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Study Guide and Review - Chapter 8
State whether each sentence is true or false . If false , replace the underlined phrase or expression to
make a true sentence.
2
1. x + 5x + 6 is an example of a prime polynomial.
SOLUTION: The statement is false. A polynomial that cannot be written as a product of two polynomials with integral coefficients
2
is called a prime polynomial. The polynomial x + 5x + 6 can be written as (x + 2)(x + 3), so it is not prime. The
2
polynomial x + 5x + 7 is an example of a prime polynomial.
2. (x + 5)(x − 5) is the factorization of a difference of squares.
SOLUTION: The factored form of the difference of squares is called the product of a sum and a difference. So, (x + 5)(x − 5) is
the factorization of a difference of squares. The statement is true.
2
3. (x + 5)(x − 2) is the factored form of x − 3x − 10.
SOLUTION: 2
(x − 5)(x + 2) is the factored form of x − 3x − 10. So, the statement is false.
4. Expressions with four or more unlike terms can sometimes be factored by grouping.
SOLUTION: Using the Distributive Property to factor polynomials with four or more terms is called factoring by grouping because
terms are put into groups and then factored. So, the statement is true.
5. The Zero Product Property states that if ab = 1, then a or b is 1.
SOLUTION: The statement is false. The Zero Product Property states that for any real numbers a and b, if ab = 0, then a = 0, b
= 0, or a and b are zero.
2
6. x − 12x + 36 is an example of a perfect square trinomial.
SOLUTION: Perfect square trinomials are trinomials that are the squares of binomials.
So, the statement is true.
2
7. x − 16 is an example of a perfect square trinomial.
SOLUTION: The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.
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Page 1
Study Guide and Review - Chapter 8
So, the statement is true.
2
7. x − 16 is an example of a perfect square trinomial.
SOLUTION: The statement is false. Perfect square trinomials are trinomials that are the squares of binomials.
2
2
x − 16 is the product of a sum and a difference. So, x − 16 an example of a difference of squares.
2
8. 4x – 2x + 7 is a polynomial of degree 2.
SOLUTION: true
2
9. The leading coefficient of 1 + 6a + 9a is 1.
SOLUTION: The standard form of a polynomial has the terms in order from greatest to least degree. In this form, the coefficient
of the first term is called the leading coefficient. For this polynomial, the leading coefficient is 9.
10. The FOIL method is used to multiply two trinomials.
SOLUTION: false; binomials
FOIL Method:
To multiply two binomials, find the sum of the products of F the First terms, O the Outer terms, I the Inner terms, L
and the Last terms.
Write each polynomial in standard form.
11. x + 2 + 3x
2
SOLUTION: 2
The greatest degree is 2. Therefore, the polynomial can be rewritten as 3x + x + 2.
12. 1 – x
4
SOLUTION: 4
The greatest degree is 4. Therefore, the polynomial can be rewritten as –x + 1.
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13. 2 + 3x + x
2
Page 2
Study
Guide
anddegree
Review
- Chapter
8 the polynomial can be rewritten as 3x2 + x + 2.
The
greatest
is 2.
Therefore,
12. 1 – x
4
SOLUTION: 4
The greatest degree is 4. Therefore, the polynomial can be rewritten as –x + 1.
13. 2 + 3x + x
2
SOLUTION: 2
The greatest degree is 2. Therefore, the polynomial can be rewritten as x + 3x + 2.
5
2
14. 3x – 2 + 6x – 2x + x
3
SOLUTION: 5
3
2
The greatest degree is 5. Therefore, the polynomial can be rewritten as 3x + x – 2x + 6x – 2.
Find each sum or difference.
3
3
15. (x + 2) + (−3x − 5)
SOLUTION: 2
2
16. a + 5a − 3 − (2a − 4a + 3)
SOLUTION: 2
2
eSolutions
Cognero
17. (4x −Manual
3x +- Powered
5) + (2xby −
5x + 1)
SOLUTION: Page 3
Study Guide and Review - Chapter 8
2
2
17. (4x − 3x + 5) + (2x − 5x + 1)
SOLUTION: 18. PICTURE FRAMES Jean is framing a painting that is a rectangle. What is the perimeter of the frame?
SOLUTION: 2
The perimeter of the frame is 4x + 4x + 8.
Solve each equation.
2
2
19. x (x + 2) = x(x + 2x + 1)
SOLUTION: 2
20. 2x(x + 3) = 2(x + 3)
SOLUTION: 2
21. 2(4w + w ) − 6 = 2w(w − 4) + 10
SOLUTION: eSolutions Manual - Powered by Cognero
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Study Guide and Review - Chapter 8
2
21. 2(4w + w ) − 6 = 2w(w − 4) + 10
SOLUTION: 22. GEOMETRY Find the area of the rectangle.
SOLUTION: 3
2
The area of the rectangle is 3x + 3x − 21x.
Find each product.
23. (x − 3)(x + 7)
SOLUTION: 24. (3a − 2)(6a + 5)
SOLUTION: 25. (3r − 7t)(2r + 5t)
SOLUTION: 26. (2x + 5)(5x + 2)
SOLUTION: eSolutions
Manual - Powered by Cognero
Page 5
SOLUTION: Study Guide and Review - Chapter 8
26. (2x + 5)(5x + 2)
SOLUTION: 27. PARKING LOT The parking lot shown is to be paved. What is the area to be paved?
SOLUTION: 2
2
The area to be paved is 10x + 7x − 12 units .
Find each product.
28. (x + 5)(x − 5)
SOLUTION: 2
29. (3x − 2)
SOLUTION: 2
30. (5x + 4)
SOLUTION: 31. (2x − 3)(2x + 3)
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SOLUTION: Page 6
SOLUTION: Study Guide and Review - Chapter 8
31. (2x − 3)(2x + 3)
SOLUTION: 32. (2r + 5t)
2
SOLUTION: 33. (3m − 2)(3m + 2)
SOLUTION: 34. GEOMETRY Write an expression to represent the area of the shaded region.
SOLUTION: Find the area of the larger rectangle.
Find the area of the smaller rectangle.
To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger
rectangle.
eSolutions
by Cognero
The Manual
area of- Powered
the shaded
region
2
is 3x − 21.
Use the Distributive Property to factor each polynomial.
Page 7
SOLUTION: Study Guide and Review - Chapter 8
34. GEOMETRY Write an expression to represent the area of the shaded region.
SOLUTION: Find the area of the larger rectangle.
Find the area of the smaller rectangle.
To find the area of the shaded region, subtract the area of the smaller rectangle from the area of the larger
rectangle.
2
The area of the shaded region is 3x − 21.
Use the Distributive Property to factor each polynomial.
35. 12x + 24y
SOLUTION: Factor
The greatest common factor of each term is 2· 3 or 12.
12x + 24y = 12(x + 2y)
2
2
36. 14x y − 21xy + 35xy
SOLUTION: Factor
The greatest common factor of each term is 7xy.
2
2
14x y − 21xy + 35xy = 7xy(2x − 3 + 5y)
3
eSolutions
37. 8xy Manual
− 16x -yPowered
+ 10y by Cognero
SOLUTION: Page 8
The
greatest
of each
Study
Guide
andcommon
Review factor
- Chapter
8 term is 7xy.
2
2
14x y − 21xy + 35xy = 7xy(2x − 3 + 5y)
3
37. 8xy − 16x y + 10y
SOLUTION: Factor.
The greatest common factor of each term is 2y.
3
3
8xy − 16x y + 10y = 2y(4x − 8x + 5)
2
38. a − 4ac + ab − 4bc
SOLUTION: Factor by grouping.
2
39. 2x − 3xz − 2xy + 3yz
SOLUTION: 40. 24am − 9an + 40bm − 15bn
SOLUTION: Solve each equation. Check your solutions.
2
41. 6x = 12x
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.
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and
Page 9
SOLUTION: Study Guide and Review - Chapter 8
Solve each equation. Check your solutions.
2
41. 6x = 12x
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.
and
The solutions are 0 and 2.
2
42. x = 3x
SOLUTION: Factor the trinomial using the Zero Product Property.
The roots are 0 and 3. Check by substituting 0 and 3 in for x in the original equation.
and
The solutions are 0 and 3.
2
43. 3x = 5x
SOLUTION: Factor the trinomial using the Zero Product Property.
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Page 10
Study Guide and Review - Chapter 8
The solutions are 0 and 3.
2
43. 3x = 5x
SOLUTION: Factor the trinomial using the Zero Product Property.
The roots are 0 and
. Check by substituting 0 and
in for x in the original equation.
and
The solutions are 0 and
.
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.
and
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The solutions are 0 and 2.
Page 11
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.
and
The solutions are 0 and 2.
3
2
45. GEOMETRY The area of the rectangle shown is x − 2x + 5x square units. What is the length?
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.
2
Because the width is x, the length must be x − 2x + 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
–1, –15
–3, –5
Sum of –8
–16
–8
The correct factors are –3 and –5.
Check using a Graphing calculator.
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Page 12
SOLUTION: 3
2
2
The
area of
theReview
rectangle
is x − 2x8 + 5x or x(x – 2x + 5). Area is found by multiplying the length by the width.
Study
Guide
and
- Chapter
2
Because the width is x, the length must be x − 2x + 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
–1, –15
–3, –5
Sum of –8
–16
–8
The correct factors are –3 and –5.
Check using a Graphing calculator.
[– 10, 10] scl: 1 by [– 10, 10] scl: 1
2
47. x + 9x + 20
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
Sum of 9
21
12
9
The correct factors are 4 and 5.
Check using a Graphing calculator.
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[– 10, 10] scl: 1 by [– 10, 10] scl: 1
2
Page 13
Study
Guide
- Chapter
[– 10,
10]and
scl: Review
1 by [– 10,
10] scl: 18
2
47. x + 9x + 20
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
Sum of 9
21
12
9
The correct factors are 4 and 5.
Check using a Graphing calculator.
[– 10, 10] scl: 1 by [– 10, 10] scl: 1
2
48. x − 5x − 6
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
The correct factors are 1 and −6.
Check using a Graphing calculator.
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[– 10, 10] scl: 1 by [– 12, 8] scl: 1
Page 14
Study Guide and Review - Chapter 8
[– 10, 10] scl: 1 by [– 10, 10] scl: 1
2
48. x − 5x − 6
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
The correct factors are 1 and −6.
Check using a Graphing calculator.
[– 10, 10] scl: 1 by [– 12, 8] scl: 1
2
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
Sum of 3
17
–17
7
–7
3
–3
The correct factors are –3 and 6.
Check using a Graphing calculator.
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Page 15
Study Guide and Review - Chapter 8
[– 10, 10] scl: 1 by [– 12, 8] scl: 1
2
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
Sum of 3
17
–17
7
–7
3
–3
The correct factors are –3 and 6.
Check using a Graphing calculator.
[– 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 solutions are –10 and 5.
Page 16
Study Guide and Review - Chapter 8
[– 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
The solutions are –10 and 5.
2
51. x − 6x + 8 = 0
SOLUTION: 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: eSolutions Manual - Powered by Cognero
Page 17
Study Guide and Review - Chapter 8
The solutions are 2 and 4.
2
52. x + 12x + 32 = 0
SOLUTION: 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
53. x − 2x − 48 = 0
SOLUTION: The roots are –6 and 8. Check by substituting –6 and 8 in for x in the original equation.
and
The solutions are –6 and 8.
2
54. x + 11x + 10 = 0
SOLUTION: eSolutions Manual - Powered by Cognero
Page 18
Study Guide and Review - Chapter 8
The solutions are –6 and 8.
2
54. x + 11x + 10 = 0
SOLUTION: The roots are –10 and –1. Check by substituting –10 and –1 in for x in the original equation.
and
The solutions are –10 and –1.
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.
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 .
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 –2, 84
82
2, –84
–82
–3, 56
53
3, –56
–53
–4,- 42
eSolutions Manual
Powered by Cognero 38
Page 19
4, –42
–38
–6, 28
22
Study Guide and Review - Chapter 8
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 .
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 –2, 84
82
2, –84
–82
–3, 56
53
3, –56
–53
–4, 42
38
4, –42
–38
–6, 28
22
The correct factors are –6 and 28.
2
So, 12x + 22x − 14 = 2(2x − 1)(3x + 7).
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
–1, 135
134
1, –135
–134
–3, 45
42
3, –45
–42
–5, 27
22
5, –27
–22
–9, 15
6
9, –15
–6
The correct factors are –15 and 9.
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Page 20
negative.
2(3) = 6
Guide and Review - Chapter 8
Study
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
–1, 135
134
1, –135
–134
–3, 45
42
3, –45
–42
–5, 27
22
5, –27
–22
–9, 15
6
9, –15
–6
The correct factors are –15 and 9.
2
So, 3x − 6x − 45 = 3(x − 5)(x + 3).
2
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 –1, 48
47
1, –48
–47
–2, 24
22
2, –24
–22
–3, 16
13
3, –16
–13
–4, 12
8
4, –12
–8
–6, 8
2
6, –8
–2
The correct factors are –3 and 16.
2
So, 2a + 13a − 24 = (2a − 3)(a + 8).
eSolutions Manual - Powered by Cognero
Solve each equation. Confirm your answers using a graphing calculator.
2
60. 40x + 2x = 24
Page 21
Study Guide and Review - Chapter 8
2
So, 3x − 6x − 45 = 3(x − 5)(x + 3).
2
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 –1, 48
47
1, –48
–47
–2, 24
22
2, –24
–22
–3, 16
13
3, –16
–13
–4, 12
8
4, –12
–8
–6, 8
2
6, –8
–2
The correct factors are –3 and 16.
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: The roots are
and or –0.80 and 0.75 . 2
Confirm the roots using a graphing calculator. Let Y1 = 40x + 2x and Y2 = –24. Use the intersect option from the
CALC menu to find the points of intersection. eSolutions Manual - Powered by Cognero
Page 22
Study Guide and Review - Chapter 8
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: The roots are
and or –0.80 and 0.75 . 2
Confirm the roots using a graphing calculator. Let Y1 = 40x + 2x and Y2 = –24. Use the intersect option from the
CALC menu to find the points of intersection. [–5, 5] scl: 1 by [–5, 25] scl: 3
The solutions are
and [–5, 5] scl: 1 by [–5, 25] scl: 3
.
2
61. 2x − 3x − 20 = 0
SOLUTION: The Manual
roots are
eSolutions
- Powered or
by −2.5 and 4. Cognero
Page 23
2
Confirm the roots using a graphing calculator. Let Y1 = 2x − 3x − 20 and Y2 = 0. Use the intersect option from
[–5, 5] scl: 1 by [–5, 25] scl: 3
[–5, 5] scl: 1 by [–5, 25] scl: 3
The
solutions
.
Study
Guide
and are
Review and - Chapter
8
2
61. 2x − 3x − 20 = 0
SOLUTION: The roots are
or −2.5 and 4. 2
Confirm the roots using a graphing calculator. Let Y1 = 2x − 3x − 20 and Y2 = 0. Use the intersect option from
the CALC menu to find the points of intersection. [–10, 10] scl: 1 by [–15, 5] scl: 1
The solutions are
[–10, 10] scl: 1 by [–15, 5] scl: 1
and 4.
2
62. −16t + 36t − 8 = 0
SOLUTION: The roots are 2 and
. 2
Page 24
calculator. Let Y1 = −16t + 36t − 8 and Y2 = 0. Use the intersect option from
the CALC menu to find the points of intersection. eSolutions
Manual
- Powered
by Cognero
Confirm
the
roots using
a graphing
[–10, 10] scl: 1 by [–15, 5] scl: 1
Study
Guide
and are
Review and 4.
- Chapter 8
The
solutions
[–10, 10] scl: 1 by [–15, 5] scl: 1
2
62. −16t + 36t − 8 = 0
SOLUTION: The roots are 2 and
. 2
Confirm the roots using a graphing calculator. Let Y1 = −16t + 36t − 8 and Y2 = 0. Use the intersect option from
the CALC menu to find the points of intersection. [–2, 3] scl: 1 by [–20, 10] scl: 6
The solutions are 2 and
[–2, 3] scl: 1 by [–20, 10] scl: 6
.
2
63. 6x − 7x − 5 = 0
SOLUTION: The roots are
and or −0.5 and 1.67. eSolutions Manual - Powered by Cognero
2
Page 25
Confirm the roots using a graphing calculator. Let Y1 = 6x − 7x − 5 and Y2 = 0. Use the intersect option from the
CALC menu to find the points of intersection. [–2, 3] scl: 1 by [–20, 10] scl: 6
Study
Guide
and are
Review
8
The
solutions
2 and- Chapter
.
[–2, 3] scl: 1 by [–20, 10] scl: 6
2
63. 6x − 7x − 5 = 0
SOLUTION: The roots are
and or −0.5 and 1.67. 2
Confirm the roots using a graphing calculator. Let Y1 = 6x − 7x − 5 and Y2 = 0. Use the intersect option from the
CALC menu to find the points of intersection. [–5, 5] scl: 0.5 by [–10, 10] scl: 1
The solutions are
and [–5, 5] scl: 0.5 by [–10, 10] scl: 1
.
2
64. GEOMETRY The area of the rectangle shown is 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
2, –21
–19
11
–3, 14
3, –14
–11
1
–6, 7
6,
–7
–1
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The correct factors are –3 and 14.
[–5, 5] scl: 0.5 by [–10, 10] scl: 1 [–5, 5] scl: 0.5 by [–10, 10] scl: 1
Study
Guide
and are
Review and - Chapter
8
The
solutions
.
2
64. GEOMETRY The area of the rectangle shown is 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
2, –21
–19
11
–3, 14
3, –14
–11
1
–6, 7
6, –7
–1
The correct factors are –3 and 14.
2
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.
Factor each polynomial.
2
65. y − 81
SOLUTION: 66. 64 − 25x
2
SOLUTION: 2
67. 16a − 21b
2
SOLUTION: 2
2
The number 21 is not a perfect square. So, 16a − 21b is prime.
2
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68. 3x −
eSolutions
Manual
SOLUTION: Page 27
2
67. 16a − 21b
2
SOLUTION: Study Guide and Review - Chapter 8
2
2
The number 21 is not a perfect square. So, 16a − 21b is prime.
2
68. 3x − 3
SOLUTION: Solve each equation by factoring. Confirm your answers using a graphing calculator.
2
69. a − 25 = 0
SOLUTION: The roots are –5 and 5. Confirm the roots using a graphing calculator. Let Y1 = a2 – 25 and Y2 = 0. Use the intersect option from the
CALC menu to find the points of intersection.
Thus, the solutions are –5 and 5.
2
70. 9x − 25 = 0
SOLUTION: eSolutions Manual - Powered by Cognero
The roots are
and or about -1.667 and 1.667. Page 28
Study
Guide
Review
Thus,
the and
solutions
are -–5Chapter
and 5. 8
2
70. 9x − 25 = 0
SOLUTION: The roots are
and or about -1.667 and 1.667. Confirm the roots using a graphing calculator. Let Y1 = 9x 2 – 25 and Y2 = 0. Use the intersect option from the
CALC menu to find the points of intersection.
Thus, the solutions are
and .
2
71. 81 − y = 0
SOLUTION: eSolutions
- Powered
by Cognero
The Manual
roots are
-9 and 9. Page 29
Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the
Study
Guide
Review
Thus,
the and
solutions
are - Chapter
and 8.
2
71. 81 − y = 0
SOLUTION: The roots are -9 and 9. Confirm the roots using a graphing calculator. Let Y1 = 81 - y 2 and Y2 = 0. Use the intersect option from the
CALC menu to find the points of intersection.
[-10, 10] scl: 1 by [-10, 90] scl:10 [-10, 10] scl: 1 by [-10, 90] scl:10
Thus, the solutions are –9 and 9.
2
72. x − 5 = 20
SOLUTION: The roots are -5 and 5. Confirm the roots using a graphing calculator. Let Y1 = x 2 - 5 and Y2 = 20. Use the intersect option from the
CALC menu to find the points of intersection.\
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eSolutions
Page 30
[-10, 10] scl: 1 by [-10, 90] scl:10 [-10, 10] scl: 1 by [-10, 90] scl:10
Guide and Review - Chapter 8
Study
Thus, the solutions are –9 and 9.
2
72. x − 5 = 20
SOLUTION: The roots are -5 and 5. Confirm the roots using a graphing calculator. Let Y1 = x 2 - 5 and Y2 = 20. Use the intersect option from the
CALC menu to find the points of intersection.\
[-10, 10] scl:1 by [-10, 40] scl: 5 [-10, 10] scl:1 by [-10, 40] scl: 5
Thus, the solutions are –5 and 5.
73. EROSION A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t
2
seconds is given by the equation d = 16t . How long does it take the boulder to hit the water?
SOLUTION: 2
The distance the boulder falls is 64 feet. So, 64 = 16t .
The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.
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Factor each polynomial, if possible. If the polynomial cannot be factored write prime .
2
74. x + 12x + 36
Page 31
[-10, 10] scl:1 by [-10, 40] scl: 5 [-10, 10] scl:1 by [-10, 40] scl: 5
Study
Guide and Review - Chapter 8
Thus, the solutions are –5 and 5.
73. EROSION A boulder falls down a mountain into water 64 feet below. The distance d that the boulder falls in t
2
seconds is given by the equation d = 16t . How long does it take the boulder to hit the water?
SOLUTION: 2
The distance the boulder falls is 64 feet. So, 64 = 16t .
The roots are –2 and 2. The time cannot be negative. So, it takes 2 seconds for the boulder to hit the water.
Factor each polynomial, if possible. If the polynomial cannot be factored write prime .
2
74. x + 12x + 36
SOLUTION: 2
75. x + 5x + 25
SOLUTION: 2
There are no factors of 25 that have a sum of 5. So, x + 5x + 25 is prime.
2
76. 9y − 12y + 4
SOLUTION: 77. 4 − 28a + 49a
2
SOLUTION: 4
78. x − 1
SOLUTION: eSolutions Manual - Powered by Cognero
Page 32
SOLUTION: Study Guide and Review - Chapter 8
4
78. x − 1
SOLUTION: 4
79. x − 16x
2
SOLUTION: Solve each equation. Confirm your answers using a graphing calculator.
2
80. (x − 5) = 121
SOLUTION: The roots are –6 and 16. 2
Confirm the roots using a graphing calculator. Let Y1 = (x − 5) and Y2 = 121. Use the intersect option from the
CALC menu to find the points of intersection. [–20, 20] scl: 3 by [–5, 125] scl: 13
[–20, 20] scl: 3 by [–5, 125] scl: 13
The Manual
solutions
are –6byand
16.
eSolutions
- Powered
Cognero
2
81. 4c + 4c + 1 = 9
Page 33
Study Guide and Review - Chapter 8
Solve each equation. Confirm your answers using a graphing calculator.
2
80. (x − 5) = 121
SOLUTION: The roots are –6 and 16. 2
Confirm the roots using a graphing calculator. Let Y1 = (x − 5) and Y2 = 121. Use the intersect option from the
CALC menu to find the points of intersection. [–20, 20] scl: 3 by [–5, 125] scl: 13
[–20, 20] scl: 3 by [–5, 125] scl: 13
The solutions are –6 and 16.
2
81. 4c + 4c + 1 = 9
SOLUTION: The roots are –2 and 1. eSolutions Manual - Powered by Cognero
2
Page 34
Confirm the roots using a graphing calculator. Let Y1 = 4c + 4c + 1 and Y2 = 9. Use the intersect option from the
CALC menu to find the points of intersection. [–20, 20] scl: 3 by [–5, 125] scl: 13
[–20, 20] scl: 3 by [–5, 125] scl: 13
Study
Guide
and are
Review
- Chapter
8
The
solutions
16.
–6 and
2
81. 4c + 4c + 1 = 9
SOLUTION: The roots are –2 and 1. 2
Confirm the roots using a graphing calculator. Let Y1 = 4c + 4c + 1 and Y2 = 9. Use the intersect option from the
CALC menu to find the points of intersection. [–5, 5] scl: 1 by [–5, 15] scl: 2
[–5, 5] scl: 1 by [–5, 15] scl: 2
Thus, the solutions are –2 and 1.
2
82. 4y = 64
SOLUTION: The roots are –4 and 4. eSolutions Manual - Powered by Cognero
2
Page 35
Confirm the roots using a graphing calculator. Let Y1 = 4y and Y2 = 64. Use the intersect option from the CALC
menu to find the points of intersection. [–5, 5] scl: 1 by [–5, 15] scl: 2
Guide and Review - Chapter 8
Study
Thus, the solutions are –2 and 1.
[–5, 5] scl: 1 by [–5, 15] scl: 2
2
82. 4y = 64
SOLUTION: The roots are –4 and 4. 2
Confirm the roots using a graphing calculator. Let Y1 = 4y and Y2 = 64. Use the intersect option from the CALC
menu to find the points of intersection. [–10, 10] scl: 1 by [–5, 75] scl: 10 [–10, 10] scl: 1 by [–5, 75] scl: 10
Thus, the solutions are –4 and 4.
2
83. 16d + 40d + 25 = 9
SOLUTION: eSolutions Manual - Powered by Cognero
The roots are –2 and
. Page 36
[–10, 10] scl: 1 by [–5, 75] scl: 10 [–10, 10] scl: 1 by [–5, 75] scl: 10
Guide and Review - Chapter 8
Study
Thus, the solutions are –4 and 4.
2
83. 16d + 40d + 25 = 9
SOLUTION: The roots are –2 and
. 2
Confirm the roots using a graphing calculator. Let Y1 = 16d + 40d + 25 and Y2 = 9. Use the intersect option from
the CALC menu to find the points of intersection. [–5, 5] scl: 0.5 by [–5, 15] scl: 5
[–5, 5] scl: 0.5 by [–5, 15] scl: 5
Thus, the solutions are –2 and
.
84. LANDSCAPING A sidewalk of equal width is being built around a square yard. What is the width of the
sidewalk?
SOLUTION: Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the
width
and length
arebythe
same.
eSolutions
Manual
- Powered
Cognero
Page 37
[–5, 5] scl: 0.5 by [–5, 15] scl: 5
[–5, 5] scl: 0.5 by [–5, 15] scl: 5
Study
Guide
Review
Thus,
the and
solutions
are -–2Chapter
and
.8
84. LANDSCAPING A sidewalk of equal width is being built around a square yard. What is the width of the
sidewalk?
SOLUTION: Let x = width of the sidewalk. Then, 2x + 25 = the width of the sidewalk and yard. Because the yard is square, the
width and length are the same.
The roots are –27.5 and 2.5. The width of the sidewalk cannot be negative. So, the width of the sidewalk is 2.5 feet.
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Page 38