9 ft
12 ft
16 ft
25 ft
yz
Chapter 5 Summary and Review: Vocabulary Reinforcement 117
32. Let s= the length of a side of the square, in ft.
34. Solve: 0 = 48t−16t2
it is launched.
36. Solve: x2+(x+2)
2= 130
42. 1.2+(−2)3+3.4=1.2−8+3.4
44. 2{x−3[4 −(x−1)] + x}=2{x−3[4 −x+1]+x}
46. From the drawing in the text we see that the length of each
Since the width of the roof cannot be negative, we use
we round up, finding that 10 squares would be needed.
50. We add labels to the drawing in the text.
(y+ 48)(y−48) = 0
tract to find z.
1521 = x2
Chapter 5 Vocabulary Reinforcement
1. To factor a polynomial is to express it as a product.
118 Chapter 5: Polynomials: Factoring
5. When factoring a polynomial with four terms, try factoring
by grouping.
Chapter 5 Concept Reinforcement
Chapter 5 Study Guide
1. 8x3y2=2·2·2·x3·y2
powers of xand of yare xand y, respectively, because 1
2. 27x5−9x3+18x2=9x2·3x3−9x2·x+9x2·2
4. x2+6x+8
5. 6z2−21z−12
Now we factor 2z2−7z−4.
(2) We can factor 2z2as
(4) Look for combinations of factors from steps (2) and
(3) such that the sum of their products is the middle
6. 6y2+7y−3
(1) There is no common factor (other than 1 or −1).
6y2+7y−3=6y2+9y−2y−3
9. 27 −125x3=3
3−(5x)3
=(3−5x)(9+15x+25x2)
10. 1
11. x2+4x=5
(x+ 5)(x−1)=0
Chapter 5 Review Exercises
Each coefficient has a factor of 5. There are no other com-
2. 12x3=2·2·3·x3
−60x2y=−1·2·2·3·5·x2·y
3. 5−20x6
Chapter 5 Summary and Review: Review Exercises 119
4. x2−3x=x(x−3)
7. x2+14x+49=x2+2·x·7+7
2=(x+7)
2
10. 6x2−5x+1
There is no common factor (other than 1). This poly-
=(3x−1)(2x−1)
14. x4+4x3−2x−8=(x4+4x3)+(−2x−8)
=x3(x+4)−2(x+4)
17. 75+12x2+60x=12x2+60x+75
19. x3−x2−30x=x(x2−x−30) = x(x−6)(x+5)
24. 2x2−7x−4=(2x+ 1)(x−4)
=(7b5−2a4)2
30. x2y2+xy −12=(xy + 4)(xy −3)
34. 5y3+40t3=5(y3+8t3)
The solutions are 1 and −3.
36. x2+2x−35=0
120 Chapter 5: Polynomials: Factoring
38. 3x2+2=5x
3and 1.
39. x2=64
40. 16 = x(x−6)
16 = x2−6x
0=x2−6x−16
41. Let y= 0 and solve for x.
42. Let y= 0 and solve for x.
x=−3
2or x =5
43. Familiarize. Let b= the length of the base, in cm. Then
b+ 1 = the height.
Solve.
0=(b+ 6)(b−5)
44. Familiarize. Let x= the smaller integer. Then x+2 is
the other integer.
Solve.
x(x+ 2) = 288
If x=−18, then x+2=−18+2=−16.
45. Familiarize. Let z= the length of the zipline, in ft. The
Translate. We use the Pythagorean theorem.
a2+b2=c2
Chapter 5 Summary and Review: Review Exercises 121
Check. The length must be positive, so −842 cannot be
46. Familiarize. We make a drawing. Let d= the distance
d
5
Solve.
2+d2=5
2
Check. A distance cannot be negative, so we check only
47. Familiarize. Let s= the length of a side of the original
Area of enlarged square
is 81 km2
.
↓↓↓
s2+6s−72=0
Check. The length of a side of the square cannot be nega-
State. The length of a side of the original square is 6 km.
50. Familiarize. Let w= the width of the margins, in cm.
2·20 ·15
Solve.
2w−5=0 or w −15=0
15 −2(2.5) = 15 −5 = 10. Thus the printed area is 15 ·10,
51. Familiarize. Let n= the number.
Translate.
Solve.
n3=2n2
122 Chapter 5: Polynomials: Factoring
0=2·0. If n= 2, then n3=2
3=8,n2=2
2= 4, and
52. Familiarize. Let w= the width of the original rectangle,
Solve.
2·6 = 12, 2w+ 20 = 12 + 20 = 32 and w−1=6−1=5.
The area of a rectangle with dimensions 32 by 5 is 32 ·5,
or 160, so the answer checks.
State. The length of the original rectangle is 12 in. and
53. Familiarize. First we can use the Pythagorean theorem
to find x, in ft. Then the height of the telephone pole is
a2+b2=c2
x2+4x+4+4x2= 4624 Multiplying by 4
5x2+4x+ 4 = 4624
2x+1 is 1
2·30+1, or
54. x2+25=0
55. (x−2)(x+ 3)(2x−5) = 0
56. (x−3)4x2+3x(x−3)−(x−3)10 = 0
Chapter 5 Discussion and Writing Exercises
1. Although x3−8x2+15xcan be factored as (x2−5x)(x−3),
look for a common factor first.
3. For x=−3:
4. The equation is not in the form ab = 0. The correct pro-
4 ft
3 ft
Chapter 5 Test 123
5. One solution of the equation is 0. Dividing both sides of
6. She could use the measuring sticks to draw a right angle as
Chapter 5 Test
2. x2−7x+10
3. x2+25−10x=x2−10x+25
4. 6y2−8y3+4y4=4y4−8y3+6y2=
5. x3+x2+2x+2
7. x3+2x2−3x
8. 28x−48+10x2
9. 4x2−9=(2x)2−32Difference of squares
10. x2−x−12
11. 6m3+9m2+3m
12. 3w2−75=3(w2−25) 3 is a common factor.
13. 60x+45x2+20
=45x2+60x+20
= 5(9x2+12x+ 4) 5 is a common factor.
14. 3x4−48
=3(x2+ 4)(x2−22) Difference of squares
124 Chapter 5: Polynomials: Factoring
16. 5x2−26x+5
There is no common factor (other than 1). This polyno-
middle term and factor by grouping:
5x2−26x+5=5x2−x−25x+5
18. 80 −5x4
= 5(16 −x4) 5 is a common factor.
19. 6t3+9t2−15t
20. 4x2−4x−15
21. 3m2−9mn −30n2
24. 2x2=32
2x2−32=0
2(x2−16)=0
25. x2−x−20=0
(x−5)(x+4)=0
26. 2x2+7x=15
2x2+7x−15=0
x2−3x=28
x2−3x−28=0
28. We let y= 0 and solve for x.
x+5=0 or x −7=0
29. We let y= 0 and solve for x.
3x−2=0 or x −1=0
Translate. We use the formula for the area of a rectangle.
48 = (w+2)w
Solve.
Chapter 5 Test 125
w+8=0 or w −6=0
31. Familiarize. Using the labels on the drawing in the
Translate. We use the formula for the area of a triangle.
32. Familiarize. Using the labels on the drawing in the text,
we let x= the distance between the two marked points, in
25 = x2
0=x2−25
33. 2y4−32 = 2(y4−16)
in meters. Then 5w= the length. The new width and
length are w+ 2 and 5w−3, respectively.
Solve.
35. (a+3)
2−2(a+3)−35
We can think of a+ 3 as the variable in this expression.
36. 20x(x+ 2)(x−1) = 5x3−24x−14x2
(20x2+40x)(x−1) = 5x3−24x−14x2
37. x2−y2=(x+y)(x−y)=4·6 = 24, so choice D is correct.
126 Chapter 5: Polynomials: Factoring
Cumulative Review Chapters 1 – 5
1. 2
3=2
3·7
7=14
21
5
3. 2.06 + (−4.79) −(−3.08)=2.06+(−4.79) + 3.08
4. We do the long division 5.652 ÷3.6 and make the answer
negative.
1.57
5. 2
9−3
86
7=−2·3·6
9·8·7=−2·3·2·3·1
3·3·2·2·2·7
=−6
5
7. [3x+2(x−1)] −[2x−(x+ 3)]
8. 1−[14+28÷7−(6+9÷3)]
=1−[14+4−(6 + 3)]
=1−[14+4−9]
=1−[9]
=−8
11. (2x2−3x3+x−4)+(x4−x−5x2)
12. (2x2y2+xy −2xy2)−(2xy −2xy2+x2y)
13.
x2+3x+2
x−1x3+2x2−x+1
14. (2t−3)2=(2t)2−2·2t·3+3
2=4t2−12t+9
18. 2y2+3y+4
2y−1
−2y2−3y−4
Cumulative Review Chapters 1 – 5 127
20. x2+2x−8
21. 4x2−25
22. 3x3−4x2+3x−4
23. x2−26x+ 169
24. 75x2−108y2
= 3(25x2−36y2) Factoring out the common
25. 6x2−13x−63
We will use the FOIL method.
(6x+)(x+ ) or (3x+ )(2x+).
−63, 1 and 21, −3 and −21, 3 and 9, −7 and −9, 7.
The factorization is (3x+ 7)(2x−9).
26. x4−2x2−3
28. 6p2+pq −q2
(2) Multiply 6 and −q2:6(−q2)=−6q2
(5) Factor by grouping.
29. 10x3+52x2+10x
10x3+52x2+10x=2x(5x2+26x+5)
(5x+)(x+).
as 5, 1.
(4) Look for combinations of factors from steps (2) and
30. 49x3−42x2+9x
31. 3x2+5x−4
32. 75x3+27x=3x(25x2+9)
33. 3x8−48y8
=3(x8−16y8) Difference of squares
=3(x4+4y4)(x4−4y4) Difference of squares
35. 2x5−2x3+x2−1=2x3(x2−1)+(x2−1)
36. 3x−5=2x+10
128 Chapter 5: Polynomials: Factoring
37. 3y+4>5y−8
x−15=0 or x +1
40. x3+x2=25x+25
x=−1or x =−5or x =5
The solutions are −1, −5 and 5.
41. 2x2=72
x+6=0 or x −6=0
43. x2+17x+70=0
(x+ 7)(x+10)=0
44. 14y2=21y
45. 1.6−3.5x=0.9
x2−x−12=8
x2−x−20=0
47. 1.5x−3.6≤1.3x+0.4
10(1.5x−3.6) ≤10(1.3x+0.4) Clearing decimals
15x−36 ≤13x+4
2x≤40 Subtracting 13xand
48. 2x−[3x−(2x+3)]=3x+[4−(2x+ 1)]
50. Familiarize. Let xand x+ 2 represent the integers.
Translate.
Cumulative Review Chapters 1 – 5 129
51. Familiarize. Let xand x+ 2 represent the integers.
Translate.
The product of the integers
is 360.
Check.−20 and −18 are consecutive even integers and
−20(−18) = 360. Also, 18 and 20 are consecutive even
52. Familiarize. Let h= the height of the window, in feet.
Then h+ 3 = the length.
Check. Since the height of the window cannot be nega-
53. Familiarize. Let w= the width of the lot, in meters.
State. The lot is 350 m by 150 m.
54. Familiarize. Let b= the amount that was borrowed. The
simple interest after 1 year is 12% ·b,or0.12b.
55. Familiarize. Let h= the length of the hypotenuse, in
substituting 15 for a,h−9 for b, and hfor c.
152+(h−9)2=h2
56. Familiarize. Let x= the length of the first piece of wire,
piece
plus Second
piece
plus Third
piece
is Total
length.
↓↓↓↓↓↓↓
130 Chapter 5: Polynomials: Factoring
Solve.
57. Familiarize. Let p= the price of the shoes before the
State. Before the reduction, the price of the shoes was
$29.
58. Familiarize. Let x= the width of the base, in ft. Then
98 = 1
2·x·x
59. 3x+4y=−12
12+4y=−12
60. (x+ 3)(x−5) ≤(x+ 2)(x−1)
x2−2x−15 ≤x2+x−2
−3x−15 ≤−2 Subtracting x2and x
61. x−3
2−2x+5
26 =4x+11
13 ,LCD is 26
Cumulative Review Chapters 1 – 5 131
62. (x+1)
2=25
63. x2(x−3) −x(x−3) −2(x−3)
64. 4a2−4a+1−9b2−24b−16
65. cx2−40x+ 16 will be the square of a binomial if it is of
66. Familiarize. Let r= the radius of the original circle, in
cm. Then r+ 2 = the radius of the new circle.
Translate.