The graphs coincide. Therefore, the trinomial has been factored

8-6 Solving x^2 + bx + c = 0
Factor each polynomial. Confirm your answers using a graphing calculator.
2
1. x + 14x + 24
SOLUTION: In this trinomial, b = 14 and c = 24, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 24 and identify the factors with a sum of 14.
Factors of 24
Sum
1, 24
25
2, 12
14
3, 8
11
4, 6
10
The correct factors are 2 and 12.
2
Confirm by graphing Y1 = x +14x + 24 and Y2 = (x + 2)(x + 12) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (x + 2)(x + 12).
2
2. y − 7y − 30
SOLUTION: In this trinomial, b = −7 and c = −30, so m + p is negative and mp is negative. Therefore, m and p must have
different signs. List the factors of −30 and identify the factors with a sum of −7.
Sum Factors of −30
−29
1, −30
−13
2, −15
−7
3, −10
−1
5, −6
1
6, −5
7
10, −3
13
15, −2
29
30, −1
The correct factors are 3 and −10.
2
Confirm by graphing Y1 = y – 7y – 30 + and Y2 = (x – 10)(x + 3) on the same screen.
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Page 1
The graphs coincide. Therefore, the trinomial has been factored correctly. √
8-6 Solving
x^2are
+ bx
+ 2)(x
c = 0+ 12).
The factors
(x +
2
2. y − 7y − 30
SOLUTION: In this trinomial, b = −7 and c = −30, so m + p is negative and mp is negative. Therefore, m and p must have
different signs. List the factors of −30 and identify the factors with a sum of −7.
Sum Factors of −30
−29
1, −30
−13
2, −15
−7
3, −10
−1
5, −6
1
6, −5
7
10, −3
13
15, −2
29
30, −1
The correct factors are 3 and −10.
2
Confirm by graphing Y1 = y – 7y – 30 + and Y2 = (x – 10)(x + 3) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (x – 10)(x + 3).
2
3. n + 4n − 21
SOLUTION: In this trinomial, b = 4 and c = −21, so m + p is positive and mp is negative. Therefore, m and p must have different
signs. List the factors of −21 and identify the factors with a sum of 4.
Sum Factors of −21
−20
1, −21
−4
3, −7
4
7, −3
21, −1
20
The correct factors are 7 and −3.
2
Confirm by graphing Y1 = n + 4n– 21 + and Y2 = (n – 3)(n + 7) on the same screen.
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Page 2
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (x – 10)(x + 3).
2
3. n + 4n − 21
SOLUTION: In this trinomial, b = 4 and c = −21, so m + p is positive and mp is negative. Therefore, m and p must have different
signs. List the factors of −21 and identify the factors with a sum of 4.
Sum Factors of −21
−20
1, −21
−4
3, −7
4
7, −3
21, −1
20
The correct factors are 7 and −3.
2
Confirm by graphing Y1 = n + 4n– 21 + and Y2 = (n – 3)(n + 7) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (n – 3)(n + 7).
2
4. m − 15m + 50
SOLUTION: In this trinomial, b = –15 and c = 50, so m + p is negative and mp is positive. Therefore, m and p must both be
negative. List the factors of 50 and identify the factors with a sum of –15.
Factors of 50
Sum –1, –50
–51
–2, –25
–27
–5, –10
–15
The correct factors are –5 and –10.
2
Confirm by graphing Y1 = m – 15m + 50 + and Y2 = (m – 5)(m – 10) on the same screen.
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The graphs coincide. Therefore, the trinomial has been factored correctly. √
Page 3
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (n – 3)(n + 7).
2
4. m − 15m + 50
SOLUTION: In this trinomial, b = –15 and c = 50, so m + p is negative and mp is positive. Therefore, m and p must both be
negative. List the factors of 50 and identify the factors with a sum of –15.
Factors of 50
Sum –1, –50
–51
–2, –25
–27
–5, –10
–15
The correct factors are –5 and –10.
2
Confirm by graphing Y1 = m – 15m + 50 + and Y2 = (m – 5)(m – 10) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (m – 5)(m – 10).
Solve each equation. Check your solutions.
2
5. x − 4x − 21 = 0
SOLUTION: List the factors of 21 and identify the factors with a sum of –4.
Factors of 21
Sum
1, –21
–20
–1, 21
19
3, –7
–4
–3, 7
4
The roots are –3 and 7. Check by substituting –3 and 7 in for x in the original equation.
and
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Page 4
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (m – 5)(m – 10).
Solve each equation. Check your solutions.
2
5. x − 4x − 21 = 0
SOLUTION: List the factors of 21 and identify the factors with a sum of –4.
Factors of 21
Sum
1, –21
–20
–1, 21
19
3, –7
–4
–3, 7
4
The roots are –3 and 7. Check by substituting –3 and 7 in for x in the original equation.
and
The solutions are –3 and 7.
2
6. n − 3n + 2 = 0
SOLUTION: List the factors of 2 and identify the factors with a sum of –3.
Factors of 2
Sum
1, 2
3
–1, –2
–3
The roots are 1 and 2. Check by substituting 1 and 2 in for n in the original equation.
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and
Page 5
8-6 Solving
x^2 +are
bx–3
+ and
c = 7.
0
The solutions
2
6. n − 3n + 2 = 0
SOLUTION: List the factors of 2 and identify the factors with a sum of –3.
Factors of 2
Sum
1, 2
3
–1, –2
–3
The roots are 1 and 2. Check by substituting 1 and 2 in for n in the original equation.
and
The solutions are 1 and 2.
2
7. x − 15x + 54 = 0
SOLUTION: List the factors of 54 and identify the factors with a sum of –15.
Factors of 54
Sum
1, 54
55
–1, –54
–55
2, 27
29
–2, –27
–29
3, 18
21
–3, –18
–21
6, 9
15
–6, –9
–15
The roots are 9 and 6. Check by substituting 9 and 6 in for x in the original equation.
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Page 6
8-6 Solving x^2 + bx + c = 0
The solutions are 1 and 2.
2
7. x − 15x + 54 = 0
SOLUTION: List the factors of 54 and identify the factors with a sum of –15.
Factors of 54
Sum
1, 54
55
–1, –54
–55
2, 27
29
–2, –27
–29
3, 18
21
–3, –18
–21
6, 9
15
–6, –9
–15
The roots are 9 and 6. Check by substituting 9 and 6 in for x in the original equation.
and
The solutions are 6 and 9.
2
8. x + 12x = −32
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 32 and identify the factors with a sum of 12.
Factors of 32
Sum
1, 32
33
2, 16
18
4, 8
12
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Page 7
8-6 Solving x^2 + bx + c = 0
The solutions are 6 and 9.
2
8. x + 12x = −32
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 32 and identify the factors with a sum of 12.
Factors of 32
Sum
1, 32
33
2, 16
18
4, 8
12
The roots are –8 and –4. Check by substituting –8 and –4 in for x in the original equation.
and
The solutions are –8 and -4.
2
9. x − x − 72 = 0
SOLUTION: List the factors of 72 and identify the factors with a sum of –1.
Factors of 72
Sum
–1, 72
71
1, –72
– 71
–2, 36
34
2, –36
– 34
–3, 24
21
3, –24
– 21
–4, 18
14
4, –18
– 14
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6
– 6, 12
6,– 12
–6
–8, 9
1
Page 8
8-6 Solving x^2 + bx + c = 0
The solutions are –8 and -4.
2
9. x − x − 72 = 0
SOLUTION: List the factors of 72 and identify the factors with a sum of –1.
Factors of 72
Sum
–1, 72
71
1, –72
– 71
–2, 36
34
2, –36
– 34
–3, 24
21
3, –24
– 21
–4, 18
14
4, –18
– 14
6,
12
6
–
6,– 12
–6
–8, 9
1
8, – 9
–1
The roots are –8 and 9. Check by substituting –8 and 9 in for x in the original equation.
and
The solutions are –8 and 9.
2
10. x − 10x = −24
SOLUTION: Rewrite the equation with 0 on the right side.
. List the factors of 24 and identify the factors with a sum of –10.
Factors of 24
Sum
–1,
–24
eSolutions Manual - Powered by Cognero–25
–2, –12
–14
–3, –8
–11
Page 9
8-6 Solving x^2 + bx + c = 0
The solutions are –8 and 9.
2
10. x − 10x = −24
SOLUTION: Rewrite the equation with 0 on the right side.
. List the factors of 24 and identify the factors with a sum of –10.
Factors of 24
Sum
–1, –24
–25
–2, –12
–14
–3, –8
–11
–4, –6
–10
The roots are 6 and 4. Check by substituting 6 and 4 in for x in the original equation.
and
The solutions are 6 and 4.
11. FRAMING Tina bought a frame for a photo, but the photo is too big for the frame. Tina needs to reduce the width
and length of the photo by the same amount. The area of the photo should be reduced to half the original area. If the
original photo is 12 inches by 16 inches, what will be the dimensions of the smaller photo?
SOLUTION: Let x be the amount that Tina should reduce the photo. So the dimensions are now (12 – x)(16 – x). The original
area was 12(16) = 192 square inches. Since the area is to be reduced by half, the new area will be 96 square inches.
Now solve the equation.
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Page 10
8-6 Solving x^2 + bx + c = 0
The solutions are 6 and 4.
11. FRAMING Tina bought a frame for a photo, but the photo is too big for the frame. Tina needs to reduce the width
and length of the photo by the same amount. The area of the photo should be reduced to half the original area. If the
original photo is 12 inches by 16 inches, what will be the dimensions of the smaller photo?
SOLUTION: Let x be the amount that Tina should reduce the photo. So the dimensions are now (12 – x)(16 – x). The original
area was 12(16) = 192 square inches. Since the area is to be reduced by half, the new area will be 96 square inches.
Now solve the equation.
The answer x = 24 does not make sense because it would result in a new length of 12 – 24, or –12 inches.
Therefore, the length and the width of the photo must both be reduced by 4 inches. So, the new dimensions are 8
inches by 12 inches.
Factor each polynomial. Confirm your answers using a graphing calculator.
2
12. x + 17x + 42
SOLUTION: In this trinomial, b = 17 and c = 42, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 42 and identify the factors with a sum of 17.
Factors of 42
Sum
1, 42
43
2, 21
23
3, 14
17
6, 7
13
The correct factors are 3 and 14.
2
Confirm by graphing Y1 = x + 17x + 42 and Y2 = (x + 3)(x + 14) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (x + 3)(x + 14).
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2
13. y − 17y + 72
Page 11
The answer x = 24 does not make sense because it would result in a new length of 12 – 24, or –12 inches.
8-6 Solving
x^2
bx + cand
= 0the width of the photo must both be reduced by 4 inches. So, the new dimensions are 8
Therefore,
the+length
inches by 12 inches.
Factor each polynomial. Confirm your answers using a graphing calculator.
2
12. x + 17x + 42
SOLUTION: In this trinomial, b = 17 and c = 42, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 42 and identify the factors with a sum of 17.
Factors of 42
Sum
1, 42
43
2, 21
23
3, 14
17
6, 7
13
The correct factors are 3 and 14.
2
Confirm by graphing Y1 = x + 17x + 42 and Y2 = (x + 3)(x + 14) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (x + 3)(x + 14).
2
13. y − 17y + 72
SOLUTION: In this trinomial, b = −17 and c = 72, so m + p is negative and mp is positive. Therefore, m and p must both be
negative. List the factors of 72 and identify the factors with a sum of −17.
Factors of 72
Sum
−73
−1, −72
−38
−2, −36
−27
−3, −24
−22
−4, −18
−18
−6, −12
−17
−8, −9
The correct factors are −8 and −9.
2
Confirm by graphing Y1 = y − 17y + 72 and Y2 = (y – 8)(y – 9) on the same screen.
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Page 12
The graphs coincide. Therefore, the trinomial has been factored correctly. √
8-6 Solving
x^2are
+ bx
+ 3)(x
c = 0+ 14).
The factors
(x +
2
13. y − 17y + 72
SOLUTION: In this trinomial, b = −17 and c = 72, so m + p is negative and mp is positive. Therefore, m and p must both be
negative. List the factors of 72 and identify the factors with a sum of −17.
Factors of 72
Sum
−73
−1, −72
−38
−2, −36
−27
−3, −24
−22
−4, −18
−18
−6, −12
−17
−8, −9
The correct factors are −8 and −9.
2
Confirm by graphing Y1 = y − 17y + 72 and Y2 = (y – 8)(y – 9) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (y – 8)(y – 9).
2
14. a + 8a − 48
SOLUTION: In this trinomial, b = 8 and c = −48, so m + p is positive and mp is negative. Therefore, m and p must have opposite
signs. List the factors of −48 and identify the factors with a sum of 8.
Sum
Factors of −48
−1, 48
47
−2, 24
22
−3, 16
13
−4, 12
8
−6, 8
2
−8, 6
−2
−12, 4
−8
−16, 3
−13
−24, 2
−22
−48, 1
−47
The correct factors are −4 and 12.
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2
Confirm by graphing Y1 = a + 8a − 48 and Y2 = (a – 4)(a + 12) on the same screen.
Page 13
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (y – 8)(y – 9).
2
14. a + 8a − 48
SOLUTION: In this trinomial, b = 8 and c = −48, so m + p is positive and mp is negative. Therefore, m and p must have opposite
signs. List the factors of −48 and identify the factors with a sum of 8.
Sum
Factors of −48
−1, 48
47
−2, 24
22
−3, 16
13
−4, 12
8
−6, 8
2
−8, 6
−2
−12, 4
−8
−16, 3
−13
−24, 2
−22
−48, 1
−47
The correct factors are −4 and 12.
2
Confirm by graphing Y1 = a + 8a − 48 and Y2 = (a – 4)(a + 12) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (a – 4)(a + 12).
2
15. n − 2n − 35
SOLUTION: In this trinomial, b = −2 and c = −35, so m + p is negative and mp is negative. Therefore, m and p must have opposite
signs. List the factors of −35 and identify the factors with a sum of −2.
Sum
Factors of −35
−1, 35
34
−5, 7
2
−7, 5
−2
−35, 1
−34
The correct factors are −7 and 5.
2
Confirm by graphing Y1 = n – 2n – 35 and Y2 = (n – 7)(n + 5) on the same screen.
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Page 14
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (a – 4)(a + 12).
2
15. n − 2n − 35
SOLUTION: In this trinomial, b = −2 and c = −35, so m + p is negative and mp is negative. Therefore, m and p must have opposite
signs. List the factors of −35 and identify the factors with a sum of −2.
Sum
Factors of −35
−1, 35
34
−5, 7
2
−7, 5
−2
−35, 1
−34
The correct factors are −7 and 5.
2
Confirm by graphing Y1 = n – 2n – 35 and Y2 = (n – 7)(n + 5) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (n – 7)(n + 5).
16. 44 + 15h + h
2
SOLUTION: 2
First rearrange the polynomial in decreasing order, h + 15h + 44.
In this trinomial, b = 15 and c = 44, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 44 and identify the factors with a sum of 15.
Factors of 44
Sum
1, 44
45
2, 22
24
4, 11
15
The correct factors are 4 and 11.
2
Confirm by graphing Y1 = 44 + 15h + h and Y2 = (h + 4)(h + 11) on the same screen.
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The graphs coincide. Therefore, the trinomial has been factored correctly. √
Page 15
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (n – 7)(n + 5).
16. 44 + 15h + h
2
SOLUTION: 2
First rearrange the polynomial in decreasing order, h + 15h + 44.
In this trinomial, b = 15 and c = 44, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 44 and identify the factors with a sum of 15.
Factors of 44
Sum
1, 44
45
2, 22
24
4, 11
15
The correct factors are 4 and 11.
2
Confirm by graphing Y1 = 44 + 15h + h and Y2 = (h + 4)(h + 11) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (h + 4)(h + 11).
17. 40 − 22x + x
2
SOLUTION: 2
First rearrange the polynomial in decreasing order, x – 22x + 40.
In this trinomial, b = −22 and c = 40, so m + p is negative and mp is positive. Therefore, m and p must both be
negative. List the factors of 40 and identify the factors with a sum of −22.
Factors of 40
Sum
−41
−1, −40
−22
−2, −20
−14
−4, −10
−13
−5, −8
The correct factors are −2 and −20.
2
Confirm by graphing Y1 = 40 – 22x + x and Y2 = (x – 2)(x – 20) on the same screen.
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Page 16
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (h + 4)(h + 11).
17. 40 − 22x + x
2
SOLUTION: 2
First rearrange the polynomial in decreasing order, x – 22x + 40.
In this trinomial, b = −22 and c = 40, so m + p is negative and mp is positive. Therefore, m and p must both be
negative. List the factors of 40 and identify the factors with a sum of −22.
Factors of 40
Sum
−41
−1, −40
−22
−2, −20
−14
−4, −10
−13
−5, −8
The correct factors are −2 and −20.
2
Confirm by graphing Y1 = 40 – 22x + x and Y2 = (x – 2)(x – 20) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (x – 2)(x – 20).
18. −24 − 10x + x
2
SOLUTION: 2
First rearrange the polynomial in decreasing order x – 10x – 24.
In this trinomial, b = –10 and c = –24, so m + p is negative and mp is negative. Therefore, m and p must have
opposite signs. List the factors of –24 and identify the factors with a sum of –10.
Factors of –24
Sum
–1, 24
23
–2, 12
10
–3, 8
5
–4, 6
2
–6, 4
–2
–8, 3
–5
–12, 2
–10
–24, 1
–23
The correct factors are –12 and 2.
2
Confirm by graphing Y1 = –24 – 10x + x and Y2 = (x + 2)(x – 12) on the same screen.
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Page 17
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (x – 2)(x – 20).
18. −24 − 10x + x
2
SOLUTION: 2
First rearrange the polynomial in decreasing order x – 10x – 24.
In this trinomial, b = –10 and c = –24, so m + p is negative and mp is negative. Therefore, m and p must have
opposite signs. List the factors of –24 and identify the factors with a sum of –10.
Factors of –24
Sum
–1, 24
23
–2, 12
10
–3, 8
5
–4, 6
2
–6, 4
–2
–8, 3
–5
–12, 2
–10
–24, 1
–23
The correct factors are –12 and 2.
2
Confirm by graphing Y1 = –24 – 10x + x and Y2 = (x + 2)(x – 12) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (x + 2)(x – 12).
2
19. −42 − m + m
SOLUTION: 2
First rearrange the polynomial in decreasing order m – m – 42.
In this trinomial, b = –1 and c = –42, so m + p is negative and mp is negative. Therefore, m and p must have opposite
signs. List the factors of –42 and identify the factors with a sum of –1.
Factors of –42
Sum
–1, 42
41
–2, 21
19
–3, 14
11
–6, 7
1
–7, 6
–1
–14, 3
–11
–21, 2
–19
–42, 1
–41
The correct factors are –7 and 6.
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Page 18
2
Confirm by graphing Y1 = –42 – m + m and Y2 = (m + 6)(m – 7) on the same screen.
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (x + 2)(x – 12).
2
19. −42 − m + m
SOLUTION: 2
First rearrange the polynomial in decreasing order m – m – 42.
In this trinomial, b = –1 and c = –42, so m + p is negative and mp is negative. Therefore, m and p must have opposite
signs. List the factors of –42 and identify the factors with a sum of –1.
Factors of –42
Sum
–1, 42
41
–2, 21
19
–3, 14
11
–6, 7
1
–7, 6
–1
–14, 3
–11
–21, 2
–19
–42, 1
–41
The correct factors are –7 and 6.
2
Confirm by graphing Y1 = –42 – m + m and Y2 = (m + 6)(m – 7) on the same screen.
The graphs coincide. Therefore, the trinomial has been factored correctly. √
The factors are (m + 6)(m – 7).
Solve each equation. Check your solutions.
2
20. x − 7x + 12 = 0
SOLUTION: List the factors of 12 and identify the factors with a sum of –7.
Factors of 12
Sum
–1, –12
–13
–2, –6
–8
–3, –4
–7
The roots are 3 and 4. Check by substituting 3 and 4 in for x in the original equation.
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Page 19
The graphs
the trinomial has been factored correctly. √
8-6 Solving
x^2coincide.
+ bx + cTherefore,
=0
The factors are (m + 6)(m – 7).
Solve each equation. Check your solutions.
2
20. x − 7x + 12 = 0
SOLUTION: List the factors of 12 and identify the factors with a sum of –7.
Factors of 12
Sum
–1, –12
–13
–2, –6
–8
–3, –4
–7
The roots are 3 and 4. Check by substituting 3 and 4 in for x in the original equation.
and
The solutions are 3 and 4.
2
21. y + y = 20
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of –20 and identify the factors with a sum of 1.
Factors of –20
Sum
1, –20
–19
–1, 20
19
–8
2,– 10
8
– 2, 10
–1
4, – 5
1
– 4, 5
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Page 20
8-6 Solving
x^2 +are
bx3+and
c =4.0
The solutions
2
21. y + y = 20
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of –20 and identify the factors with a sum of 1.
Factors of –20
Sum
1, –20
–19
–1, 20
19
–8
2,– 10
8
– 2, 10
–1
4, – 5
4,
5
1
–
The roots are –5 and 4. Check by substituting –5 and 4 in for y in the original equation.
and
The solutions are –5 and 4.
2
22. x − 6x = 27
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of –27 and identify the factors with a sum of –6.
Factors of –27
Sum
1,
–27
eSolutions Manual - Powered by Cognero–26
–1, 27
26
3, –9
–6
Page 21
8-6 Solving x^2 + bx + c = 0
The solutions are –5 and 4.
2
22. x − 6x = 27
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of –27 and identify the factors with a sum of –6.
Factors of –27
Sum
1, –27
–26
–1, 27
26
3, –9
–6
–3, 9
6
The roots are –3 and 9. Check by substituting –3 and 9 in for x in the original equation.
and
The solutions are –3 and 9.
2
23. a + 11a = −18
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 18 and identify the factors with a sum of 11.
Factors of 18
Sum
1, 18
19
2, 9
11
3, 6
19
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Page 22
8-6 Solving x^2 + bx + c = 0
The solutions are –3 and 9.
2
23. a + 11a = −18
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 18 and identify the factors with a sum of 11.
Factors of 18
Sum
1, 18
19
2, 9
11
3, 6
19
The roots are –2 and –9. Check by substituting –2 and –9 in for a in the original equation.
and
The solutions are –2 and –9.
2
24. c + 10c + 9 = 0
SOLUTION: List the factors of 9 and identify the factors with a sum of 10.
Factors of 9
Sum
1, 9
10
3, 3
6
The roots are –1 and –9. Check by substituting –1 and –9 in for c in the original equation.
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Page 23
8-6 Solving x^2 + bx + c = 0
The solutions are –2 and –9.
2
24. c + 10c + 9 = 0
SOLUTION: List the factors of 9 and identify the factors with a sum of 10.
Factors of 9
Sum
1, 9
10
3, 3
6
The roots are –1 and –9. Check by substituting –1 and –9 in for c in the original equation.
and
The solutions are –1 and –9.
2
25. x − 18x = −32
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 32 and identify the factors with a sum of –18.
Factors of 32
Sum
–1, –32
–33
–2, –16
–18
–4, –8
–12
The roots are 16 and 2. Check by substituting 16 and 2 in for x in the original equation.
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Page 24
8-6 Solving x^2 + bx + c = 0
The solutions are –1 and –9.
2
25. x − 18x = −32
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 32 and identify the factors with a sum of –18.
Factors of 32
Sum
–1, –32
–33
–2, –16
–18
–4, –8
–12
The roots are 16 and 2. Check by substituting 16 and 2 in for x in the original equation.
and
The solutions are 2 and 16.
2
26. n − 120 = 7n
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of −120 and identify the factors with a sum of −7.
Factors of −120
Sum
−1, 120
119
1, −120
−119
−2, 60
58
2, −60
−58
3,
40
37
−
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−37
3, −40
26
−4, 30
−26
4, −30
Page 25
List the factors of −120 and identify the factors with a sum of −7.
Factors of −120
Sum
−1, 120
119
1,
−120
−119
8-6 Solving x^2 + bx + c = 0
−2, 60
58
2, −60
−58
37
−3, 40
−37
3, −40
26
−4, 30
−26
4, −30
19
−5, 24
−19
5, −24
14
−6, 20
−14
6, −20
7
−8, 15
−7
8, −15
−2
−12, 10
12, −10
2
The roots are 15 and −8. Check by substituting 15 and −8 in for n in the original equation.
and
The solutions are −8 and 15.
2
27. d + 56 = −18d
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 56 and identify the factors with a sum of 18.
Factors of 56
Sum
1, 56
57
2, 38
40
4, 14
18
7, 8- Powered by Cognero15
eSolutions Manual
Page 26
8-6 Solving x^2 + bx + c = 0
The solutions are −8 and 15.
2
27. d + 56 = −18d
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 56 and identify the factors with a sum of 18.
Factors of 56
Sum
1, 56
57
2, 38
40
4, 14
18
7, 8
15
The roots are −4 and −14. Check by substituting −4 and −14 in for d in the original equation.
and
The solutions are −4 and −14.
2
28. y − 90 = 13y
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of −90 and identify the factors with a sum of −13.
Factors of 90
Sum
1, −90
−89
−1, 90
89
2, −45
−43
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−2, 45
43
−27
3, −30
3, 30
27
Page 27
8-6 Solving x^2 + bx + c = 0
The solutions are −4 and −14.
2
28. y − 90 = 13y
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of −90 and identify the factors with a sum of −13.
Factors of 90
Sum
1, −90
−89
−1, 90
89
2, −45
−43
−2, 45
43
−27
3, −30
27
−3, 30
−13
5 ,−18
13
−5, 18
−9
6, −15
9
−6, 15
−1
9, −10
1
−9, 10
The roots are −5 and 18. Check by substituting −5 and 18 in for y in the original equation.
and
The solutions are −5 and 18.
2
29. h + 48 = 16h
SOLUTION: Rewrite the equation with 0 on the right side.
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Page 28
8-6 Solving x^2 + bx + c = 0
The solutions are −5 and 18.
2
29. h + 48 = 16h
SOLUTION: Rewrite the equation with 0 on the right side.
List the factors of 48 and identify the factors with a sum of –16.
Factors of 48
Sum
–1, –48
–49
–2, –24
–26
–3, –16
–19
–4, –12
–16
–14
– 6, – 8
The roots are 4 and 12. Check by substituting 4 and 12 in for h in the original equation.
and
The solutions are 4 and 12.
30. GEOMETRY A triangle has an area of 36 square feet. If the height of the triangle is 6 feet more than its base,
what are its height and base?
SOLUTION: Let b represent the base. Then b + 6 is the height.
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List the factors of –72 and identify the factors with a sum of 6.
Page 29
8-6 Solving x^2 + bx + c = 0
The solutions are 4 and 12.
30. GEOMETRY A triangle has an area of 36 square feet. If the height of the triangle is 6 feet more than its base,
what are its height and base?
SOLUTION: Let b represent the base. Then b + 6 is the height.
List the factors of –72 and identify the factors with a sum of 6.
Factors of –72 Sum
1, –72
–71
–1, 72
71
2, –36
–34
–2, 36
34
–21
3, – 24
21
– 3, 24
–14
4, – 18
14
– 4, 18
–6
6, – 12
6
– 6, 12
–1
8, – 9
1
– 8, 9
However, the base cannot be negative, so the base is 6 feet. The height is b + 6 = 6 + 6 = 12 feet.
2
31. GEOMETRY A rectangle has an area represented by x − 4x − 12 square feet. If the length is x + 2 feet, what is
the width of the rectangle?
SOLUTION: 2
The area of the rectangle is x − 4x − 12.
List the factors of −12 and identify the factors with a sum of −4.
Factors of −12
Sum
1, −12
−11
−1, 12
11
2, −6
−4
−2, -6Powered by Cognero 4
eSolutions Manual
−1
3, −4
1
−3, 4
Page 30
8-6 Solving
x^2 + bx + c = 0
However, the base cannot be negative, so the base is 6 feet. The height is b + 6 = 6 + 6 = 12 feet.
2
31. GEOMETRY A rectangle has an area represented by x − 4x − 12 square feet. If the length is x + 2 feet, what is
the width of the rectangle?
SOLUTION: 2
The area of the rectangle is x − 4x − 12.
List the factors of −12 and identify the factors with a sum of −4.
Factors of −12
Sum
1, −12
−11
−1, 12
11
2, −6
−4
−2, 6
4
−1
3, −4
1
−3, 4
Then area of the rectangle is (x + 2)(x – 6). Area is found by multiplying the length by the width. Because the length
is x + 2, the width must be x − 6.
32. SOCCER The width of a high school soccer field is 45 yards shorter than its length.
a. Define a variable, and write an expression for the area of the field.
b. The area of the field is 9000 square yards. Find the dimensions.
SOLUTION: a. Let = length. The area of the field is the length times the width, or (
b.
− 45).
Then length cannot be negative, so it is 120 yd and the width is 120 – 45, or 75 yards.
CCSS STRUCTURE Factor each polynomial.
2
33. q + 11qr + 18r
2
SOLUTION: In this trinomial, b = 11 and c = 18, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 18 and identify the factors with a sum of 11.
Factors of 18
1, 18
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2, 9
3, 6
Sum 19
11
9
Page 31
8-6 Solving x^2 + bx + c = 0
Then length cannot be negative, so it is 120 yd and the width is 120 – 45, or 75 yards.
CCSS STRUCTURE Factor each polynomial.
2
33. q + 11qr + 18r
2
SOLUTION: In this trinomial, b = 11 and c = 18, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of 18 and identify the factors with a sum of 11.
Factors of 18
1, 18
2, 9
3, 6
Sum 19
11
9
The correct factors are 2 and 9.
The trinomial has two variables q and r. The first term in each binomial will have q's, the second term will have the
r along with the factors. 2
34. x − 14xy − 51y
2
SOLUTION: In this trinomial, b = −14 and c = −51, so m + p is negative and mp is negative. Therefore, m and p must have
different signs. List the factors of −51 and identify the factors with a sum of −14.
Sum Factors of −51
−1, 51
50
−3, 17
14
−17, 3
−14
−51, 1
−50
The correct factors are 3 and −17. The trinomial has two variables x and y . The first term in each binomial will have x's, the second term will have the
y along with the factors. 2
35. x − 6xy + 5y
2
SOLUTION: In this trinomial, b = −6 and c = 5, so m + p is negative and mp is positive. Therefore, m and p must both be negative.
List the negative factors of 5 and identify the factors with a sum of −6.
Factors of 5
Sum
−6
−1, −5
The correct factors are −1 and −6. The trinomial has two variables x and y . The first term in each binomial will have x's, the second term will have the
y along with the factors. eSolutions Manual - Powered by Cognero
2
36. a + 10ab − 39b
2
Page 32
The trinomial has two variables x and y . The first term in each binomial will have x's, the second term will have the
y along with the factors. 8-6 Solving x^2 + bx + c = 0
2
35. x − 6xy + 5y
2
SOLUTION: In this trinomial, b = −6 and c = 5, so m + p is negative and mp is positive. Therefore, m and p must both be negative.
List the negative factors of 5 and identify the factors with a sum of −6.
Factors of 5
Sum
−6
−1, −5
The correct factors are −1 and −6. The trinomial has two variables x and y . The first term in each binomial will have x's, the second term will have the
y along with the factors. 2
36. a + 10ab − 39b
2
SOLUTION: In this trinomial, b = 10 and c = −39, so m + p is positive and mp is negative. Therefore, m and p must have different
signs. List the factors of −39 and identify the factors with a sum of 10.
Sum Factors of −39
−38
1, −39
−10
3, −13
−3, 13
10
−39, 1
−38
The correct factors are −3 and 13. The trinomial has two variables a and b. The first term in each binomial will have as, the second term will have the
b along with the factors. 37. SWIMMING The length of a rectangular swimming pool is 20 feet greater than its width. The area of the pool is
525 square feet.
a. Define a variable and write an equation for the area of the pool.
b. Solve the equation.
c. Interpret the solutions. Do both solutions make sense? Explain.
SOLUTION: a. Sample answer: Let w = width. Then the length is 20 feet greater than its width, so = w + 20. The area, which is 525 square feet, is found by multiplying the length times the width. Therefore the area is (w + 20)w = 525.
b. List the factors of −525 and identify the factors with a sum of 20.
Factors of −525
Sum
1, −525
−524
−1, 525
524
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5, −105
−100
−5, 105
100
−68
7, −75
Page 33
The trinomial has two variables a and b. The first term in each binomial will have as, the second term will have the
b along with the factors. 8-6 Solving x^2 + bx + c = 0
37. SWIMMING The length of a rectangular swimming pool is 20 feet greater than its width. The area of the pool is
525 square feet.
a. Define a variable and write an equation for the area of the pool.
b. Solve the equation.
c. Interpret the solutions. Do both solutions make sense? Explain.
SOLUTION: a. Sample answer: Let w = width. Then the length is 20 feet greater than its width, so = w + 20. The area, which is 525 square feet, is found by multiplying the length times the width. Therefore the area is (w + 20)w = 525.
b. List the factors of −525 and identify the factors with a sum of 20.
Factors of −525
Sum
1, −525
−524
−1, 525
524
5, −105
−100
−5, 105
100
−68
7, −75
68
−7, 75
−20
15, −35
20
−15, 35
−4
21, −25
4
−21, 25
The width cannot be negative, therefore it is 15 feet. The length is 15 + 20, 35 feet.
c. The solution of 15 means that the width is 15 ft. The solution −35 does not make sense because the width cannot
be negative.
GEOMETRY Find an expression for the perimeter of a rectangle with the given area.
2
38. A = x + 24x − 81
SOLUTION: 2
The area of the rectangle is x + 24x − 81.
In this trinomial, b = 24 and c = 24, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of − 81 and identify the factors with a sum of 24.
Sum
Factors of − 81
1,−81
−80
−1, 81
80
3, −27
−24
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Page 34
−3, 27
24
9,−9
0
2
The width cannot be negative, therefore it is 15 feet. The length is 15 + 20, 35 feet.
c. The solution
8-6 Solving
x^2 + bx
+ cmeans
= 0 that the width is 15 ft. The solution −35 does not make sense because the width cannot
of 15
be negative.
GEOMETRY Find an expression for the perimeter of a rectangle with the given area.
2
38. A = x + 24x − 81
SOLUTION: 2
The area of the rectangle is x + 24x − 81.
In this trinomial, b = 24 and c = 24, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of − 81 and identify the factors with a sum of 24.
Sum
Factors of − 81
1,−81
−80
−1, 81
80
3, −27
−24
−3, 27
24
9,−9
0
2
Then area of the rectangle of x + 24x − 81 can be factored to (x + 27)(x – 3). Area is found by multiplying the length by the width, so the length is x + 27 and the width is x − 3.
So, an expression for the perimeter of the rectangle is 4x + 48.
2
39. A = x + 13x − 90
SOLUTION: 2
The area of the rectangle is x + 13x − 90.
In this trinomial, b = 13 and c = − 90, so m + p is positive and mp is positive. Therefore, m and p must both be
positive. List the factors of − 90 and identify the factors with a sum of 13.
Sum
Factors of − 90
1, −90
−89
−1, 90
89
2, −45
−43
−2, 45
43
−27
3, −30
27
−3, 30
−13
5, −18
5,
18
13
−
−9
6, −15
9
−6, 15
−1
9, −10
9,
10
1
−
2
The area of the rectangle x + 13x − 90 factors to (x + 18)(x – 5). Area is found by multiplying the length by the width, so the length is x + 18 and the width is x − 5.
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Page 35
8-6 Solving
x^2 + bxfor
+ cthe
= 0perimeter of the rectangle is 4x + 48.
So, an expression
2
39. A = x + 13x − 90
SOLUTION: 2
The area of the rectangle is x + 13x − 90.
In this trinomial, b = 13 and c = − 90, so m + p is positive and mp is positive. Therefore, m and p must both be
positive. List the factors of − 90 and identify the factors with a sum of 13.
Sum
Factors of − 90
1, −90
−89
−1, 90
89
2, −45
−43
−2, 45
43
−27
3, −30
27
−3, 30
−13
5, −18
13
−5, 18
−9
6, −15
9
−6, 15
−1
9, −10
1
−9, 10
2
The area of the rectangle x + 13x − 90 factors to (x + 18)(x – 5). Area is found by multiplying the length by the width, so the length is x + 18 and the width is x − 5.
So, an expression for the perimeter of the rectangle is 4x + 26.
40. MULTIPLE REPRESENTATIONS In this problem, you will explore factoring when the leading coefficient is
not 1.
a. TABULAR Copy and complete the table below.
b. ANALYTICAL How are m and p related to a and c?
c. ANALYTICAL How are m and p related to b?
2
d. VERBAL Describe a process you can use for factoring a polynomial of the form ax + bx + c.
SOLUTION: a.
Use FOIL or the Distributive Property to find the product of the binomials. eSolutions Manual - Powered by Cognero
Page 36
8-6 Solving x^2 + bx + c = 0
So, an expression for the perimeter of the rectangle is 4x + 26.
40. MULTIPLE REPRESENTATIONS In this problem, you will explore factoring when the leading coefficient is
not 1.
a. TABULAR Copy and complete the table below.
b. ANALYTICAL How are m and p related to a and c?
c. ANALYTICAL How are m and p related to b?
2
d. VERBAL Describe a process you can use for factoring a polynomial of the form ax + bx + c.
SOLUTION: a.
Use FOIL or the Distributive Property to find the product of the binomials. b. ac is the product of the coefficients of the first and last terms. mp is the product of the coefficients of the middle
two terms when foiling. Looking at the last two columns and you will see that mp = ac.
c. Look at the second and third columns and you will see that m + p = b. The coefficients of the middle terms equal
b. d. When factoring trinomials , we look for two integers, m and p , for which mp = ac and m + p = b.
2
41. ERROR ANALYSIS Jerome and Charles have factored x + 6x − 16. Is either of them correct? Explain your
reasoning.
SOLUTION: Charles is correct. In this trinomial, b = 6 and c = − 16, so m + p is positive and mp is positive. Therefore, m and p must both be positive.
List the factors of − 16 and identify the factors with a sum of − 6.
Sum
Factors of − 16
1, −16
−15
−1, 16
15
2, −8
−6
−2, 8
6
4, −4
0
2
The correct factors are −2, 8. Jerome’s answer once multiplied is x − 6x − 16. The middle term should be positive.
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CCSS ARGUMENTS Find all values of k so that each polynomial can be factored using integers.
2
42. x + k x − 19
Page 37
2, −8
−6
−2, 8
6
4,
4
− + bx + c = 0 0
8-6 Solving x^2
2
The correct factors are −2, 8. Jerome’s answer once multiplied is x − 6x − 16. The middle term should be positive.
CCSS ARGUMENTS Find all values of k so that each polynomial can be factored using integers.
2
42. x + k x − 19
SOLUTION: In this trinomial, b = k, c = −19, and mp is negative. Therefore, m + p must equal the sum of the factors of −19. List
the factors of −19. The sum of m + p will equal k.
Sum of m + p
Factors of −19
18
−1, 19
−19, 1
−18
Therefore, k could have values of 18 or −18.
2
43. x + k x + 14
SOLUTION: In this trinomial, b = k, c = 14 and mp is positive. Therefore, m + p must equal the sum of the factors of 14. List the
factors of 14. The sum of m + p will equal k.
Factors of 14
Sum of m + p
1, 14
15
2, 7
9
−15
−1, −14
−9
−2, −7
Therefore, k could have values of 9, 15, −9 or −15.
2
44. x − 8x + k, k > 0
SOLUTION: In this trinomial, b = −8, c = k and m + p is negative. Therefore, mp must equal the product of the two numbers that
add to −8. Because k > 0, this means that both m and p are negative. List the numbers that add to −8. The product of
mp will equal k.
Two numbers that add
Product of mp
to −8
7
−1, −7
12
−2, −6
15
−3, −5
16
−4, −4
Therefore, k could have values of 7, 12, 15, or 16.
2
45. x − 5x + k, k > 0
SOLUTION: In this trinomial, b = −5, c = k and m + p is negative. Therefore, mp must equal the product of the two numbers that
add to −5. Because k > 0, this means that both m and p are negative. List the numbers that add to −5. The product of
mp will equal k.
Two numbers that add
Product of mp
to −5
4
−1, −4
6
−2, −3
Therefore, k could have values of 4 or 6.
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2
Page 38
46. REASONING For any factorable trinomial, x + bx + c, will the absolute value of b sometimes, always, or never
be less than the absolute value of c? Explain.
Two numbers that add
Product of mp
to −5
4
−1, −4
8-6 Solving x^2
bx + c = 0
6
−2,+−3
Therefore, k could have values of 4 or 6.
2
46. REASONING For any factorable trinomial, x + bx + c, will the absolute value of b sometimes, always, or never
be less than the absolute value of c? Explain.
SOLUTION: 2
The absolute value of b will sometimes be less than the absolute value of c. Sample answer: The trinomial x + 10x +
2
9 = (x + 1)(x + 9) and 10 > 9. The trinomial x + 7x + 10 = (x + 2)(x + 5) and 7 < 10.
47. OPEN ENDED Give an example of a trinomial that can be factored using the factoring techniques presented in this
lesson. Then factor the trinomial.
SOLUTION: 2
Students’ answers will vary. Sample answer: x + 19x − 20
In this trinomial, b = 19 and c = −20, so m + p is positive and mp is negative. Therefore, m and p must have different
signs. List the factors of −20, and look for the pair of factors with a sum of 19.
Sum Factors of −20
19
−1, 20
8
−2, 10
1
−4, 5
−5, 4
−1
−10, 2
−8
−20, 1
−19
The correct factors are −1 and 20.
2
48. CHALLENGE Factor (4y − 5) + 3(4y − 5) − 70.
SOLUTION: 2
2
The trinomial is written in x + bx + c = 0 form.We can substitute x for 4y – 5 to get x + 3x – 70 = 0.
In this trinomial, b = 3 and c = −70, so m + p is positive and mp is negative. Therefore, m and p must have different
signs. List the factors of −70, and look for the pair of factors with a sum of 3.
Sum Factors of −70
69
–1, 70
33
−2, 35
9
−5, 14
3
−7, 10
−10, 7
−3
−14, 5
−9
−35, 2
−33
−70, 1
−69
The correct factors are –7 and 10.
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2
49. WRITING IN MATH Explain how to factor trinomials of the form x + bx + c and how to determine the signs of
−20, 1
The correct factors are −1 and 20.
−19
8-6 Solving x^2 + bx + c = 0
2
48. CHALLENGE Factor (4y − 5) + 3(4y − 5) − 70.
SOLUTION: 2
2
The trinomial is written in x + bx + c = 0 form.We can substitute x for 4y – 5 to get x + 3x – 70 = 0.
In this trinomial, b = 3 and c = −70, so m + p is positive and mp is negative. Therefore, m and p must have different
signs. List the factors of −70, and look for the pair of factors with a sum of 3.
Sum Factors of −70
69
–1, 70
33
−2, 35
9
−5, 14
3
−7, 10
−10, 7
−3
−14, 5
−9
−35, 2
−33
−70, 1
−69
The correct factors are –7 and 10.
2
49. WRITING IN MATH Explain how to factor trinomials of the form x + bx + c and how to determine the signs of
the factors of c.
SOLUTION: Find factors m and p such that m + p = b and mp = c. If b and c are positive, then m and p are positive. For example:
If b is negative and c is positive, then m and p are negative. For example:
When c is negative, m and p have different signs and the factor with the greatest absolute value has the same sign as
b.
For example:
and
50. Which inequality is shown in the graph?
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When c is negative, m and p have different signs and the factor with the greatest absolute value has the same sign as
b.
8-6 Solving x^2 + bx + c = 0
For example:
and
50. Which inequality is shown in the graph?
A B C D SOLUTION: The line is dashed which means that choices A and D can be eliminated. So check (0, 0) in the inequalities given in
choices B and C to determine which is the correct choice.
True
False
So, the correct choice is C.
51. SHORT RESPONSE Olivia must earn more than $254 from selling candy bars in order to go on a trip with the
National Honor Society. If each candy bar is sold for $1.25, what is the fewest candy bars she must sell?
SOLUTION: Let x represent the number of candy bars Olivia must sell.
So, Olivia must sell at least 204 candy bars to go on the trip.
52. GEOMETRY Which expression represents the length of the rectangle?
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8-6 Solving x^2 + bx + c = 0
So, Olivia must sell at least 204 candy bars to go on the trip.
52. GEOMETRY Which expression represents the length of the rectangle?
F x + 5
G x + 6
H x − 6
J x − 5
SOLUTION: 2
The area of the rectangle is x − 3x − 18.
In this trinomial, b = − 3and c = − 18, so m + p is positive and mp is positive. Therefore, m and p must both be
positive. List the factors of − 18 and identify the factors with a sum of − 3.
Sum
Factors of − 18
1,−18
−17
−1, 18
17
2, −9
−6
6
−2, 9
−3
3, −6
3
−3, 6
2
The area of the rectangle x − 3x − 18 factors to (x + 3)(x – 6). Because the width is x + 3, the length must be x − 6. So, the correct choice is H.
53. The difference of 21 and a number n is 6. Which equation shows the relationship?
A 21 − n = 6
B 21 + n = 6
C 21n = 6
D 6n = −21
SOLUTION: The phrase “is 6” means “= 6”, so choice D can be eliminated. The phrase “difference of” means “subtraction”, so
A is the correct choice.
Factor each polynomial.
2
54. 10a + 40a
SOLUTION: The greatest common factor is
2
10a + 40a = 10a(a + 4)
or 10a.
2
55. 11x + 44x y
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The greatest common factor is
2
8-6 Solving
x^2= +10a(a
bx ++c 4)
=0
10a + 40a
or 10a.
2
55. 11x + 44x y
SOLUTION: The greatest common factor is
2
11x + 44x y = 11x(1 + 4xy)
3 2
or 11x.
2
56. 2m p − 16mp + 8mp
SOLUTION: The greatest common factor is
or 2mp.
3 2
2
2
2m p − 16mp + 8mp = 2mp(m p − 8p + 4)
57. 2ax + 6xc + ba + 3bc
SOLUTION: Factor by grouping.
58. 8ac − 2ad + 4bc − bd
SOLUTION: Factor by grouping.
2
59. x − xy − xy + y
2
SOLUTION: Factor by grouping.
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a polynomial
that
represents
60. Write
the area of the shaded region in the figure.
Page 43
8-6 Solving x^2 + bx + c = 0
60. Write a polynomial that represents the area of the shaded region in the figure.
SOLUTION: Use elimination to solve each system of equations.
61. SOLUTION: Because –x and x have opposite coefficients, add the equations.
Now, substitute 13 for y in either equation to find the value of x.
Check the solution in each equation.
Therefore, the solution is (4, 13).
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SOLUTION: 8-6 Solving
x^2 + bx + c = 0
Use elimination to solve each system of equations.
61. SOLUTION: Because –x and x have opposite coefficients, add the equations.
Now, substitute 13 for y in either equation to find the value of x.
Check the solution in each equation.
Therefore, the solution is (4, 13).
62. SOLUTION: Because 5a and –5a have opposite coefficients, add the equations.
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Therefore, the solution is (4, 13).
8-6 Solving x^2 + bx + c = 0
62. SOLUTION: Because 5a and –5a have opposite coefficients, add the equations.
Now, substitute –3 for b in either equation to find the value of a.
Check the solution in each equation.
Therefore, the solution is (2, –3).
63. SOLUTION: Because d and −d have opposite coefficients, add the equations.
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Now, substitute 3 for c in either equation to find the value of d.
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Therefore, the solution is (2, –3).
8-6 Solving
x^2 + bx + c = 0
63. SOLUTION: Because d and −d have opposite coefficients, add the equations.
Now, substitute 3 for c in either equation to find the value of d.
Check the solution in each equation.
Therefore, the solution is (3, 6).
64. SOLUTION: Because 2y and −2y have opposite coefficients, add the equations.
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Now, substitute 2 for x in either equation to find the value of y.
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8-6 Solving x^2 + bx + c = 0
Therefore, the solution is (3, 6).
64. SOLUTION: Because 2y and −2y have opposite coefficients, add the equations.
Now, substitute 2 for x in either equation to find the value of y.
Check the solution in each equation.
Therefore, the solution is (2, 1).
65. LANDSCAPING Kendrick is planning a circular flower garden with a low fence around the border. He has 38
feet of fence. What is the radius of the largest garden he can make? (Hint: C = 2πr)
SOLUTION: If he has 38 feet of fence, use the formula, C = 2πr, with C = 38.
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So, the largest radius he can make is about 6 feet.
Factor each polynomial.
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8-6 Solving
x^2 + bx + c = 0
Therefore, the solution is (2, 1).
65. LANDSCAPING Kendrick is planning a circular flower garden with a low fence around the border. He has 38
feet of fence. What is the radius of the largest garden he can make? (Hint: C = 2πr)
SOLUTION: If he has 38 feet of fence, use the formula, C = 2πr, with C = 38.
So, the largest radius he can make is about 6 feet.
Factor each polynomial.
66. 6mx − 4m + 3rx − 2r
SOLUTION: 67. 3ax − 6bx + 8b − 4a
SOLUTION: 2
2
68. 2d g + 2fg + 4d h + 4fh
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