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1: Factoring by Using the Distributive Property

6.1 • Factoring by Using the Distributive Property

231

The next four examples further illustrate this process of factoring out the greatest common

monomial factor.

Classroom Example

Factor 18x 4 Ϫ 24x 2.

EXAMPLE 3

Factor 12x3 Ϫ 8x2.

Solution

12x3 Ϫ 8x2 ϭ 4x2 (3x) Ϫ 4x2 (2)

ϭ 4x2 (3x Ϫ 2)

Classroom Example

Factor 15ab3 ϩ 27a2b.

EXAMPLE 4

ab Ϫ ac ϭ a(b Ϫ c)

Factor 12x2y ϩ 18x y2.

Solution

12x2y ϩ 18xy2 ϭ 6xy(2x) ϩ 6xy(3y)

ϭ 6xy(2x ϩ 3y)

Classroom Example

Factor 12k 2 Ϫ 33k 3 ϩ 51k 5.

EXAMPLE 5

Factor 24x3 ϩ 30x4 Ϫ 42x5.

Solution

24x3 ϩ 30x4 Ϫ 42x5 ϭ 6x3 (4) ϩ 6x3 (5x) Ϫ 6x3 (7x2 )

ϭ 6x3 (4 ϩ 5x Ϫ 7x2 )

Classroom Example

Factor 5y 4 ϩ 5y 3.

EXAMPLE 6

Factor 9x2 ϩ 9x.

Solution

9x2 ϩ 9x ϭ 9x(x) ϩ 9x(1)

ϭ 9x(x ϩ 1)

We want to emphasize the point made just before Example 3. It is important to realize

that we are factoring out the greatest common monomial factor. We could factor an expression such as 9x2 ϩ 9x in Example 6 as 9(x2 ϩ x) , 3(3x2 ϩ 3x) , 3x(3x ϩ 3) , or even

1

(18x2 ϩ 18x) , but it is the form 9x(x ϩ 1) that we want. We can accomplish this by factoring

2

out the greatest common monomial factor; we sometimes refer to this process as factoring

completely. A polynomial with integral coefficients is in completely factored form if these

conditions are met:

1. It is expressed as a product of polynomials with integral coefficients.

2. No polynomial, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients.

Thus 9(x2 ϩ x), 3(3x2 ϩ 3x), and 3x(3x ϩ 3) are not completely factored because they violate

1

condition 2. The form (18x2 ϩ 18x) violates both conditions 1 and 2.

2

Sometimes there may be a common binomial factor rather than a common monomial

factor. For example, each of the two terms of x(y ϩ 2) ϩ z(y ϩ 2) has a binomial factor of

(y ϩ 2). Thus we can factor (y ϩ 2) from each term and get

x(y ϩ 2) ϩ z(y ϩ 2) ϭ (y ϩ 2)(x ϩ z)

232

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Consider a few more examples involving a common binomial factor.

a(b ϩ c) Ϫ d(b ϩ c) ϭ (b ϩ c) (a Ϫ d)

x(x ϩ 2) ϩ 3(x ϩ 2) ϭ (x ϩ 2) (x ϩ 3)

x(x ϩ 5) Ϫ 4(x ϩ 5) ϭ (x ϩ 5) (x Ϫ 4)

It may be that the original polynomial exhibits no apparent common monomial or binomial factor, which is the case with

ab ϩ 3a ϩ bc ϩ 3c

However, by factoring a from the first two terms and c from the last two terms, we see that

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

Now a common binomial factor of (b ϩ 3) is obvious, and we can proceed as before.

a(b ϩ 3) ϩ c(b ϩ 3) ϭ (b ϩ 3)(a ϩ c)

This factoring process is called factoring by grouping. Let’s consider two more examples of

factoring by grouping.

x2 Ϫ x ϩ 5x Ϫ 5 ϭ x(x Ϫ 1) ϩ 5(x Ϫ 1)

ϭ (x Ϫ 1)(x ϩ 5)

Factor x from first two terms and 5 from

last two terms

Factor common binomial factor of

(x Ϫ 1) from both terms

6x2 Ϫ 4x Ϫ 3x ϩ 2 ϭ 2x(3x Ϫ 2) Ϫ 1(3x Ϫ 2) Factor 2x from first two terms and Ϫ1

from last two terms

ϭ (3x Ϫ 2) (2x Ϫ 1)

Factor common binomial factor of

(3x Ϫ 2) from both terms

Back to Solving Equations

Suppose we are told that the product of two numbers is 0. What do we know about the numbers? Do you agree with our conclusion that at least one of the numbers must be 0? The next

property formalizes this idea.

Property 6.1

For all real numbers a and b,

ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

Property 6.1 provides us with another technique for solving equations.

Classroom Example

Solve m2 Ϫ 8m ϭ 0.

EXAMPLE 7

Solve x2 ϩ 6x ϭ 0.

Solution

To solve equations by applying Property 6.1, one side of the equation must be a product, and

the other side of the equation must be zero. This equation already has zero on the right-hand

side of the equation, but the left-hand side of this equation is a sum. We will factor the left-hand

side, x 2 ϩ 6x, to change the sum into a product.

x2 ϩ 6x ϭ 0

x(x ϩ 6) ϭ 0

x ϭ 0 or x ϩ 6 ϭ 0

x ϭ Ϫ6

x ϭ 0 or

Factor

ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

The solution set is {Ϫ6, 0}. (Be sure to check both values in the original equation.)

6.1 • Factoring by Using the Distributive Property

Classroom Example

Solve x2 ϭ 17x.

EXAMPLE 8

233

Solve x2 ϭ 12x.

Solution

In order to solve this equation by Property 6.1, we will first get zero on the right-hand side of

the equation by adding Ϫ12x to each side. Then we factor the expression on the left-hand side

of the equation.

x2 ϭ 12x

x Ϫ 12x ϭ 0

x(x Ϫ 12) ϭ 0

x ϭ 0 or x Ϫ 12 ϭ 0

x ϭ 0 or x ϭ 12

2

Added Ϫ12x to both sides

ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

The solution set is {0, 12}.

Remark: Notice in Example 8 that we did not divide both sides of the original equation by x.

This would cause us to lose the solution of 0.

Classroom Example

Solve 5d 2 Ϫ 7d ϭ 0.

EXAMPLE 9

Solve 4x2 Ϫ 3x ϭ 0.

Solution

4x2 Ϫ 3x ϭ 0

x(4x Ϫ 3) ϭ 0

x ϭ 0 or 4x Ϫ 3 ϭ 0

x ϭ 0 or

4x ϭ 3

3

x ϭ 0 or

x ϭ

4

ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

3

The solution set is e0, f .

4

Classroom Example

Solve w(w Ϫ 8)ϩ11(w Ϫ 8) ϭ 0.

EXAMPLE 10

Solve x(x ϩ 2) ϩ 3(x ϩ 2) ϭ 0.

Solution

In order to solve this equation by Property 6.1, we will factor the left-hand side of the equation. The greatest common factor of the terms is (x + 2).

x(x ϩ 2) ϩ 3(x ϩ 2) ϭ 0

(x ϩ 2)(x ϩ 3) ϭ 0

x ϩ 2 ϭ 0 or x ϩ 3 ϭ 0

x ϭ Ϫ3

x ϭ Ϫ2 or

ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

The solution set is {Ϫ3, Ϫ2} .

Each time we expand our equation-solving capabilities, we also acquire more techniques for solving problems. Let’s solve a geometric problem with the ideas we learned in

this section.

234

Chapter 6 • Factoring, Solving Equations, and Problem Solving

EXAMPLE 11

Classroom Example

The area of a square is numerically

equal to three times its perimeter.

Find the length of a side of the

square.

The area of a square is numerically equal to twice its perimeter. Find the length of a side of

the square.

Solution

Sketch a square and let s represent the length of each side (see Figure 6.1). Then the area is

represented by s2 and the perimeter by 4s. Thus

s

s

s2 ϭ 2(4s)

s2 ϭ 8s

s2 Ϫ 8s ϭ 0

s(s Ϫ 8) ϭ 0

s ϭ 0 or s Ϫ 8 ϭ 0

s

s

Figure 6.1

s ϭ 0 or

sϭ8

Since 0 is not a reasonable answer to the problem, the solution is 8. (Be sure to check this

solution in the original statement of the problem!)

Concept Quiz 6.1

For Problems 1–10, answer true or false.

1. The greatest common factor of 6x2y3 Ϫ 12x3y2 ϩ 18x4y is 2x2y.

2. If the factored form of a polynomial can be factored further, then it has not met the conditions to be considered “factored completely.”

3. Common factors are always monomials.

4. If the product of x and y is zero, then x is zero or y is zero.

5. The factored form 3a(2a2 ϩ 4) is factored completely.

6. The solutions for the equation x(x ϩ 2) ϭ 7 are 7 and 5.

7. The solution set for x2 ϭ 7x is {7}.

8. The solution set for x(x Ϫ 2) Ϫ 3(x Ϫ 2) ϭ 0 is {2, 3}.

9. The solution set for Ϫ3x ϭ x2 is {Ϫ3, 0}.

10. The solution set for x(x ϩ 6) ϭ 2(x ϩ 6) is {Ϫ6}.

Problem Set 6.1

For Problems 1–10, find the greatest common factor of the

given expressions. (Objective 1)

For Problems 11– 46, factor each polynomial completely.

(Objective 2)

11. 8x ϩ 12y

12. 18x ϩ 24y

13. 14xy Ϫ 21y

14. 24x Ϫ 40xy

15. 18x ϩ 45x

16. 12x ϩ 28x3

7. 6x3, 8x, and 24x2

17. 12xy2 Ϫ 30x2y

18. 28x2y2 Ϫ 49x2y

8. 72xy, 36x2y, and 84xy2

19. 36a2b Ϫ 60a3b4

20. 65ab3 Ϫ 45a2b2

9. 16a2b2, 40a2b3, and 56a3b4

21. 16xy3 ϩ 25x2y2

22. 12x2y2 ϩ 29x2y

23. 64ab Ϫ 72cd

24. 45xy Ϫ 72zw

1. 24y and 30xy

2. 32x and 40xy

3. 60x2y and 84xy2

4. 72x3 and 63x2

3

2 2

5. 42ab and 70a b

10. 70a3b3, 42a2b4, and 49ab5

2 2

6. 48a b and 96ab

4

2

6.1 • Factoring by Using the Distributive Property

25. 9a2b4 Ϫ 27a2b

26. 7a3b5 Ϫ 42a2b6

63. x2 ϩ x ϭ 0

64. x2 ϩ 7x ϭ 0

27. 52x4y2 ϩ 60x6y

28. 70x5y3 Ϫ 42x8y2

65. n2 ϭ 5n

66. n2 ϭ Ϫ2n

29. 40x2y2 ϩ 8x2y

30. 84x2y3 ϩ 12xy3

67. 2y2 Ϫ 3y ϭ 0

68. 4y2 Ϫ 7y ϭ 0

31. 12x ϩ 15xy ϩ 21x2

69. 7x2 ϭ Ϫ3x

70. 5x2 ϭ Ϫ2x

32. 30x2y ϩ 40xy ϩ 55y

71. 3n2 ϩ 15n ϭ 0

72. 6n2 Ϫ 24n ϭ 0

33. 2x3 Ϫ 3x2 ϩ 4x

73. 4x2 ϭ 6x

74. 12x2 ϭ 8x

34. x4 ϩ x3 ϩ x2

75. 7x Ϫ x2 ϭ 0

76. 9x Ϫ x2 ϭ 0

77. 13x ϭ x2

78. 15x ϭ Ϫx2

79. 5x ϭ Ϫ2x2

80. 7x ϭ Ϫ5x2

35. 44y5 Ϫ 24y3 Ϫ 20y2

36. 14a Ϫ 18a Ϫ 26a

3

5

37. 14a b ϩ 35ab Ϫ 49a b

2 3

2

3

38. 24a3b2 ϩ 36a2b4 Ϫ 60a4b3

39. x(y ϩ 1) ϩ z(y ϩ 1)

235

81. x(x ϩ 5) Ϫ 4(x ϩ 5) ϭ 0

82. x(3x Ϫ 2) Ϫ 7(3x Ϫ 2) ϭ 0

83. 4(x Ϫ 6) Ϫ x(x Ϫ 6) ϭ 0

40. a(c ϩ d) ϩ 2(c ϩ d)

84. x(x ϩ 9) ϭ 2(x ϩ 9)

41. a(b Ϫ 4) Ϫ c(b Ϫ 4)

42. x(y Ϫ 6) Ϫ 3(y Ϫ 6)

For Problems 85–91, set up an equation and solve each problem. (Objective 4)

43. x(x ϩ 3) ϩ 6(x ϩ 3)

44. x(x Ϫ 7) ϩ 9(x Ϫ 7)

85. The square of a number equals nine times that number.

Find the number.

45. 2x(x ϩ 1) Ϫ 3(x ϩ 1)

46. 4x(x ϩ 8) Ϫ 5(x ϩ 8)

86. Suppose that four times the square of a number equals

20 times that number. What is the number?

For Problems 47–60, use the process of factoring by grouping to factor each polynomial. (Objective 3)

47. 5x ϩ 5y ϩ bx ϩ by

88. The area of a square is 14 times as large as the area of a

triangle. One side of the triangle is 7 inches long, and the

altitude to that side is the same length as a side of the

square. Find the length of a side of the square. Also find

the areas of both figures, and be sure that your answer

checks.

48. 7x ϩ 7y ϩ zx ϩ zy

49. bx Ϫ by Ϫ cx ϩ cy

50. 2x Ϫ 2y Ϫ ax ϩ ay

51. ac ϩ bc ϩ a ϩ b

52. x ϩ y ϩ ax ϩ ay

89. Suppose that the area of a circle is numerically equal to

the perimeter of a square, and that the length of a radius

of the circle is equal to the length of a side of the square.

Find the length of a side of the square. Express your answer in terms of .

53. x ϩ 5x ϩ 12x ϩ 60

2

54. x2 ϩ 3x ϩ 7x ϩ 21

55. x2 Ϫ 2x Ϫ 8x ϩ 16

56. x2 Ϫ 4x Ϫ 9x ϩ 36

90. One side of a parallelogram, an altitude to that side, and

one side of a rectangle all have the same measure. If an

adjacent side of the rectangle is 20 centimeters long, and

the area of the rectangle is twice the area of the parallelogram, find the areas of both figures.

57. 2x2 ϩ x Ϫ 10x Ϫ 5

58. 3x2 ϩ 2x Ϫ 18x Ϫ 12

59. 6n2 Ϫ 3n Ϫ 8n ϩ 4

60. 20n2 ϩ 8n Ϫ 15n Ϫ 6

For Problems 61–84, solve each equation. (Objective 4)

61. x2 Ϫ 8x ϭ 0

87. The area of a square is numerically equal to five times its

perimeter. Find the length of a side of the square.

62. x2 Ϫ 12x ϭ 0

91. The area of a rectangle is twice the area of a square. If

the rectangle is 6 inches long, and the width of the rectangle is the same as the length of a side of the square,

find the dimensions of both the rectangle and the

square.

236

Chapter 6 • Factoring, Solving Equations, and Problem Solving

Thoughts Into Words

92. Suppose that your friend factors 24x2y ϩ 36xy like this:

24x2y ϩ 36xy ϭ 4xy(6x ϩ 9)

ϭ (4xy)(3)(2x ϩ 3)

ϭ 12xy(2x ϩ 3)

Is this correct? Would you make any suggestions for

changing her method?

93. The following solution is given for the equation

x(x Ϫ 10) ϭ 0.

x(x Ϫ 10) ϭ 0

x2 Ϫ 10x ϭ 0

x(x Ϫ 10) ϭ 0

x ϭ 0 or x Ϫ 10 ϭ 0

x ϭ 0 or

x ϭ 10

The solution set is {0,10}. Is this a correct solution?

Would you suggest any changes to the method?

Further Investigations

94. The total surface area of a right circular cylinder is

given by the formula A ϭ 2 r2 ϩ 2 rh, where r represents the radius of a base, and h represents the height of

the cylinder. For computational purposes, it may be

more convenient to change the form of the right side of

the formula by factoring it.

A ϭ 2 r2 ϩ 2 rh ϭ 2 r(r ϩ h)

Use A ϭ 2 r(r ϩ h) to find the total surface area of

22

each of the following cylinders. Use

as an approx7

imation for .

A ϭ P ϩ Prt ϭ P(1 ϩ rt)

Use A ϭ P(1 ϩ rt) to find the total amount of money

accumulated for each of the following investments.

(a) $100 at 8% for 2 years

(b) $200 at 9% for 3 years

(c) $500 at 10% for 5 years

(d) $1000 at 10% for 10 years

(a) r ϭ 7 centimeters and h ϭ 12 centimeters

For Problems 96–99, solve for the indicated variable.

(b) r ϭ 14 meters and h ϭ 20 meters

96. ax ϩ bx ϭ c for x

(c) r ϭ 3 feet and h ϭ 4 feet

97. b2x2 Ϫ cx ϭ 0 for x

(d) r ϭ 5 yards and h ϭ 9 yards

95. The formula A ϭ P ϩ Prt yields the total amount of

money accumulated (A) when P dollars is invested at

Answers to the Concept Quiz

1. False

2. True

3. False

4. True

9. True

10. False

6.2

r percent simple interest for t years. For computational

purposes, it may be convenient to change the right side

of the formula by factoring.

5. False

98. 5ay2 ϭ by for y

99. y ϩ ay Ϫ by Ϫ c ϭ 0 for y

6. False

7. False

8. True

Factoring the Difference of Two Squares

OBJECTIVES

1

Factor the difference of two squares

2

Solve equations by factoring the difference of two squares

In Section 5.3 we noted some special multiplication patterns. One of these patterns was

(a Ϫ b)(a ϩ b) ϭ a2 Ϫ b2

We can view this same pattern as follows:

6.2 • Factoring the Difference of Two Squares

237

Difference of Two Squares

a2 Ϫ b2 ϭ (a Ϫ b)(a ϩ b)

To apply the pattern is a fairly simple process, as these next examples illustrate. The steps

inside the box are often performed mentally.

x2 Ϫ 36 ϭ

4x2 Ϫ 25 ϭ

9x2 Ϫ 16y2 ϭ

64 Ϫ y2 ϭ

(x) 2 Ϫ (6) 2

(2x) 2 Ϫ (5) 2

(3x) 2 Ϫ (4y) 2

(8) 2 Ϫ (y) 2

ϭ

ϭ

ϭ

ϭ

(x Ϫ 6)(x ϩ 6)

(2x Ϫ 5)(2x ϩ 5)

(3x Ϫ 4y)(3x ϩ 4y)

(8 Ϫ y)(8 ϩ y)

Since multiplication is commutative, the order of writing the factors is not important. For

example, (x Ϫ 6)(x ϩ 6) can also be written as (x ϩ 6)(x Ϫ 6).

You must be careful not to assume an analogous factoring pattern for the sum of two

squares; it does not exist. For example, x2 ϩ 4 ϶ (x ϩ 2)(x ϩ 2) because (x ϩ 2) (x ϩ 2) ϭ

x2 ϩ 4x ϩ 4. We say that the sum of two squares is not factorable using integers. The

1

phrase “using integers” is necessary because x2 ϩ 4 could be written as (2x2 ϩ 8) , but such

2

factoring is of no help. Furthermore, we do not consider (1)(x2 ϩ 4) as factoring x2 ϩ 4.

It is possible that both the technique of factoring out a common monomial factor and the

pattern difference of two squares can be applied to the same polynomial. In general, it is best

to look for a common monomial factor first.

Classroom Example

Factor 7x 2 Ϫ 28.

EXAMPLE 1

Factor 2x2 Ϫ 50.

Solution

2x2 Ϫ 50 ϭ 2(x2 Ϫ 25)

ϭ 2(x Ϫ 5)(x ϩ 5)

Common factor of 2

Difference of squares

In Example 1, by expressing 2x2 Ϫ 50 as 2(x Ϫ 5)(x ϩ 5), we say that the algebraic expression has been factored completely. That means that the factors 2, x Ϫ 5, and x ϩ 5 cannot

be factored any further using integers.

Classroom Example

Factor completely 32m 3 Ϫ 50m.

EXAMPLE 2

Factor completely 18y3 Ϫ 8y.

Solution

18y3 Ϫ 8y ϭ 2y(9y2 Ϫ 4)

ϭ 2y(3y Ϫ 2)(3y ϩ 2)

Common factor of 2y

Difference of squares

Sometimes it is possible to apply the difference-of-squares pattern more than once.

Consider the next example.

## Elementary algebra

## 1: Numerical and Algebraic Expressions

## 2: Prime and Composite Numbers

## 3: Integers: Addition and Subtraction

## 4: Integers: Multiplication and Division

## 1: Rational Numbers: Multiplication and Division

## 2: Addition and Subtraction of Rational Numbers

## 3: Real Numbers and Algebraic Expressions

## 5: Translating from English to Algebra

## 2: Equations and Problem Solving

## 3: More on Solving Equations and Problem Solving

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1: Factoring by Using the Distributive Property