wrote that the latter had encouraged him to
compose a short work on “Calculating by (the rules of) Completion and
Reduction, confining it to what is easiest and most useful in arithmetic, such
as men constantly require in cases of inheritance, legacies, partitions, lawsuits, and trade, and in all their dealings with one another, or where the
measuring of lands, the digging of canals, geometrical computation, and
other objects of various sorts and kinds are concerned” [Karpinski, 1915,
p 15]
It is not certain just what the terms al-jabr and muqabalah mean, but the
usual interpretation is similar to that implied in the translation above. The
word al-jabr presumably meant something like “restoration” or
“completion” and seems to refer to the transposition of subtracted terms to
the other side in an equation; the word muqabalah is said to refer to
“reduction” or “balancing”—that is, the cancellation of like terms on
opposite sides of the equation. Arabic influence in Spain long after the time
of al-Khwarizmi is found in Don Quixote, where the word algebrista is used
for a bone-setter, that is, a “restorer.”
The Latin translation of al-Khwarizmi’s Algebra opens with a brief
introductory statement of the positional system for numbers and then
proceeds to the solution, in six short chapters, of the six types of equations
made up of the three kinds of quantities: roots, squares, and numbers (that is
x, x2, and numbers). Chapter I, in three short paragraphs, covers the case of
x
squares equal to roots, expressed in modern notation as x2 = 5x,
= 4x,
3
and 5x2 = 10x, giving the answers x = 5, x = 12 and x = 2 respectively. (The
root x = 0 was not recognised.) Chapter II covers the case of squares equal to
numbers, and Chapter III solves the case of roots equal to numbers, again
with three illustrations per chapter to cover the cases in which the coefficient
of the variable term is equal to, more than, or less than one. Chapter IV, V
and VI are more interesting, for they cover in turn three classical cases of
three-term quadratic equations: (1) squares and roots equal to numbers, (2)
squares and numbers equal to roots, and (3) roots and numbers equal to
squares. The solutions are “cookbook” rules for “completing the square”
applied to specific instances. In each case only the positive answer is given.
So systematic and exhaustive was al-Khwarizmi’s exposition that the readers
must have had little difficulty in mastering the solutions. In this sense, then,
al-Khwarizmi is entitled to be known as “the father of algebra.” However, no
Module 4
120
Appendices
branch of mathematics springs up fully grown, and we cannot help but ask
where the inspiration for Arabic algebra came from. To this question no
categorical answer can be given; but the arbitrariness of the rules and the
strictly numerical form of the six chapters remind us of ancient Babylonian
and medieval Indian mathematics. The exclusion of indeterminate analysis, a
favourite Hindu topic, and the avoidance of any syncopation, such as found
in Brahmagupta, might suggest Mesopotamia as more likely a source than
India. As we read beyond the sixth chapter, however, an entirely new light
is thrown on the question. Al-Khwarizmi continued:
“We have said enough so far as numbers are concerned, about the six types
of equations. Now, however, it is necessary that we should demonstrate
geometrically the truth of the same problems which we have explained in
numbers.”
The ring in this passage is obviously Greek rather than Babylonian or Indian.
There are, therefore, three main schools of thought on the origin of Arabic
algebra: one emphasises Hindu influences, another stresses the
Mesopotamian, or Syriac-Persian, tradition and the third points to Greek
inspiration. The truth is probably approached if we combine the three
theories. The philosophers of Islam admired Aristotle to the point of aping
him, but eclectic Mohammedan mathematicians seem to have chosen
appropriate elements from various sources.
The Algebra of al-Khwarizmi betrays unmistakable Hellenic elements, but
the first geometric demonstrations have little in common with classical
Greek mathematics.
[Extracts from Carl B. Boyer A History of Mathematics (2nd Edition) p 227 232]
====================================================
Abu Jafar Muhammad ibn Musa al-Khwarizmi (to give him his full name,
which means Muhammad, the father of Jafar and the son of Musa, from
Khwarizm) was born in about 780. The name ‘al-Khwarismi’ suggest that
either he or his family came from Khwarizm, east of the Caspian Sea in what
is today Soviet Central Asia. Little is known of his early life. There is a
reference to al-Khwarizmi as ‘al-Majusi’ in a book entitled History of Kings
and Envoys by al-Tabari an Arab historian of the ninth century. In the
Pahlavi language, a Zoroastrian was sometimes referred to as ‘magus’, from
which comes our words ‘magi’ and ‘magician’. There is therefore, the view
that al’Khwarizmi may have been of Zoroastrian heritage and had acquired
his early knowledge of Indian mathematics and astronomy from Zoroastrian
clergy, some of whom were reputed to be well acquainted with these
subjects.
In about 820, after establishing a reputation as a talented scientist in Merv,
the capital of the Eastern provinces of the Abbasid caliphate, he was invited
by Caliph al-Mamum to move to Baghdad, where he was appointed first
astronomer and then head of the library at the House of Wisdom. He
continued to serve other caliphs, including al-Wathiq during his short rule
from 842 to 847. There is a story, told by the historian al-Tabari, that when
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Caliph al-Wathiq lay seriously ill he asked al-Khwarizmi to cast his
horoscope and find out whether he would live. Al-Khwarizmi assured the
caliph that he would live another fifty years, but al-Wathiq died within ten
days. Whether this story illustrates al-Khwarizmi’s highly developed sense
of survival or his ineptness as a fortune teller, it is difficult to say. We know
little else of his later life except that he died in about 850.
Al-Khwarizmi also constructed a zij (a set of astronomical tables) that was to
remain important in astronomy for the next five centuries. The origin of this
zij is interesting, for it shows how Indian mathematics and astronomy
entered the Arab world directly for the first time. An Arab historian al -Qifti
(c 1270) reported that in the year AH 156 (AD 773) a man well versed in
astronomy, by the name of Kanaka, came to Baghdad as a member of a
diplomatic mission from Sind, in northern India. He brought with him Indian
astronomical texts, including Surya Siddhanta and the works of
Brahmagupta. Caliph al-Mansur ordered that some of these texts should be
translated into Arabic and, according to the principles given in them, that a
handbook be constructed for use by Arab astronomers. The task was
delegated to al-Fazari, who produced a text which came to be known by later
astronomers as the Great Siindhind. The word sinshindis derived from the
Sanskrit word siddhanta, meaning an astronomical text. It was mainly on the
bias of this text, as well as some other elements from Babylonian and
Ptolemaic astronomy, that al-Khwarizmi constructed his zij. Unfortunately,
the original Arabic text is no longer extant. But a Latin translation, made in
1126 from the edited version produced by Maslama al-Majriti (a Spanish
astronomer who lived in Cordoba in about the year 1000), became one of the
most influential astronomical texts in medieval Europe.
[Extract from: Tauris, I.B., 1991, The Crest of the Peacock, page 304 - 306]
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Appendix 2
Diophantus of Alexandria (c AD 250)
Greek mathematician and the author of the Arithmetica, of which ten of the
original books are extant. About 130 problems are considered, some of
which are surprisingly hard, in the field of what have since become known as
Diophantine equations.
Diophantine equation
Any equation usually in several unknowns, that is studied in a problem
whose solutions are required to be integers, or sometimes more general
rational numbers. Examples of such problems are
(1) To find integers x, y that satisfy 11x + 3y = 1.
(2) To find rational numbers x, y, z such that x3 + y3 = z3
From: The Penguin Dictionary of Mathematics, 1989, page 99.
====================================================
The Greek Diophantus (c A.D. 75 or 250) author of the classic Arithmetica,
also wrote a treatise on polygonal numbers, only a fragment of which is
known. In it he generalises Statement 4 above. [Eight times any triangular
number plus 1 is a square number]. He also solves this problem: Given a
number, find in how many ways it can be a polygonal number. (Note, in the
table given earlier, that 55 is both a triangular and a heptagonal number; 81
is both a square and heptagonal.) The manuscript breaks off in the middle of
this problem.
From: NCTM Historical Topics for the Mathematics Classroom, 1989, page
57
====================================================
For some common fractions the Greeks had special symbols, but for others
the fractions were written with one accent on the numerator and two on the
13
.
denominator, which was written twice. Thus we have ιγ‘κθ“κθ“ =
20
Diophantus sometimes used a form similar to ours today, but he wrote the
denominator above the numerator and without the horizontal line.
From: NCTM Historical Topics for the Mathematics Classroom, 1989,
page 96
====================================================
Diophantus introduced the syncopated style of writing equations. He studied
and worked at the University of Alexandria, Egypt, where Euclid had once
taught. Very little else is known about his life except what is claimed in the
following algebraic puzzle rhyme from the Anthologia Palatine:
“Here lies Diophantus.” The wonder behold—
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Through art algebraic, the stone tells old:
“God gave him his boyhood one-sixth of his life,
One-twelfth more as youth while whiskers grew rife;
And yet one-seventh ere marriage begun;
In five years there came a bouncing new son.
Alas, the dear child of master and sage
Met fate at just half his dad’s final age.
Four years yet his studies gave solace from grief;
Then leaving scenes earthly he, too, found relief.
(What answer do you get for Diaophantus’ age? Eighty-four?)
Diophantus’ claim to fame rest on his Arithmetica, in which he gives an
ingenious treatment of indeterminate equations—usually two or more
equations in several variables that have an infinite number of rational
solutions. These are often called Diophantine equations, although he was not
the first to solve such systems. His approach is clever, but he does not
envelop a systematic method for finding general solutions. His approach is
along Babylonian lines in the sense that he expresses all unknowns in terms
of a parameter and then obtains an equation containing only the parameter.
From: NCTM Historical Topics for the Mathematics Classroom, 1989,
page 240
====================================================
At the beginning of this period [the “Silver Age” about AD 250 to 350], also
known as the Later Alexandrian Age, we find the leading Greek algebraist,
Diophantus of Alexandria, and towards its close there appeared the last
significant Greek geometer, Pappus of Alexandria. No other city has been
the centre of mathematical activity for so long a period as was Alexandria
from the days of Euclid (ca. 300 BC) to the time of Hypatia († AD 415). It
was a very cosmopolitan centre, and the mathematics that resulted from
Alexandrian scholarship was not all of the same type. The results of Heron
were markedly different from those of Euclid or Apollonius or Archimedes,
and again there is an abrupt departure from the classical Greek tradition in
the extant work of Diophantus. Uncertainty about the life of Diophantus is so
great that we do not know definitely in which century he lived. Generally he
is assumed to have flourished about AD 250, but dates a century or more
earlier or later are sometimes suggested. According to a tradition that is
reported in a collection of problems dating from the fifth or sixth century,
known as the “Greek Anthology”, if it is historically accurate, Diophantus
lived to be eighty-four years old.
Diophantus is often called the father of algebra, but we shall see that such a
designation is not to be taken literally. His work is not all the type of
material forming the basis of modern elementary algebra; nor is it yet similar
to the geometric algebra found in Euclid. The chief Diophantine work known
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to us is the Arithmetica , a treatise originally in thirteen books, only the first
six of which have survived. It should be recalled that in ancient Greece the
word arithmetic meant theory of numbers rather than computation. Often
Greek arithmetic had more in common with philosophy than with what we
think if as mathematics; hence, the subject had played a large role in
Neoplatonism during the Later Alexandrian Age. This had been particularly
true of the Introductio arithmeticae of Nicomachus of Gerasa, a NeoPythagorean who lived not far from Jerusalem about the year 100.
Quite different from the work of Nocomachus, Theon and Boethius was the
Arithmetica of Diophantus, a treatise characterized by a high degree of
mathematical skill and ingenuity. In this respect the book can be compared
with the great classics of the earlier Alexandrian Age; yet it has practically
nothing in common with these or, in fact, with any traditional Greek
mathematics. It represents essentially a new branch and makes use of a
different approach. Being divorced from geometric methods, it resembles
Babylonian algebra to a large extent. But whereas Babylonian
mathematicians had been concerned primarily with the approximate solution
of determinate equations as far as the third degree, the arithmetica of
Diophantus (such as we have it) is almost entirely devoted to the exact
solution of equations, both determinate and indeterminate. Because of the
emphasis given in the Arithmetica to the solution of indeterminate problems,
the name dealing with this topic, sometimes known indeterminate analysis,
has since become Diophantine analysis. Since this type of work today is
generally part of courses in theory of numbers, rather than elementary
algebra, it is not an appropriate basis for regarding Diophantus as the father
of algebra. There is another respect, however, in which such a paternity is
justified. Algebra now is based almost exclusively on symbolic forms of
statement, rather than on the customary written language of ordinary
communication in which earlier Greek mathematics, as well as Greek
literature, had been expressed. It has been said that three stages in the
historical development of algebra can be recognised: (1) the rhetorical or
early stage, in which everything is written out fully in words: (2) a
syncopated or intermediate stage, in which some abbreviations are adopted;
and (3) a symbolic or final stage. Such an arbitrary division of the
development of algebra into three stags is, of course, a facile oversimplification; but it can serve effectively as a first approximation to what
happened, and within such a framework the Arithmetica of Diophantus is to
be placed in the second category.
Throughout the six surviving books of the Arithmetica there is a systematic
use of abbreviations for powers of numbers and for relationships and
operations. An unknown number is represented by a symbol resembling the
Greek letter ς (perhaps for the last letter of arithmos); the square of this
appears as ∆γ , the cube as Kγ, the fourth power, called square-square, as
∆γ∆, the fifth power or square-cube as ∆ Kγ, and the sixth power or cubecube as KγK. Diophantus was, of course, familiar with the rules of
combination equivalent to our laws of exponents, and he had special names
for the reciprocals of the first six powers of the unknowns, quantities
equivalent to our negative powers. Numerical coefficients were written after
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Appendices
the symbols for the powers with which they were associated; addition of
terms was understood in the appropriate juxtaposition of the symbols for the
terms, and subtraction was represented by a single letter abbreviation placed
before the terms to be subtracted. With such a notation Diophantus was in a
position to write polynomials in a single unknown almost as concisely as we
do today. The expression 2x4 + 3x3 – 4x2 + 5x – 6, for example, might
appear in a form equivalent to SS2 C3 x5 M S4 u6, where the English letters
S, C, x, M and u have been used for ‘square’, ‘cube’, the ‘unknown’,
‘mimus’ and ‘unit’, and with our present numerals in place of the Greek
alphabetic notation that was used in the days of Diophantus. Greek algebra
now no longer was restricted to the first three powers or dimensions, and the
identities
(a2 + b2)(c2 + d2) = (ac + bd)2 + (ad – bc)2 = (ac – bd)2 + (ad + bc)2,
which played important roles in medieval algebra and modern trigonometry,
appear in the work of Diophantus. The chief difference between the
Diophantine syncopation and the modern algebraic notation is in the lack of
special symbols for operations and relations, as well as of the exponential
notation. These missing elements of notation were largely contributions of
the period from the late fifteenth to the early seventeenth centuries in
Europe.
If we think primarily of matters of notation, Diophantus has a good claim to
be known as the father of algebra, but in terms of motivation and concepts
the claim is less appropriate. The Arithmetica is not a systematic exposition
of the algebraic operations or the algebraic functions or of the solution of
algebraic equations. It is instead a collection of some 150 problems, all
worked out in terms of specific numerical examples, although perhaps
generality of method was intended. There is no postulational development,
nor is an effort made to find all possible solutions. In the case of quadratic
equations with two positive roots, only the larger is given, and negative roots
are not recognised. No clear-cut distinction is made between determinate and
indeterminate problems, and even for the latter, for which the number of
solutions generally is unlimited, only a single answer is given. Diophantus
solved problems involving several unknown numbers by skilfully expressing
all unknown quantities, where possible, in terms of only one of them. Two
problems from the Arithmetica will serve to illustrate the Diophantine
approach. In finding two numbers such that their sum is 20 and the sum of
their squares is 208, the numbers are not designated as x and y, but as 10 + x
and 10 – x (in terms of our modern notation). Then (10 + x)2 + (10 – x)2 =
208, hence x = 2; so the numbers sought are 8 and 12. Diophantus handled
also the analogous problem in which the sum of the two numbers and the
sum of the cubes of the numbers are given as 10 and 370 respectively.
In these problems Diophantus is dealing with a determined equation, but he
used much the same approach in indeterminate analysis. In one problem it is
required to find two numbers such that either when added to the square of
the other will yield a perfect square. This is a typical instance of diophantine
analysis in which only rational numbers are acceptable as answers. In
solving the problem Diophantus did not call the numbers x and y, but rather x
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and 2x + 1. Here the second when added to the square of the first, will yield
a perfect square no matter what value one choose for x. Now it is required
also that (2x + 1)2 + x must be a perfect square. Here Diophantus does not
point out the infinity of possible answers. He is satisfied to choose a
particular case of a perfect square, in this instance the number (2x – 2)2, such
that when equated to (2x + 1)2 + x an equation that is linear in x results. Here
3
19
, so that the other number, 2x + 1, is
. One could, of
the result is x =
13
13
course, have used (2x – 3)2 or (2x – 4)2, or expressions of similar form,
instead of (2x – 2)2, to arrive at other pairs of numbers having the desired
property. Here we see an approach that comes close to a ‘method’ in
Diophantus’ work: When two conditions are to be satisfied by two numbers,
the two numbers are so chosen that one of the two conditions is satisfied; and
then one turns to the problem of satisfying the second condition. That is,
instead of handling simultaneous equations on two unknowns, Diophantus
operates with successive conditions so that only a single unknown number
appears in the work.
Among the indeterminate problems in the Arithmetica are some involving
equations such as x2 = 1 + 30y2 and x2 = 1 + 26y2, which are instances of
the so-called “Pell-equation” x2 = 1 + py2; again a single answer is though to
suffice. In a sense it is not fair to criticise Diophantus for being satisfied with
a single answer, for he was solving problems, not equations. In a sense the
Arithmetica is not an algebra textbook, but a problem collection in the
application of algebra. In this respect Diophantus is like the Babylonian
algebraists; and his work sometimes is regarded as “the finest flowering of
Babylonian algebra.” (Swift 1956). To some extent such a characterisation is
unfair to Diophantus, for his numbers are entirely abstract and do not refer to
measures of grain or dimensions of fields or monetary units, as was the case
in Egyptian and Mesopotamian algebra. Moreover, he is interested only in
exact rational solutions, whereas the Babylonians were computationally
inclined and were willing to accept approximations to irrational solutions to
equations. Hence, cubic equations seldom enter the work of Diophantus,
whereas among the Babylonians attention had been given to the reduction of
cubics to the standard form n3 + n2 = a in order to solve approximately
through interpolation in a table of values of n3 + n2.
We do not know how many of the problem in the Arithmetica were original
or whether Diophantus had borrowed from other similar collections. Possibly
some of the problems or methods are traceable back to Babylonian sources,
for puzzles and exercises have a way of reappearing generation after
generation. To us today the Arithmetica of Diophantus looks strikingly
original, but possibly this impression results from the loss of rival problem
collections. Our view of Greek mathematics is derived from a relatively
small number of surviving works, and conclusions derived from these
necessarily are precarious. Indications that Diophantus may have been less
isolated a figure than has been supposed are found in a collection of
problems from about the early second century of our era (hence presumably
antedating the Arithmetica) in which some Diophantine symbols appear.
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Nevertheless, Diophantus has had a greater influence on modern number
theory than any other nongeometric Greek algebraist. In particular, Fermat
was led to his celebrated “great” or “last” theorem when he sought to
generalise a problem that he had read in the Arithmetica of Diophantus
(II:8): to divide a given square into two squares.
[Extracts from Carl B. Boyer A History of Mathematics (2nd Edition)
p 178 - 183]
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Appendix 3
François Viète (Franciscus Vieta) (1540 - 1603)
French mathematician noted for his In artum analyticam isagoge (1591;
Introduction to the Analytical Arts), one of the earliest Western works on
algebra. In it he denoted unknowns by vowels and known quantities by
consonants, and also introduced an improved notation for squares, cubes and
other powers. With his new algebraic techniques Viète succeeded in solving
a number of problems classical authors had found unyielding to geometrical
attacks. He developed new methods of solving of solving equations in his De
aequantionum recognitione et emendatione (1615; On the Recognition and
Emendation of Equations). He was the first, in his Canon mathematicus seu
ad triangula (1579: The Mathematical Canon Applied to Triangles), to
tackle the problem of solving plane and spherical triangles with the help of
the six main trigonometrical functions.
From: The Penguin Dictionary of Mathematics, 1989, page 339
====================================================
The Frenchman François Viète was the first in his ‘logistica speciosa’ to
introduce letters as general (positive) coefficients and to put some other
finishing touches on symbolism, which was finally up to date by the time of
Isaac Newton (1642 - 1727). But Viète’s most significant contributions were
contained in De aequantionum recognitione et emendatione, published
posthumously in 1615. In this work he:
1. Gave transformations for increasing or multiplying the roots of an
equation by a constant.
2. Indicated awareness of relations between the roots and coefficients of
polynomial equations.
3. Stated a transformation that rids a polynomial of its next-to-highestdegree term.
Viète’s inability to accept negative numbers (not to mention imaginary
numbers) prevented him from attaining the generality he sought (and partly
comprehended) in giving, for example, relations between the roots and the
coefficients of a polynomial equation.
An example of the notation used by Viète
I QC - 15 QQ + 85C - 225Q + 274N aequatur 120
In our notation this means:
-15x4 + 85x3 – 225x2 + 274x = 120
From: NCTM Historical Topics for the Mathematics Classroom, 1989, page
245/263
====================================================
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In 1560, François Viète, a student of law in France, served Henri IV of
Navarra in the war against Spain by decoding intercepted letters which were
written in code. He went on to develop interest in astronomy and plane and
spherical trigonometry, and published a book in which he introduced the use
of letters not only for unknowns but also to represent known quantities
From: Open University, PM 753C, Preparing to teach Equations, page 32.
====================================================
Without doubt it was in algebra that Viète made his most estimable
contributions, for it was here that he came close to modern views.
Mathematics is a form of reasoning, and not a bag of tricks, such as
Diophantus had possessed; yet algebra, during the Arabic and early modern
periods, had not gone far from freeing itself from the treatment of special
cases. There could be little advance in algebraic theory so long as the chief
occupation was with finding “the thing” in an equation with specific
numerical coefficients. Symbols and abbreviations for an unknown, and for
powers of the unknown, as well as for operations and for the relationship of
equality had to be developed
Here Viète introduced a convention as simple as it was fruitful. He used a
vowel to represent the quantity in algebra that was assumed to be unknown
or underdetermined and a consonant to represent a magnitude or number
assumed to be known or given. Here we find for the first time in algebra a
clear-cut distinction between the important concept of a parameter and the
idea of an unknown quantity.
Had Viète adapted other symbolisms extant in his day, he might have written
all quadratic equations in the single form BA2 + CA + D = 0, where A is the
unknown and B, C and D are parameters; but unfortunately he was modern
only in some ways and ancient and medieval in others. His algebra is
fundamentally syncopated rather than symbolic, for although he wisely
adopted the Germanic symbols for addition and subtraction and, still more
wisely, used differing symbols for parameters and unknowns, the remainder
of his algebra consisted of words and abbreviations. The third power of the
unknown was not A3, or even AAA, but A cubus, and the second power was
A quadratus. Multiplication was signified by the Latin word in, division was
indicated by the fraction line, and for equality Viète used an abbreviation for
the Latin aequalis. It is not given for one man to make the whole of a given
change; it must come in steps.
One of the steps beyond the work of Viète was taken by Harriot when he
revived the idea Stidel had had of writing the cube of the unknown as AAA.
This notation was used systematically by Harriot in his posthumous book
entitles Artis analyticae praxis and printed in 1631. Its title had been
suggested by the earlier work of Viète, who had disliked the Arabic name
algebra.
In view of the type of reasoning so frequently used in algebra, Viète called
the subject “the analytic art”. Moreover, he had a clear awareness of the
broad scope of the subject, realising that the unknown quantity need not be
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either a number or a geometric line. Algebra reasons about “types” or
species, hence Viète contrasted logista speciosa with logista numerosa. His
algebra was presented in the Isagoge (or Introduction), printed in 1591, but
his several other algebraic works did not appear until many years after his
death. In all of these he maintained a principle of homogeneity in equations,
so that in an equation such as x3 + 3ax = b the a is designated as planum and
the b as solidum. This suggest a certain inflexibility, which Descartes
removed a generation later; but homogeneity also has certain advantages, as
Viète undoubtedly saw.
The algebra of Viète is noteworthy for the generality of its expressions, but
there are also other novel aspects. For one thing, Viète suggested a new
approach to the solution of the cubic. Having reduced it to the standard
equivalent of x3 + 3ax = b, he introduced a new unknown quantity y that was
related to x through the equation in y3, for which the solution is readily
obtained. Moreover, Viète was aware of some the relations between roots
and coefficients of an equation, although here he was hampered by his
failure to allow the coefficients and roots to be negative.
Extracts from: A History of Mathematics (2nd Edition), C.B. Boyer, page
303-305
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