Download Factor the trinomial using the Zero Product Property. or

Study Guide and Review - Chapter 8
The solutions are 0 and 3.
Solve each equation. Check your solutions.
2
41. 6x = 12x
SOLUTION: Factor the trinomial using the Zero Product Property.
2
43. 3x = 5x
SOLUTION: Factor the trinomial using the Zero Product Property.
or The roots are 0 and 2. Check by substituting 0 and 2
in for x in the original equation.
The roots are 0 and
. Check by substituting 0 and
in for x in the original equation.
and
and
The solutions are 0 and 2.
2
42. x = 3x
SOLUTION: Factor the trinomial using the Zero Product Property.
The solutions are 0 and
.
44. x(3x − 6) = 0
The roots are 0 and 3. Check by substituting 0 and 3
in for x in the original equation.
SOLUTION: Factor the trinomial using the Zero Product Property.
x(3x − 6) = 0
x = 0 or and
The roots are 0 and 2. Check by substituting 0 and 2
in for x in the original equation.
The solutions are 0 and 3.
2
43. 3x = 5x
SOLUTION: Factor the trinomial using the Zero Product Property.
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and
Page 1
3
2
2
The area of the rectangle is x − 2x + 5x or x(x –
2x + 5). Area is found by multiplying the length by
the width. Because the width is x, the length must be
2
x − 2x + 5.
Study
Guide
and are
Review
8
The
solutions
0 and- Chapter
.
44. x(3x − 6) = 0
SOLUTION: Factor the trinomial using the Zero Product Property.
x(3x − 6) = 0
x = 0 or The roots are 0 and 2. Check by substituting 0 and 2
in for x in the original equation.
Factor each trinomial. Confirm your answers
using a graphing calculator.
2
46. x − 8x + 15
SOLUTION: In this trinomial, b = –8 and c = 15, so m + p is
negative and mp is positive. Therefore, m and p must
both be negative. List the negative factors of 15, and
look for the pair of factors with a sum of –8.
Factors of 15
–1, –15
–3, –5
Sum of –8
–16
–8
The correct factors are –3 and –5.
and
Check using a Graphing calculator.
The solutions are 0 and 2.
45. GEOMETRY The area of the rectangle shown is
3
2
x − 2x + 5x square units. What is the length?
[– 10, 10] scl: 1 by [– 10, 10] scl: 1
2
47. x + 9x + 20
SOLUTION: 3
2
2
The area of the rectangle is x − 2x + 5x or x(x –
2x + 5). Area is found by multiplying the length by
the width. Because the width is x, the length must be
2
x − 2x + 5.
SOLUTION: In this trinomial, b = 9 and c = 20, so m + p is positive
and mp is positive. Therefore, m and p must both be
positive. List the positive factors of 20, and look for
the pair of factors with a sum of 9.
Factors of 20
1, 20
2, 10
4, 5
Factor each trinomial. Confirm your answers
using a graphing calculator.
2
46. x − 8x + 15
SOLUTION: In this trinomial, b = –8 and c = 15, so m + p is
negative and mp is positive. Therefore, m and p must
both be negative. List the negative factors of 15, and
look for the pair of factors with a sum of –8.
Factors of 15
Sum of –8
–1, –15
–16
–3,
–5
–8
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The correct factors are –3 and –5.
Sum of 9
21
12
9
The correct factors are 4 and 5.
Check using a Graphing calculator.
Page 2
Study
Guide
- Chapter
[– 10,
10]and
scl: Review
1 by [– 10,
10] scl: 18
2
[– 10, 10] scl: 1 by [– 10, 10] scl: 1
2
47. x + 9x + 20
48. x − 5x − 6
SOLUTION: In this trinomial, b = 9 and c = 20, so m + p is positive
and mp is positive. Therefore, m and p must both be
positive. List the positive factors of 20, and look for
the pair of factors with a sum of 9.
SOLUTION: In this trinomial, b = –5 and c = –6, so m + p is
negative and mp is negative. Therefore, m and p
must have different signs. List the factors of –6, and
look for the pair of factors with a sum of –5.
Factors of 20
1, 20
2, 10
4, 5
Sum of 9
21
12
9
The correct factors are 4 and 5.
Factors of –6
–1, 6
1, –6
2, –3
–2, 3
Sum of –5
5
–5
–1
1
The correct factors are 1 and −6.
Check using a Graphing calculator.
Check using a Graphing calculator.
[– 10, 10] scl: 1 by [– 10, 10] scl: 1
[– 10, 10] scl: 1 by [– 12, 8] scl: 1
2
48. x − 5x − 6
2
SOLUTION: In this trinomial, b = –5 and c = –6, so m + p is
negative and mp is negative. Therefore, m and p
must have different signs. List the factors of –6, and
look for the pair of factors with a sum of –5.
Factors of –6
–1, 6
1, –6
2, –3
–2, 3
Sum of –5
5
–5
–1
1
49. x + 3x − 18
SOLUTION: In this trinomial, b = 3 and c = –18, so m + p is
positive and mp is negative. Therefore, m and p must
have different signs. List the factors of –18, and look
for the pair of factors with a sum of 3.
Factors of –18
–1, 18
1, –18
–2, 9
2, –9
–3, 6
3, –6
The correct factors are 1 and −6.
Check using a Graphing calculator.
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Sum of 3
17
–17
7
–7
3
–3
The correct factors are –3 and 6.
Check using a Graphing calculator.
Page 3
Study Guide and Review - Chapter 8
[– 10, 10] scl: 1 by [– 12, 8] scl: 1
[– 10, 10] scl: 1 by [– 14, 6] scl: 1
2
Solve each equation. Check your solutions.
49. x + 3x − 18
2
SOLUTION: In this trinomial, b = 3 and c = –18, so m + p is
positive and mp is negative. Therefore, m and p must
have different signs. List the factors of –18, and look
for the pair of factors with a sum of 3.
50. x + 5x − 50 = 0
SOLUTION: Factors of –18
–1, 18
1, –18
–2, 9
2, –9
–3, 6
3, –6
Sum of 3
17
–17
7
–7
3
–3
The correct factors are –3 and 6.
The roots are –10 and 5. Check by substituting –10
and 5 in for x in the original equation.
and
Check using a Graphing calculator.
The solutions are –10 and 5.
2
51. x − 6x + 8 = 0
SOLUTION: [– 10, 10] scl: 1 by [– 14, 6] scl: 1
Solve each equation. Check your solutions.
2
50. x + 5x − 50 = 0
SOLUTION: The roots are –10 and 5. Check by substituting –10
and 5 in for x in the original equation.
and
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The roots are 2 and 4. Check by substituting 2 and 4
in for x in the original equation.
and
The solutions are 2 and 4.
2
52. x + 12x + 32 = 0
SOLUTION: Page 4
Study Guide and Review - Chapter 8
The solutions are 2 and 4.
2
The solutions are –6 and 8.
2
52. x + 12x + 32 = 0
54. x + 11x + 10 = 0
SOLUTION: SOLUTION: The roots are –8 and –4. Check by substituting –8
and –4 in for x in the original equation.
The roots are –10 and –1. Check by substituting –10
and –1 in for x in the original equation.
and
and
The solutions are –8 and –4.
The solutions are –10 and –1.
2
53. x − 2x − 48 = 0
SOLUTION: 55. ART An artist is working on a painting that is 3
inches longer than it is wide. The area of the painting
is 154 square inches. What is the length of the
painting?
SOLUTION: Let x = the width of the painting. Then, x + 3 = the
length of the painting.
The roots are –6 and 8. Check by substituting –6 and
8 in for x in the original equation.
and
Because a painting cannot have a negative
dimension, the width is 11 inches and the length is 11
+ 3, or 14 inches.
Factor each trinomial, if possible. If the
trinomial cannot be factored, write prime .
The solutions are –6 and 8.
2
54. x + 11x + 10 = 0
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SOLUTION: 2
56. 12x + 22x − 14
SOLUTION: In this trinomial, a = 12, b = 22 and c = –14, so m + p
is positive and mp is negative. Therefore, m and pPage 5
must have different signs. List the factors of 12(–14)
or –168 and identify the factors with a sum of 22.
2(3) = 6
Because a painting cannot have a negative
dimension,
theReview
width is -11
inches and
Study
Guide and
Chapter
8 the length is 11
+ 3, or 14 inches.
Factor each trinomial, if possible. If the
trinomial cannot be factored, write prime .
2
56. 12x + 22x − 14
SOLUTION: In this trinomial, a = 12, b = 22 and c = –14, so m + p
is positive and mp is negative. Therefore, m and p
must have different signs. List the factors of 12(–14)
or –168 and identify the factors with a sum of 22.
Factors of –168
Sum 82
–2, 84
2, –84
–82
53
–3, 56
3, –56
–53
38
–4, 42
4, –42
–38
22
–6, 28
The correct factors are –6 and 28.
2
So, 12x + 22x − 14 = 2(2x − 1)(3x + 7).
There are no factors of 6 with a sum of –9. So, this
trinomial is prime.
2
58. 3x − 6x − 45
SOLUTION: In this trinomial, a = 3, b = –6 and c = –45, so m + p
is negative and mp is negative. Therefore, m and p
must have different signs. List the factors of 3(–45)
or –135 and identify the factors with a sum of –6.
Factors of –135
Sum
134
–1, 135
1, –135
–134
42
–3, 45
3, –45
–42
22
–5, 27
5, –27
–22
6
–9, 15
9, –15
–6
The correct factors are –15 and 9.
2
So, 3x − 6x − 45 = 3(x − 5)(x + 3).
2
2
57. 2y − 9y + 3
SOLUTION: In this trinomial, a = 2, b = –9 and c = 3, so m + p is
negative and mp is positive. Therefore, m and p must
both be negative.
2(3) = 6
There are no factors of 6 with a sum of –9. So, this
trinomial is prime.
2
58. 3x − 6x − 45
SOLUTION: In this trinomial, a = 3, b = –6 and c = –45, so m + p
is negative and mp is negative. Therefore, m and p
must have different signs. List the factors of 3(–45)
or –135 and identify the factors with a sum of –6.
Factors of –135
Sum
134
–1, 135
1, –135
–134
–3,- 45
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3, –45
–42
22
–5, 27
59. 2a + 13a − 24
SOLUTION: In this trinomial, a = 2, b = 13 and c = –24, so m + p
is positive and mp is negative. Therefore, m and p
must have different signs. List the factors of 2(–24)
or –48 and identify the factors with a sum of 13.
Factors of –48
Sum 47
–1, 48
1, –48
–47
22
–2, 24
2, –24
–22
13
–3, 16
3, –16
–13
8
–4, 12
4, –12
–8
2
–6, 8
6, –8
–2
The correct factors are –3 and 16.
Page 6
Study Guide and Review - Chapter 8
2
So, 3x − 6x − 45 = 3(x − 5)(x + 3).
2
2
So, 2a + 13a − 24 = (2a − 3)(a + 8).
Solve each equation. Confirm your answers
using a graphing calculator.
59. 2a + 13a − 24
SOLUTION: In this trinomial, a = 2, b = 13 and c = –24, so m + p
is positive and mp is negative. Therefore, m and p
must have different signs. List the factors of 2(–24)
or –48 and identify the factors with a sum of 13.
Factors of –48
Sum 47
–1, 48
1, –48
–47
22
–2, 24
2, –24
–22
13
–3, 16
3, –16
–13
8
–4, 12
4, –12
–8
2
–6, 8
6, –8
–2
The correct factors are –3 and 16.
2
60. 40x + 2x = 24
SOLUTION: The roots are
and or –0.80 and 0.75 . Confirm the roots using a graphing calculator. Let Y1
and Y2 = –24. Use the intersect option from the C
find the points of intersection. 2
So, 2a + 13a − 24 = (2a − 3)(a + 8).
Solve each equation. Confirm your answers
using a graphing calculator.
2
60. 40x + 2x = 24
SOLUTION: [–5, 5] scl: 1 by [–5, 25] scl: 3
The solutions are
and [–5, 5] scl: 1 by
.
2
61. 2x − 3x − 20 = 0
SOLUTION: The roots are
and or –0.80 and 0.75 . Confirm the roots using a graphing calculator. Let Y1
and Y2 = –24. Use the intersect option from the C
find the points of intersection. eSolutions Manual - Powered by Cognero
Page 7
The roots are
or −2.5 and 4. [–5, 5] scl: 1 by [–5, 25] scl: 3
[–5, 5] scl: 1 by
The
solutions
.
Study
Guide
and are
Review and - Chapter
8
2
[–10, 10] scl: 1 by [–15, 5] scl: 1 [–10, 10] scl: 1 by
and 4.
The solutions are
2
61. 2x − 3x − 20 = 0
62. −16t + 36t − 8 = 0
SOLUTION: SOLUTION: The roots are
or −2.5 and 4. Confirm the roots using a graphing calculator. Let Y1
20 and Y2 = 0. Use the intersect option from the C
find the points of intersection. The roots are 2 and
. Confirm the roots using a graphing calculator. Let Y1
− 8 and Y2 = 0. Use the intersect option from the
find the points of intersection. [–10, 10] scl: 1 by [–15, 5] scl: 1 [–10, 10] scl: 1 by
The solutions are
and 4.
[–2, 3] scl: 1 by [–20, 10] scl: 6
2
The solutions are 2 and
62. −16t + 36t − 8 = 0
SOLUTION: [–2, 3] scl: 1 by [
.
2
63. 6x − 7x − 5 = 0
SOLUTION: The roots are 2 and
. and or −0.5 and 1.67. The roots are
Confirm the roots using a graphing calculator. Let Y1
− 8 and Y2 = 0. Use the intersect option from the
Confirm the roots using a graphing calculator. Let Y1
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[–2, 3] scl: 1 by [–20, 10] scl: 6 [–2, 3] scl: 1 by [
Study
Guide
and are
Review
8
The
solutions
2 and- Chapter
.
2
[–5, 5] scl: 0.5 by [–10, 10] scl: 1 [–5, 5] scl: 0.5 by
The solutions are
and .
64. GEOMETRY The area of the rectangle shown is
63. 6x − 7x − 5 = 0
2
SOLUTION: 6x + 11x − 7 square units. What is the width of the
rectangle?
The roots are
SOLUTION: To find the width, factor the area of the rectangle.
In the area trinomial, a = 6, b = 11 and c = –7, so m
+ p is positive and mp is negative. Therefore, m and
p must have different signs. List the factors of 6(–7)
or –42 and identify the factors with a sum of 11.
Factors of –42
Sum
41
–1, 42
1, –42
–41
19
–2, 21
2, –21
–19
11
–3, 14
3, –14
–11
1
–6, 7
6, –7
–1
The correct factors are –3 and 14.
and or −0.5 and 1.67. Confirm the roots using a graphing calculator. Let Y1
5 and Y2 = 0. Use the intersect option from the CA
find the points of intersection. [–5, 5] scl: 0.5 by [–10, 10] scl: 1 [–5, 5] scl: 0.5 by
The solutions are
and .
2
64. GEOMETRY The area of the rectangle shown is
2
6x + 11x − 7 square units. What is the width of the
rectangle?
SOLUTION: To find the width, factor the area of the rectangle.
In the area trinomial, a = 6, b = 11 and c = –7, so m
+ p is positive and mp is negative. Therefore, m and
p must have different signs. List the factors of 6(–7)
or –42 and identify the factors with a sum of 11.
Factors of –42
Sum
41
–1, 42
1, –42
–41
19
–2, 21
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2, –21
–19
11
–3, 14
3, –14
–11
So, 6x + 11x − 7 = (2x – 1)(3x + 7). The area of a
rectangle is found by multiplying the length by the
width. Because the length of the rectangle is 2x – 1,
the width must be 3x + 7.
Page 9