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5: Factoring, Solving Equations, and Problem Solving

# 5: Factoring, Solving Equations, and Problem Solving

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256

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Likewise, 25x2 Ϫ 40xy ϩ 16y2 is a perfect square trinomial for these reasons:

1. The first term is a square: (5x) 2.

2. The last term is a square: (4y) 2.

3. The middle term is twice the product of the quantities being squared in the first and

last terms: 2(5x)(4y).

Once we know that we have a perfect square trinomial, the factoring process follows immediately from the two basic patterns.

9x2 ϩ 30x ϩ 25 ϭ (3x ϩ 5) 2

25x2 Ϫ 40xy ϩ 16y2 ϭ (5x Ϫ 4y) 2

Here are some additional examples of perfect square trinomials and their factored

form.

x2 Ϫ 16x ϩ 64 ϭ

16x2 Ϫ 56x ϩ 49 ϭ

25x2 ϩ 20xy ϩ 4y2 ϭ

1 ϩ 6y ϩ 9y2 ϭ

4m2 Ϫ 4mn ϩ n2 ϭ

(x) 2 Ϫ 2(x) (8) ϩ (8) 2

(4x) 2 Ϫ 2(4x) (7) ϩ (7) 2

(5x) 2 ϩ 2(5x) (2y) ϩ (2y) 2

(1) 2 ϩ 2(1) (3y) ϩ (3y) 2

(2m) 2 Ϫ 2(2m)(n) ϩ (n) 2

ϭ

ϭ

ϭ

ϭ

ϭ

(x Ϫ 8) 2

(4x Ϫ 7) 2

(5x ϩ 2y) 2

(1 ϩ 3y) 2

(2m Ϫ n) 2

You may want to do this step mentally after

you feel comfortable with the process

We have considered some basic factoring techniques in this chapter one at a time, but you

must be able to apply them as needed in a variety of situations. So, let’s first summarize the

techniques and then consider some examples.

In this chapter we have discussed these techniques:

1. Factoring by using the distributive property to factor out the greatest common monomial or binomial factor

2. Factoring by grouping

3. Factoring by applying the difference-of-squares pattern

4. Factoring by applying the perfect-square-trinomial pattern

5. Factoring trinomials of the form x2 ϩ bx ϩ c into the product of two binomials

6. Factoring trinomials of the form ax2 ϩ bx ϩ c into the product of two binomials

As a general guideline, always look for a greatest common monomial factor first, and then

proceed with the other factoring techniques.

In each of the following examples we have factored completely whenever possible. Study

them carefully, and notice the factoring techniques we used.

1. 2x2 ϩ 12x ϩ 10 ϭ 2(x2 ϩ 6x ϩ 5) ϭ 2(x ϩ 1)(x ϩ 5)

2. 4x2 ϩ 36 ϭ 4(x2 ϩ 9)

Remember that the sum of two squares is not factorable

using integers unless there is a common factor

3. 4t2 ϩ 20t ϩ 25 ϭ (2t ϩ 5) 2

If you fail to recognize a perfect trinomial square, no

harm is done; simply proceed to factor into the product

of two binomials, and then you will recognize that the

two binomials are the same

6.5 • Factoring, Solving Equations, and Problem Solving

257

4. x2 Ϫ 3x Ϫ 8 is not factorable using integers. This becomes obvious from the table.

Product

1(Ϫ8)

Ϫ1(8)

2(Ϫ4)

Ϫ2(4)

Sum

ϭ Ϫ8   1 ϩ (Ϫ8) ϭ Ϫ7

ϭ Ϫ8

Ϫ1 ϩ 8 ϭ 7

ϭ Ϫ8 2 ϩ (Ϫ4) ϭ Ϫ2

ϭ Ϫ8

Ϫ2 ϩ 4 ϭ 2

No two factors of Ϫ8 produce a sum of Ϫ3.

5. 6y2 Ϫ 13y Ϫ 28 ϭ (2y Ϫ 7)(3y ϩ 4) . We found the binomial factors as follows:

(y ϩ _____)(6y Ϫ _____)

or

(y Ϫ _____)(6y ϩ _____)

or

(2y Ϫ _____)(3y ϩ _____)

or

(2y ϩ _____)(3y Ϫ _____)

Factors of 28

1 и 28

2 и 14

4и7

or

or

or

28 и 1

14 и 2

7и4

6. 32x2 Ϫ 50y2 ϭ 2(16x2 Ϫ 25y2 ) ϭ 2(4x ϩ 5y)(4x Ϫ 5y)

Solving Equations by Factoring

As stated in the preface, there is a common thread that runs throughout this text: namely, learn

a skill; next, use the skill to help solve equations; and then use equations to help solve application problems. This thread becomes evident in this chapter. After presenting a factoring

technique, we immediately solved some equations using this technique, and then considered

some applications involving such equations. The following steps summarize the equation

solving process in this chapter.

1. Organize all terms of the polynomial on the same side of the equation with zero on

the other side.

2. Factor the polynomial. This will involve a variety of factoring techniques presented

in this chapter.

3. Set each factor equal to zero and solve for the unknown.

4. Check your solutions back into the original equation.

Let’s consider some examples.

Classroom Example

Solve x2 ϭ 64x.

EXAMPLE 1

Solve x2 ϭ 25x.

Solution

x2 ϭ 25x

x Ϫ 25x ϭ 0

x(x Ϫ 25) ϭ 0

x ϭ 0       or        x Ϫ 25 ϭ 0

x ϭ 25

x ϭ 0       or

2

The solution set is {0, 25}. Check it!

258

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Classroom Example

Solve m3 Ϫ 144m ϭ 0.

Solve x3 Ϫ 36x ϭ 0.

EXAMPLE 2

Solution

x3 Ϫ 36x ϭ 0

x(x2 Ϫ 36) ϭ 0

x(x ϩ 6)(x Ϫ 6) ϭ 0

x ϭ 0 or x ϩ 6 ϭ 0  or  x Ϫ 6 ϭ 0 If abc ϭ 0, then a ϭ 0 or b ϭ 0 or c ϭ 0

xϭ6

x ϭ 0 or x ϭ Ϫ6   or

The solution set is {Ϫ6, 0, 6}. Does it check?

Classroom Example

Solve 8x2 ϩ 10x Ϫ 7 ϭ 0.

Solve 10x2 Ϫ 13x Ϫ 3 ϭ 0.

EXAMPLE 3

Solution

10x2 Ϫ 13x Ϫ 3 ϭ 0

(5x ϩ 1)(2x Ϫ 3) ϭ 0

5x ϩ 1 ϭ 0   or  2x Ϫ 3 ϭ 0

5x ϭ Ϫ1  or

2x ϭ 3

1

3

x ϭ Ϫ   or

5

2

1 3

The solution set is eϪ , f . Does it check?

5 2

Classroom Example

Solve 9x2 Ϫ 24x ϩ 16 ϭ 0.

Solve 4x2 Ϫ 28x ϩ 49 ϭ 0.

EXAMPLE 4

Solution

4x2 Ϫ 28x ϩ 49 ϭ 0

(2x Ϫ 7) 2 ϭ 0

(2x Ϫ 7)(2x Ϫ 7) ϭ 0

2x Ϫ 7 ϭ 0  or  2x Ϫ 7 ϭ 0

2x ϭ 7

2x ϭ 7  or

7

7

x ϭ   or

2

2

7

The solution set is e f .

2

Pay special attention to the next example. We need to change the form of the original

equation before we can apply the property ab ϭ 0 if and only if a ϭ 0 or b ϭ 0. A necessary

condition of this property is that an indicated product is set equal to zero.

Classroom Example

Solve (x ϩ 6)(x ϩ 3) ϭ 4.

Solve (x ϩ 1) (x ϩ 4) ϭ 40.

EXAMPLE 5

Solution

(x ϩ 1)(x ϩ 4)

x2 ϩ 5x ϩ 4

x2 ϩ 5x Ϫ 36

(x ϩ 9)(x Ϫ 4)

ϭ 40

ϭ 40

ϭ0

ϭ0

6.5 • Factoring, Solving Equations, and Problem Solving

259

x ϩ 9 ϭ 0   or   x Ϫ 4 ϭ 0

x ϭ Ϫ9  or

xϭ4

The solution set is {Ϫ9, 4}. Check it!

Classroom Example

Solve 3a2 Ϫ 6a Ϫ 45 ϭ 0.

EXAMPLE 6

Solve 2n2 ϩ 16n Ϫ 40 ϭ 0.

Solution

2n2 ϩ 16n Ϫ 40

2(n2 ϩ 8n Ϫ 20)

n2 ϩ 8n Ϫ 20

(n ϩ 10)(n Ϫ 2)

n ϩ 10 ϭ 0

ϭ0

ϭ0

1

ϭ 0 Multiplied both sides by

2

ϭ0

or  n Ϫ 2 ϭ 0

nϭ2

n ϭ Ϫ10  or

The solution set is {Ϫ10, 2}. Does it check?

Problem Solving

Reminder: Throughout this book we highlight the need to learn a skill, to use that skill to

help solve equations, and then to use equations to help solve problems. Our new factoring

skills have provided more ways of solving equations, which in turn gives us more power to

solve word problems. We conclude the chapter by solving a few more problems.

Classroom Example

Find two numbers whose product is

70 if one of the numbers is 1 less

than three times the other number.

EXAMPLE 7

Find two numbers whose product is 65 if one of the numbers is 3 more than twice the other

number.

Solution

Let n represent one of the numbers; then 2n ϩ 3 represents the other number. Since their product is 65, we can set up and solve the following equation:

n(2n ϩ 3) ϭ 65

2n ϩ 3n Ϫ 65 ϭ 0

(2n ϩ 13)(n Ϫ 5) ϭ 0

2n ϩ 13 ϭ 0   or  n Ϫ 5 ϭ 0

2n ϭ Ϫ13  or

nϭ5

13

n ϭ Ϫ   or

nϭ5

2

2

13

13

, then 2n ϩ 3 ϭ 2aϪ b ϩ 3 ϭ Ϫ10. If n ϭ 5, then 2n ϩ 3 ϭ 2(5) ϩ 3

2

2

13

ϭ 13. Thus the numbers are Ϫ and Ϫ10, or 5 and 13.

2

If n ϭ Ϫ

Classroom Example

The area of a rectangular poster is

48 square centimeters. The length is

8 centimeters longer than the width

of the poster. Find the length and

width of the rectangular poster.

EXAMPLE 8

The area of a triangular sheet of paper is 14 square inches. One side of the triangle is 3 inches longer than the altitude to that side. Find the length of the one side and the length of the

altitude to that side.

260

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Solution

Let h represent the altitude to the side. Then h ϩ 3 represents the side of the triangle (see

Figure 6.5).

h

h+3

Figure 6.5

1

Since the formula for finding the area of a triangle is A ϭ bh, we have

2

1

h(h ϩ 3) ϭ 14

2

h(h ϩ 3) ϭ 28 Multiplied both sides by 2

h2 ϩ 3h ϭ 28

h2 ϩ 3h Ϫ 28 ϭ 0

(h ϩ 7)(h Ϫ 4) ϭ 0

h ϩ 7 ϭ 0   or  h Ϫ 4 ϭ 0

hϭ4

h ϭ Ϫ7  or

The solution of Ϫ7 is not reasonable. Thus the altitude is 4 inches and the length of the side

to which that altitude is drawn is 7 inches.

Classroom Example

A photograph measures 14 inches

wide and 19 inches long. A strip of

uniform width is to be cut off from

both ends and both sides of the

photograph in order to reduce the

area of the photograph to 176 square

inches. Find the width of the strip.

EXAMPLE 9

A strip with a uniform width is shaded along both sides and both ends of a rectangular poster

with dimensions 12 inches by 16 inches. How wide is the strip if one-half of the area of the

Solution

Let x represent the width of the shaded strip of the poster in Figure 6.6. The area of the strip

1

is one-half of the area of the poster; therefore, it is (12)(16) ϭ 96 square inches. Furthermore,

2

we can represent the area of the strip around the poster by the words the area of the poster

minus the area of the unshaded portion.

x

x

H

MAT 16 − 2x

N

ART OSITIO12 − 2x

EXP

2010

x

16 inches

Figure 6.6

x

12 inches

6.5 • Factoring, Solving Equations, and Problem Solving

261

Thus we can set up and solve the following equation:

Area of poster Ϫ Area of unshaded portion ϭ Area of strip

16(12)

Ϫ

(16 Ϫ 2x)(12 Ϫ 2x)

ϭ

96

192 Ϫ (192 Ϫ 56x ϩ 4x 2 ) ϭ 96

192 Ϫ 192 ϩ 56x Ϫ 4x 2 ϭ 96

Ϫ4x 2 ϩ 56x Ϫ 96 ϭ 0

x 2 Ϫ 14x ϩ 24 ϭ 0

(x Ϫ 12)(x Ϫ 2) ϭ 0

x Ϫ 12 ϭ 0      or      x Ϫ 2 ϭ 0

x ϭ 12      or

x ϭ2

Obviously, the strip cannot be 12 inches wide because the total width of the poster is 12 inches.

Thus we must disregard the solution of 12 and conclude that the strip is 2 inches wide.

Concept Quiz 6.5

For Problems 1–7, match each factoring problem with the name of the type of pattern that

would be used to factor the problem.

1.

2.

3.

4.

5.

6.

7.

x2 ϩ 2xy ϩ y2

x2 Ϫ y2

ax ϩ ay ϩ bx ϩ by

x2 ϩ bx ϩ c

ax2 ϩ bx ϩ c

ax2 ϩ ax ϩ a

(a ϩ b)x ϩ (a ϩ b)y

A.

B.

C.

D.

E.

F.

G.

Trinomial with an x-squared coeffcient of one

Common binomial factor

Difference of two squares

Common factor

Factor by grouping

Perfect-square trinomial

Trinomial with an x-squared coefficient of not one

Problem Set 6.5

For Problems 1–12, factor each of the perfect square trinomials. (Objective 1)

1. x 2 ϩ 4x ϩ 4

2. x 2 ϩ 18x ϩ 81

3. x 2 Ϫ 10x ϩ 25

4. x 2 Ϫ 24x ϩ 144

5. 9n ϩ 12n ϩ 4

6. 25n ϩ 30n ϩ 9

7. 16a Ϫ 8a ϩ 1

8. 36a2 Ϫ 84a ϩ 49

2

2

2

9. 4 ϩ 36x ϩ 81x 2

10. 1 Ϫ 4x ϩ 4x 2

11. 16x 2 Ϫ 24xy ϩ 9y 2

12. 64x ϩ 16xy ϩ y

2

19. 3a2 Ϫ 7a Ϫ 4

20. a2 ϩ 7a Ϫ 30

21. 8x 2 ϩ 72

22. 3y 3 Ϫ 36y 2 ϩ 96y

23. 9x 2 ϩ 30x ϩ 25

24. 5x 2 Ϫ 5x Ϫ 6

25. 15x 2 ϩ 65x ϩ 70

27. 24x 2 ϩ 2x Ϫ 15

26. 4x 2 Ϫ 20xy ϩ 25y 2

28. 9x 2y Ϫ 27xy

29. xy ϩ 5y Ϫ 8x Ϫ 40

30. xy Ϫ 3y ϩ 9x Ϫ 27

31. 20x 2 ϩ 31xy Ϫ 7y 2

2

For Problems 13– 40, factor each polynomial completely.

Indicate any that are not factorable using integers. (Objective 2)

32. 2x 2 Ϫ xy Ϫ 36y 2

33. 24x 2 ϩ 18x Ϫ 81

34. 30x 2 ϩ 55x Ϫ 50

35. 12x 2 ϩ 6x ϩ 30

36. 24x 2 Ϫ 8x ϩ 32

37. 5x 4 Ϫ 80

13. 2x 2 ϩ 17x ϩ 8

15. 2x 3 Ϫ 72x

14. x 2 ϩ 19x

16. 30x 2 Ϫ x Ϫ 1

38. 3x 5 Ϫ 3x

17. n2 Ϫ 7n Ϫ 60

18. 4n3 Ϫ 100n

40. 4x 2 Ϫ 28xy ϩ 49y 2

39. x 2 ϩ 12xy ϩ 36y 2

262

Chapter 6 • Factoring, Solving Equations, and Problem Solving

For Problems 41– 70, solve each equation. (Objective 3)

41. 4x Ϫ 20x ϭ 0

42. Ϫ3x Ϫ 24x ϭ 0

43. x 2 Ϫ 9x Ϫ 36 ϭ 0

44. x 2 ϩ 8x Ϫ 20 ϭ 0

45. Ϫ2x3 ϩ 8x ϭ 0

46. 4x 3 Ϫ 36x ϭ 0

2

2

47. 6n2 Ϫ 29n Ϫ 22 ϭ 0

77. In an office building, a room contains 54 chairs.

The number of chairs per row is three less than twice the

number of rows. Find the number of rows and the number of chairs per row.

78. An apple orchard contains 85 trees. The number of trees

in each row is three less than four times the number of

rows. Find the number of rows and the number of trees

per row.

48. 30n2 Ϫ n Ϫ 1 ϭ 0

49. (3n Ϫ 1)(4n Ϫ 3) ϭ 0

50. (2n Ϫ 3)(7n ϩ 1) ϭ 0

51. (n Ϫ 2)(n ϩ 6) ϭ Ϫ15

52. (n ϩ 3)(n Ϫ 7) ϭ Ϫ25

53. 2x 2 ϭ 12x

54. Ϫ3x2 ϭ 15x

55. t3 Ϫ 2t2 Ϫ 24t ϭ 0

56. 2t3 Ϫ 16t2 Ϫ 18t ϭ 0

57. 12 Ϫ 40x ϩ 25x 2 ϭ 0

58. 12 Ϫ 7x Ϫ 12x 2 ϭ 0

79. Suppose that the combined area of two squares

is 360 square feet. Each side of the larger square is three

times as long as a side of the smaller square. How big is

each square?

80. The area of a rectangular slab of sidewalk is 45 square

feet. Its length is 3 feet more than four times its width.

Find the length and width of the slab.

81. The length of a rectangular sheet of paper is 1 centimeter

more than twice its width, and the area of the rectangle is

55 square centimeters. Find the length and width of the

rectangle.

59. n2 Ϫ 28n ϩ 192 ϭ 0

60. n2 ϩ 33n ϩ 270 ϭ 0

61. (3n ϩ 1)(n ϩ 2) ϭ 12

62. (2n ϩ 5)(n ϩ 4) ϭ Ϫ1

63. x 3 ϭ 6x 2

76. A number is one less than three times another num-ber. If

the product of the two numbers is 102, find the numbers.

64. x3 ϭ Ϫ 4x2

65. 9x 2 Ϫ 24x ϩ 16 ϭ 0

66. 25x 2 ϩ 60x ϩ 36 ϭ 0

67. x 3 ϩ 10x 2 ϩ 25x ϭ 0

68. x 3 Ϫ 18x 2 ϩ 81x ϭ 0

69. 24x 2 ϩ 17x Ϫ 20 ϭ 0

70. 24x 2 ϩ 74x Ϫ 35 ϭ 0

For Problems 71– 88, set up an equation and solve each

problem. (Objective 4)

71. Find two numbers whose product is 15 such that one

of the numbers is seven more than four times the other

number.

72. Find two numbers whose product is 12 such that one of

the numbers is four less than eight times the other

number.

73. Find two numbers whose product is Ϫ1. One of the

numbers is three more than twice the other number.

74. Suppose that the sum of the squares of three consecutive

integers is 110. Find the integers.

75. A number is one more than twice another number. The

sum of the squares of the two numbers is 97. Find the

numbers.

82. Suppose that the length of a certain rectangle is three

times its width. If the length is increased by 2 inches,

and the width increased by 1 inch, the newly formed

rectangle has an area of 70 square inches. Find the

length and width of the original rectangle.

83. The area of a triangle is 51 square inches. One side

of the triangle is 1 inch less than three times the length

of the altitude to that side. Find the length of that side

and the length of the altitude to that side.

84. Suppose that a square and a rectangle have equal areas.

Furthermore, suppose that the length of the rectangle is

twice the length of a side of the square, and the width of

the rectangle is 4 centimeters less than the length of a

side of the square. Find the dimensions of both figures.

85. A strip of uniform width is to be cut off of both sides

and both ends of a sheet of paper that is 8 inches by

11 inches, in order to reduce the size of the paper to an

area of 40 square inches. Find the width of the strip.

86. The sum of the areas of two circles is 100␲ square centimeters. The length of a radius of the larger circle is

2 centimeters more than the length of a radius of the

smaller circle. Find the length of a radius of each circle.

87. The sum of the areas of two circles is 180␲ square inches.

The length of a radius of the smaller circle is 6 inches

less than the length of a radius of the larger circle. Find

the length of a radius of each circle.

88. A strip of uniform width is shaded along both sides and

both ends of a rectangular poster that is 18 inches by

14 inches. How wide is the strip if the unshaded portion

of the poster has an area of 165 square inches?

6.5 • Factoring, Solving Equations, and Problem Solving

263

Thoughts Into Words

91. Explain how you would solve (x ϩ 2)(x ϩ 3) ϭ (x ϩ 2)

(3x Ϫ 1). Do you see more than one approach to this

problem?

89. When factoring polynomials, why do you think that it is

best to look for a greatest common monomial factor

first?

90. Explain how you would solve (4x Ϫ 3)(8x ϩ 5) ϭ 0 and

also how you would solve (4x Ϫ 3)(8x ϩ 5) ϭ Ϫ9.

1. F or A

2. C

3. E

4. A

5. G

6. D

7. B

Chapter 6 Summary

OBJECTIVE

SUMMARY

EXAMPLE

Find the greatest common

factor.

By “the greatest common factor of two or

more monomials” we mean the monomial

with the largest numerical coefficient and

the highest power of the variables, which is

a factor of each given monomial.

Find the greatest common factor of 12a3b3,

18a2b2, and 54ab4.

The distributive property in the form

ab ϩ ac ϭ a(b ϩ c) provides the basis for

factoring out a greatest common monomial

or binomial factor.

Factor 36x2 y ϩ 18x Ϫ 27xy.

(Section 6.1/Objective 1)

Factor out the greatest

common factor.

(Section 6.1/Objective 2)

Factor by grouping.

(Section 6.1/Objective 3)

Factor the difference of two

squares.

(Section 6.2/Objective 1)

Factor trinomials of the form

x2 ϩ bx ϩ c.

(Section 6.3/Objective 1)

Rewriting an expression such as

ab ϩ 3a ϩ bc ϩ 3c as a(b ϩ 3) ϩ c(b ϩ 3)

and then factoring out the common

binomial factor of b ϩ 3 so that

a(b ϩ 3) ϩ c(b ϩ 3) becomes (b ϩ 3)

(a ϩ c), is called factoring by grouping.

The factoring pattern, the difference of two

squares, is a2 Ϫ b2 ϭ (a Ϫ b)(a ϩ b). Be

careful not to apply the pattern to the sum

of two squares, such as a2 ϩ b2. There is

no factoring pattern for the sum of two

squares.

The following multiplication pattern

provides a technique for factoring trinomials

of the form x2 ϩ bx ϩ c.

(x ϩ r)(x ϩ s) ϭ x2 ϩ rx ϩ sx ϩ rs

ϭ x2 ϩ (r ϩ s)x ϩ rs

Solution

The largest numerical coefficient that is a

factor of all three terms is 6. The highest

exponent for a that is a factor of all three

terms is 1. The highest exponent for b that

is a factor of all three terms is 2. So the

greatest common factor is 6ab2.

Solution

36x2 y ϩ 18x Ϫ 27xy

ϭ 9x(4xy) ϩ 9x(2)Ϫ9x(3y)

ϭ 9x(4xy ϩ 2 Ϫ 3y)

Factor 15xy ϩ 6x ϩ 10y2 ϩ 4y.

Solution

15xy ϩ 6x ϩ 10y2 ϩ 4y

ϭ 3x(5y ϩ 2) ϩ 2y(5y ϩ 2)

ϭ (5y ϩ 2)(3x ϩ 2y)

Factor 4x2 Ϫ 81y2.

Solution

4x2 Ϫ 81y2 ϭ (2x ϩ 9y)(2x Ϫ 9y)

Factor x2 ϩ 2x Ϫ 35.

Solution

The factors of Ϫ35 that sum to 2 are 7

and Ϫ5.

x2 ϩ 2x Ϫ 35 ϭ (x ϩ 7)(x Ϫ 5)

For trinomials of the form x2 ϩ bx ϩ c, we

want two factors of c whose sum will be

equal to b.

Factor trinomials in which the

(Section 6.4/Objective 1)

We presented two different techniques

for factoring trinomials of the form

ax2 ϩ bx ϩ c. To review these techniques,

turn to Section 6.4 and study the

examples. The examples here show

the two techniques.

Factor:

(a) 3x2 ϩ 7x ϩ 4

(b) 2x2 Ϫ 5x ϩ 3

Solution

(a) To factor 3x2 ϩ 7x ϩ 4, we can look at

the first term and the sign situation to

(continued)

264

Chapter 6 • Summary

OBJECTIVE

SUMMARY

265

EXAMPLE

determine that the factors are of the form

(3x ϩ

) and (x ϩ

). By trial and

error we can arrive at the correct factors.

3x2 ϩ 7x ϩ 4 ϭ (3x ϩ 4)(x ϩ 1)

(b) 2x2 Ϫ 5x ϩ 3 sum of Ϫ5

product of 2(3) ϭ 6

So we want two numbers whose sum is

Ϫ5 and whose product is 6. The numbers

are Ϫ2 and Ϫ3. Therefore, we rewrite the

middle term as Ϫ2x Ϫ3x.

2x2 Ϫ 5x ϩ 3 ϭ 2x2 Ϫ 2x Ϫ 3x ϩ 3

ϭ 2x(x Ϫ 1) Ϫ 3(x Ϫ 1)

ϭ (x Ϫ 1)(2x Ϫ 3)

Factor perfect-square

trinomials.

(Section 6.5/Objective 1)

Recognize the different types

of factoring.

(Section 6.5/Objective 2)

Perfect-square trinomials are easy to

recognize because of the nature of their

terms. The first term and the last term will

be the squares of a quantity. The middle

term is twice the product of the quantities

being squared in the first and last terms.

Factor 16x2 ϩ 56x ϩ 49.

As a general guideline for factoring

completely, always look for a greatest

common factor first, and then proceed with

one or more of the following techniques.

Factor 3x2 ϩ 12xy ϩ 12y2.

1. Apply the difference-of-squares pattern.

Solution

16x2 ϩ 56x ϩ 49

ϭ (4x2 ) ϩ 2(4x)(7) ϩ (7)2

ϭ (4x ϩ 7)2

Solution

3x2 ϩ 12xy ϩ 12y2

ϭ 3(x2 ϩ 4xy ϩ 4y2)

ϭ 3(x ϩ 2y)2

2. Apply the perfect-square pattern.

3. Factor a trinomial of the form

x2 ϩ bx ϩ c into the product of two

binomials.

4. Factor a trinomial of the form

ax2 ϩ bx ϩ c into the product of two

binomials.

Use factoring to solve

equations.

(Section 6.1/Objective 4;

Section 6.2/Objective 2;

Section 6.3/Objective 2;

Section 6.4/Objective 2;

Section 6.5/Objective 3)

Property 6.1 states that for all real numbers

a and b, ab ϭ 0 if and only if a ϭ 0 or

b ϭ 0. To solve equations by applying this

property, first set the equation equal to zero.

Proceed by factoring the other side of the

equation. Then set each factor equal to 0

and solve the equations.

Solve 2x2 ϭ Ϫ10x.

Solution

2x2 ϭ Ϫ10x

2x2 ϩ 10x ϭ 0

2x(x ϩ 5) ϭ 0

2x ϭ 0

or

xϩ5ϭ0

xϭ0

or

x ϭ Ϫ5

The solution is {Ϫ5, 0}.

(continued)

266

Chapter 6 • Factoring, Solving Equations, and Problem Solving

OBJECTIVE

SUMMARY

EXAMPLE

Solve word problems that

involve factoring.

Knowledge of factoring has expanded

the techniques available for solving word

problems. This chapter introduced the

Pythagorean theorem. The theorem pertains

to right triangles and states that in any right

triangle, the square of the longest side is

equal to the sum of the squares of the other

two sides. The formula for the theorem is

written as a2 ϩ b2 ϭ c2, where a and b are

the legs of the triangle and c is the

hypotenuse.

The length of one leg of a right triangle is

1 inch more than the length of the other

leg. The length of the hypotenuse is

5 inches. Find the length of the two legs.

(Section 6.3/Objectives 3 and

4; Section 6.5/Objective 4)

Solution

Let x represent the length of one leg of

the triangle. Then x ϩ 1 will represent the

length of the other leg. We know the

hypotenuse is equal to 5. Apply

Pythagorean theorem.

x2 ϩ (x ϩ 1)2 ϭ 5

x2 ϩ x2 ϩ 2x ϩ 1 ϭ 25

2x2 ϩ 2x Ϫ 24 ϭ 0

2(x2 ϩ x Ϫ 12) ϭ 0

x2 ϩ x Ϫ 12 ϭ 0

(x ϩ 4)(x Ϫ 3) ϭ 0

xϩ4ϭ0

or

xϪ3ϭ0

xϭϪ4

or

xϭ3

Because the length of a side of the triangle cannot be negative, the only viable

answer is 3. Therefore, the length of one

leg of the triangle is 3 inches, and the

length of the other leg is 4 inches.

Chapter 6 Review Problem Set

For Problems 1– 24, factor completely. Indicate any polynomials that are not factorable using integers.

1. x Ϫ 9x ϩ 14

2. 3x ϩ 21x

3. 9x Ϫ 4

4. 4x2 ϩ 8x Ϫ 5

5. 25x Ϫ 60x ϩ 36

6. n ϩ 13n ϩ 40n

7. y ϩ 11y Ϫ 12

8. 3xy2 ϩ 6x2y

2

2

2

2

2

9. x Ϫ 1

3

2

10. 18n ϩ 9n Ϫ 5

4

2

11. x2 ϩ 7x ϩ 24

12. 4x2 Ϫ 3x Ϫ 7

13. 3n2 ϩ 3n Ϫ 90

14. x3 Ϫ xy2

15. 2x ϩ 3xy Ϫ 2y

2

2

16. 4n Ϫ 6n Ϫ 40

2

17. 5x ϩ 5y ϩ ax ϩ ay

27. 2x2 ϩ 3x Ϫ 20 ϭ 0

28. 9n2 ϩ 21n Ϫ 8 ϭ 0

29. 6n2 ϭ 24

30. 16y2 ϩ 40y ϩ 25 ϭ 0

31. t3 Ϫ t ϭ 0

32. 28x2 ϩ 71x ϩ 18 ϭ 0

33. x2 ϩ 3x Ϫ 28 ϭ 0

34. (x Ϫ 2)(x ϩ 2) ϭ 21

35. 5n2 ϩ 27n ϭ 18

36. 4n2 ϩ 10n ϭ 14

37. 2x3 Ϫ 8x ϭ 0

18. 21t Ϫ 5t Ϫ 4

19. 2x Ϫ 2x

38. x2 Ϫ 20x ϩ 96 ϭ 0

20. 3x Ϫ 108x

21. 16x ϩ 40x ϩ 25

39. 4t2 ϩ 17t Ϫ 15 ϭ 0

2

3

3

2

22. xy Ϫ 3x Ϫ 2y ϩ 6

23. 15x2 Ϫ 7xy Ϫ 2y2

40. 3(x ϩ 2) Ϫ x(x ϩ 2) ϭ 0

24. 6n4 Ϫ 5n3 ϩ n2

41. (2x Ϫ 5)(3x ϩ 7) ϭ 0

For Problems 25– 44, solve each equation.

42. (x ϩ 4)(x Ϫ 1) ϭ 50

25. x ϩ 4x Ϫ 12 ϭ 0

43. Ϫ7n Ϫ 2n2 ϭ Ϫ15

2

26. x ϭ 11x

2

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