Factor 6x4 5x2 12x2 10 by grouping

Factoring The sport of skydiving was born in the 1930s soon after the military began using

parachutes as a means of deploying troops. Today, skydiving is a popular sport

around the world.

With as little as 8 hours of ground instruction, first-time jumpers can be ready

to make a solo jump. Without the assistance of oxygen, skydivers can jump from

as high as 14,000 feet and reach speeds of more than 100 miles per hour as they

fall toward the earth. Jumpers usually open their parachutes between 2000 and

3000 feet and then gradually glide down to their landing area. If the jump and the

parachute are handled correctly, the landing can be as gentle as jumping off two

steps.

Making a jump and floating to earth are only part of the sport of skydiving. For

example, in an activity called “relative work skydiving,” a team of as many as 920

free-falling skydivers join together to make geometrically shaped formations. In a

related exercise called “canopy relative work,” the team members form geometric

patterns after their parachutes or canopies have opened. This kind of skydiving

takes skill and practice, and teams are not always successful in their attempts.

The amount of time a skydiver has for a free fall depends on the height of the

jump and how much the skydiver uses the air to slow the fall.

5C h a p te r

In Exercises 85 and 86 of Section 5.6 we find the amount of time that it takes a skydiver

to fall from a given height.

5.1 Factoring Out Common Factors

5.2 Special Products and Grouping

5.3 Factoring the Trinomial ax2 � bx � c with a � 1

5.4 Factoring the Trinomial ax2 � bx � c with a � 1

5.5 Difference and Sum of Cubes and a Strategy

5.6 Solving Quadratic Equations by Factoring

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322 Chapter 5 Factoring 5-2

In This Section

U1V Prime Factorization of Integers

U2V Greatest Common Factor

U3V Greatest Common Factor for Monomials

U4V Factoring Out the Greatest Common Factor

U5V Factoring Out the Opposite of the GCF

U6V Applications

5.1 Factoring Out Common Factors

In Chapter 4, you learned how to multiply a monomial and a polynomial. In this section, you will learn how to reverse that multiplication by finding the greatest common factor for the terms of a polynomial and then factoring the polynomial.

U1V Prime Factorization of Integers To factor an expression means to write the expression as a product. For example, if we start with 12 and write 12 � 4 � 3, we have factored 12. Both 4 and 3 are factors or divisors of 12. There are other factorizations of 12:

12 � 2 � 6 12 � 1 � 12 12 � 2 � 2 � 3 � 22 � 3

The one that is most useful to us is 12 � 22 � 3, because it expresses 12 as a product of prime numbers.

The numbers 2, 3, 5, 7, 11, 13, 17, 19, and 23 are the first nine prime numbers. A positive integer larger than 1 that is not a prime is a composite number. The numbers 4, 6, 8, 9, 10, and 12 are the first six composite numbers. Every composite number is a product of prime numbers. The prime factorization for 12 is 22 � 3.

Prime Number

A positive integer larger than 1 that has no positive integral factors other than itself and 1 is called a prime number.

E X A M P L E 1

U Helpful Hint V

The prime factorization of 36 can be found also with a factoring tree:

36

2 18

2 9

3 3 So 36 � 2 � 2 � 3 � 3.

Prime factorization Find the prime factorization for 36.

Solution We start by writing 36 as a product of two integers:

36 � 2 � 18 Write 36 as 2 � 18.

� 2 � 2 � 9 Replace 18 by 2 � 9.

� 2 � 2 � 3 � 3 Replace 9 by 3 � 3.

� 22 � 32 Use exponential notation.

The prime factorization for 36 is 22 � 32.

Now do Exercises 1–6

For larger integers, it is better to use the method shown in Example 2 and to recall some divisibility rules. Even numbers are divisible by 2. If the sum of the digits of a number is divisible by 3, then the number is divisible by 3. Numbers that end in 0 or 5 are divisible by 5. Two-digit numbers with repeated digits (11, 22, 33, . . .) are divisible by 11.

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5-3 5.1 Factoring Out Common Factors 323

E X A M P L E 2

U Helpful Hint V

The fact that every composite number has a unique prime factori- zation is known as the fundamental theorem of arithmetic.

Factoring a large number Find the prime factorization for 420.

Solution Start by dividing 420 by the smallest prime number that will divide into it evenly (without remainder). The smallest prime divisor of 420 is 2.

210 2�4�2�0�

Now find the smallest prime that will divide evenly into the quotient, 210. The smallest prime divisor of 210 is 2. Continue this procedure, as follows, until the quotient is a prime number:

2 420___

2 210 420 � 2 � 210___

3 105 210 � 2 � 105___

5 35 105 � 3 � 35__

7

The product of all of the prime numbers in this procedure is 420:

420 � 2 � 2 � 3 � 5 � 7

So the prime factorization of 420 is 22 � 3 � 5 � 7. Note that it is not necessary to divide by the smallest prime divisor at each step. We get the same factorization if we divide by any prime divisor.

Now do Exercises 7–12

� � � �

Strategy for Finding the GCF for Positive Integers

1. Find the prime factorization for each integer.

2. The GCF is the product of the common prime factors using the smallest exponent that appears on each of them.

If two integers have no common prime factors, then their greatest common factor is 1, because 1 is a factor of every integer. For example, 6 and 35 have no common prime

U2V Greatest Common Factor The largest integer that is a factor of two or more integers is called the greatest com- mon factor (GCF) of the integers. For example, 1, 2, 3, and 6 are common factors of 18 and 24. Because 6 is the largest, 6 is the GCF of 18 and 24. We can use prime factoriza- tions to find the GCF. For example, to find the GCF of 8 and 12, we first factor 8 and 12:

8 � 2 � 2 � 2 � 23 12 � 2 � 2 � 3 � 22 � 3

We see that the factor 2 appears twice in both 8 and 12. So 22, or 4, is the GCF of 8 and 12. Notice that 2 is a factor in both 23 and 22 � 3 and that 22 is the smallest power of 2 in these factorizations. In general, we can use the following strategy to find the GCF.

U Helpful Hint V

Note that the division in Example 2 can be done also as follows:

7 5 35

3 105 2 210 2 420 � � � �

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324 Chapter 5 Factoring 5-4

E X A M P L E 3 Greatest common factor Find the GCF for each group of numbers.

a) 150, 225 b) 216, 360, 504 c) 55, 168

Solution a) First find the prime factorization for each number:

2 150 3 225___

3 75 3 75 5 25 5 25

5 5

150 � 2 � 3 � 52 225 � 32 � 52

Because 2 is not a factor of 225, it is not a common factor of 150 and 225. Only 3 and 5 appear in both factorizations. Looking at both 2 � 3 � 52 and 32 � 52, we see that the smallest power of 5 is 2 and the smallest power of 3 is 1. So the GCF for 150 and 225 is 3 � 52, or 75.

b) First find the prime factorization for each number:

216 � 23 � 33 360 � 23 � 32 � 5 504 � 23 � 32 � 7

The only common prime factors are 2 and 3. The smallest power of 2 in the factorizations is 3, and the smallest power of 3 is 2. So the GCF is 23 � 32, or 72.

c) First find the prime factorization for each number:

55 � 5 � 11 168 � 23 � 3 � 7

Because there are no common factors other than 1, the GCF is 1.

Now do Exercises 13–22

�� ��

��

U3V Greatest Common Factor for Monomials To find the GCF for a group of monomials, we use the same procedure as that used for integers.

Strategy for Finding the GCF for Monomials

1. Find the GCF for the coefficients of the monomials.

2. Form the product of the GCF for the coefficients and each variable that is common to all of the monomials, where the exponent on each variable is the smallest power of that variable in any of the monomials.

factors because 6 � 2 � 3 and 35 � 5 � 7. However, because 6 � 1 � 6 and 35 � 1 � 35, the GCF for 6 and 35 is 1.

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5-5 5.1 Factoring Out Common Factors 325

E X A M P L E 4 Greatest common factor for monomials Find the greatest common factor for each group of monomials.

a) 15x2, 9x3 b) 12x2y2, 30x2yz, 42x3y

Solution a) Since 15 � 3 � 5 and 9 � 32, the GCF for 15 and 9 is 3. Since the smallest power

of x in 15x2 and 9x3 is 2, the GCF is 3x2. If we write these monomials as

15x2 � 5 � 3 � x � x and 9x3 � 3 � 3 � x � x � x,

we can see that 3x2 is the GCF.

b) Since 12 � 22 � 3, 30 � 2 � 3 � 5, and 42 � 2 � 3 � 7, the GCF for 12, 30, and 42 is 2 � 3 or 6. For the common variables x and y, 2 is the smallest power of x and 1 is the smallest power of y. So the GCF for the three monomials is 6x2y. Note that z is not in the GCF because it is not in all three monomials.

Now do Exercises 23–34

E X A M P L E 5 Factoring out the greatest common factor Factor the following polynomials by factoring out the GCF.

a) 25a2 � 40a b) 6x4 � 12x3 � 3x2 c) x2y5 � x6y3

Solution a) The GCF for the coefficients 25 and 40 is 5. Because the smallest power of the

common factor a is 1, we can factor 5a out of each term:

25a2 � 40a � 5a � 5a � 5a � 8

� 5a(5a � 8)

b) The GCF for 6, 12, and 3 is 3. We can factor x2 out of each term, since the smallest power of x in the three terms is 2. So factor 3x2 out of each term as follows:

6x4 � 12x3 � 3x2 � 3x2 � 2x2 � 3x2 � 4x � 3x2 � 1

� 3x2(2x2 � 4x � 1)

Check by multiplying: 3x2(2x2 � 4x � 1) � 6x4 � 12x3 � 3x2.

U4V Factoring Out the Greatest Common Factor In Chapter 4, we used the distributive property to multiply monomials and polynomials. For example,

6(5x � 3) � 30x � 18.

If we start with 30x � 18 and write

30x � 18 � 6(5x � 3),

we have factored 30x � 18. Because multiplication is the last operation to be performed in 6(5x � 3), the expression 6(5x � 3) is a product. Because 6 is the GCF for 30 and 18, we have factored out the GCF.

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326 Chapter 5 Factoring 5-6

E X A M P L E 6 A binomial factor Factor out the greatest common factor.

a) (a � b)w � (a � b)6 b) x(x � 2) � 3(x � 2)

c) y(y � 3) � (y � 3)

Solution a) The greatest common factor is a � b:

(a � b)w � (a � b)6 � (a � b)(w � 6)

b) The greatest common factor is x � 2:

x(x � 2) � 3(x � 2) � (x � 3)(x � 2)

c) The greatest common factor is y � 3:

y(y � 3) � (y � 3) � y(y � 3) � 1(y � 3)

� (y � 1)(y � 3)

Now do Exercises 63–70

c) The GCF for the numerical coefficients is 1. Both x and y are common to each term. Using the lowest powers of x and y, we get

x2y5 � x6y3 � x2y3 � y2 � x2y3 � x4

� x2y3(y2 � x4). Check by multiplying.

Now do Exercises 35-62

Because of the commutative property of multiplication, the common factor can be placed on either side of the other factor. So in Example 5, the answers could be written as (5a � 8)5a, (2x2 � 4x � 1)3x2, and (y2 � x4)x2y3.

If the GCF is one of the terms of the polynomial, then you must remem- ber to leave a 1 in place of that term when the GCF is factored out. For example,

ab � b � a � b � 1 � b � b(a � 1).

You should always check your answer by multiplying the factors.

In Example 6, the greatest common factor is a binomial. This type of factoring will be used in factoring trinomials by grouping in Section 5.2.

CAUTION

U5V Factoring Out the Opposite of the GCF The greatest common factor for �4x � 2xy is 2x. Note that you can factor out the GCF (2x) or the opposite of the GCF (�2x):

�4x � 2xy � 2x(�2 � y) �4x � 2xy � �2x(2 � y)

It is useful to know both of these factorizations. Factoring out the opposite of the GCF will be used in factoring by grouping in Section 5.2 and in factoring trinomials with negative leading coefficients in Section 5.4. Remember to check all factoring by multiplying the factors to see if you get the original polynomial.

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5-7 5.1 Factoring Out Common Factors 327

E X A M P L E 7 Factoring out the opposite of the GCF Factor each polynomial twice. First factor out the greatest common factor, and then factor out the opposite of the GCF.

a) 3x � 3y b) a � b

c) �x3 � 2x2 � 8x

Solution a) 3x � 3y � 3(x � y) Factor out 3.

� �3(�x � y) Factor out �3.

Note that the signs of the terms in parentheses change when �3 is factored out. Check the answers by multiplying.

b) a � b � 1(a � b) Factor out 1, the GCF of a and b.

� �1(�a � b) Factor out �1, the opposite of the GCF.

We can also write a � b � �1(b � a).

c) �x3 � 2x2 � 8x � x(�x2 � 2x � 8) Factor out x. � �x(x2 � 2x � 8) Factor out �x.

Now do Exercises 71–86

Be sure to change the sign of each term in parentheses when you factor out the opposite of the greatest common factor.

U6V Applications

CAUTION

Warm-Ups ▼

Fill in the blank. 1. To means to write as a product. 2. A number is an integer greater than 1 that has no

factors besides itself and 1. 3. The of two numbers is the

largest number that is a factor of both. 4. All factoring can be checked by the factors.

True or false? 5. There are only nine prime numbers. 6. The prime factorization of 32 is 23 � 3. 7. The integer 51 is a prime number. 8. The GCF for 12 and 16 is 4. 9. The GCF for x5y3 � x4y7 is x4y3.

10. We can factor out 2xy or �2xy from 2x2y � 6xy2.

Area of a rectangular garden The area of a rectangular garden is x2 � 8x � 15 square feet. If the length is x � 5 feet, then what binomial represents the width?

Solution Note that the area of a rectangle is the product of the length and width. Since x2 � 8x � 15 � (x � 5)(x � 3) and the length is x � 5 feet, the width must be x � 3 feet.

Now do Exercises 87–90

E X A M P L E 8

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U1V Prime Factorization of Integers

Find the prime factorization of each integer. See Examples 1 and 2.

1. 18 2. 20

3. 52 4. 76

5. 98 6. 100

7. 216 8. 248

9. 460 10. 345

11. 924 12. 585

U2V Greatest Common Factor

Find the greatest common factor for each group of integers. See Example 3. See the Strategy for Finding the GCF for Positive Integers box on page 323.

13. 8, 20 14. 18, 42

15. 36, 60 16. 42, 70

17. 40, 48, 88 18. 15, 35, 45

19. 76, 84, 100 20. 66, 72, 120

21. 39, 68, 77 22. 81, 200, 539

U3V Greatest Common Factor for Monomials

Find the greatest common factor for each group of monomials. See Example 4. See the Strategy for Finding the GCF for Monomials box on page 324.

23. 6x, 8x3 24. 12x2, 4x3

25. 12x3, 4x2, 6x 26. 3y5, 9y4, 15y3

27. 3x2y, 2xy2 28. 7a2x3, 5a3x

29. 24a2bc, 60ab2 30. 30x2yz3, 75x3yz6

31. 12u3v2, 25s2t4 32. 45m2n5, 56a4b8

33. 18a3b, 30a2b2, 54ab3 34. 16x2z, 40xz2, 72z3

U4V Factoring Out the Greatest Common Factor

Complete the factoring of each monomial.

35. 27x � 9( ) 36. 51y � 3y( )

37. 24t2 � 8t( ) 38. 18u2 � 3u( )

39. 36y5 � 4y2( ) 40. 42z4 � 3z2( ) 41. u4v3 � uv( ) 42. x5y3 � x2y( ) 43. �14m4n3 � 2m4( ) 44. �8y3z4 � 4z3( ) 45. �33x4y3z2 � �3x3yz( ) 46. �96a3b4c5 � �12ab3c3( )

Factor out the GCF in each expression. See Example 5.

47. 2w � 4t

48. 6y � 3

49. 12x � 18y

50. 24a � 36b

51. x3 � 6x

52. 10y4 � 30y2

53. 5ax � 5ay

54. 6wz � 15wa

55. h5 � h3

56. y6 � y5

57. �2k7m4 � 4k3m6

58. �6h5t2 � 3h3t6

59. 2x3 � 6x2 � 8x

60. 6x3 � 18x2 � 24x

61. 12x4t � 30x3t � 24x2t2

62. 15x2y2 � 9xy2 � 6x2y

Factor out the GCF in each expression. See Example 6.

63. (x � 3)a � (x � 3)b

64. (y � 4)3 � (y � 4)z

65. x(x � 1) � 5(x � 1)

66. a(a � 1) � 3(a � 1)

67. m(m � 9) � (m � 9)

68. (x � 2)x � (x � 2)

69. a(y � 1)2 � b(y � 1)2

70. w(w � 2)2 � 8(w � 2)2

Exercises

U Study Tips V • To get the big picture, survey the chapter that you are studying. Read the headings to get the general idea of the chapter content. • Read the chapter summary several times while you are working in a chapter to see what’s important in the chapter.

5 .1

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5-9 5.1 Factoring Out Common Factors 329

U5V Factoring Out the Opposite of the GCF

First factor out the GCF, and then factor out the opposite of the GCF. See Example 7.

71. 8x � 8y 72. 2a � 6b 73. �4x � 8x2

74. �5x2 � 10x 75. x � 5 76. a � 6 77. 4 � 7a 78. 7 � 5b 79. �24a3 � 16a2

80. �30b4 � 75b3

81. �12x2 � 18x 82. �20b2 � 8b 83. �2x3 � 6x2 � 14x 84. �8x4 � 6x3 � 2x2

85. 4a3b � 6a2b2 � 4ab3

86. 12u5v6 � 18u2v3 � 15u4v5

U6V Applications

Solve each problem by factoring. See Example 8.

87. Uniform motion. Helen traveled a distance of 20x � 40 miles at 20 miles per hour on the Yellowhead Highway. Find a binomial that represents the time that she traveled.

88. Area of a painting. A rectangular painting with a width of x centimeters has an area of x2 � 50x square centimeters. Find a binomial that represents the length. See the accom- panying figure.

Figure for Exercise 88

?

Area � x 2 � 50x cm2

x cm

Figure for Exercise 89

200

100

0 1 2 3

Radius (inches)

Su rf

ac e

ar ea

( in

.2 )

89. Tomato soup. The amount of metal S (in square inches) that it takes to make a can for tomato soup depends on the radius r and height h:

S � 2�r2 � 2�rh

a) Rewrite this formula by factoring out the greatest common factor on the right-hand side.

b) Let h � 5 in. and write a formula that expresses S in terms of r.

c) The accompanying graph shows S for r between 1 in. and 3 in. (with h � 5 in.). Which of these r-values gives the maximum surface area?

90. Amount of an investment. The amount of an investment of P dollars for t years at simple interest rate r is given by A � P � Prt.

a) Rewrite this formula by factoring out the greatest common factor on the right-hand side.

b) Find A if $8300 is invested for 3 years at a simple interest rate of 15%.

Getting More Involved

91. Discussion

Is the greatest common factor of �6x2 � 3x positive or negative? Explain.

92. Writing

Explain in your own words why you use the smallest power of each common prime factor when finding the GCF of two or more integers.

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Kayaks have been built by the Aleut and Inuit peoples for the past 4000 years. Today’s builders have access to materials and techniques unavailable to the orig- inal kayak builders. Modern kayakers incorporate hydrodynamics and materials technology to create designs that are efficient and stable. Builders measure how well their designs work by calculating indicators such as prismatic coefficient, block coefficient, and the midship area coefficient, to name a few.

Even the fitting of a kayak to the paddler is done scientifically. For example, the formula

PL � 2 � BL � BS �0.38 � EE � 1.2 ���B2W� �� �S2W��2�� (SL)�2�� can be used to calculate the appropriate paddle length. BL is the length of the pad- dle’s blade. BS is a boating style factor, which is 1.2 for touring, 1.0 for river run- ning, and 0.95 for play boating. EE is the elbow to elbow distance with the paddler’s arms straight out to the sides. BW is the boat width and SW is the shoulder width. SL is the spine length, which is the distance measured in a sitting position from the chair seat to the top of the paddler’s shoulder. All lengths are in centimeters.

The degree of control a kayaker exerts over the kayak depends largely on the body con- tact with it. A kayaker wears the kayak. So the choice of a kayak should hinge first on the right body fit and comfort and second on the skill level or intended paddling style. So design- ing, building, and even fitting a kayak is a blend of art and science.

Math at Work Kayak Design

In This Section

U1V Factoring by Grouping

U2V Factoring a Difference of Two Squares

U3V Factoring a Perfect Square Trinomial

U4V Factoring Completely

5.2 Special Products and Grouping

In Section 5.1 you learned how to factor out the greatest common factor from all of the terms of a polynomial. In this section you will learn to factor a four-term polynomial by factoring out a common factor from the first two terms and then a common factor from the last two terms.

U1V Factoring by Grouping The product of two binomials may have four terms. For example,

(x � a)(x � 3) � (x � a)x � (x � a)3 � x2 � ax � 3x � 3a.

To factor x2 � ax � 3x � 3a, we simply reverse the steps we used to find the product. Factor out the common factor x from the first two terms and the common factor 3 from the last two terms:

x2 � ax � 3x � 3a � x(x � a) � 3(x � a) Factor out x and 3.

� (x � 3)(x � a) Factor out x � a. It does not matter whether you take out the common factor to the right or left. So (x � a)(x � 3) is also correct and we could have factored as follows:

x2 � ax � 3x � 3a � (x � a)x � (x � a)3

� (x � a)(x � 3)

330 Chapter 5 Factoring 5-10

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5-11 5.2 Special Products and Grouping 331

This method of factoring is called factoring by grouping.

Strategy for Factoring a Four-Term Polynomial by Grouping

1. Factor out the GCF from the first group of two terms.

2. Factor out the GCF from the last group of two terms.

3. Factor out the common binomial.

Factoring by Grouping Use grouping to factor each polynomial.

a) xy � 2y � 5x � 10 b) x2 � wx � x � w

Solution a) The first two terms have a common factor of y, and the last two terms have a

common factor of 5:

xy � 2y � 5x � 10 � y(x � 2) � 5(x � 2) Factor out y and 5.

� (y � 5)(x � 2) Factor out x � 2.

Check by using FOIL.

b) The first two terms have a common factor of x, and the last two have a common factor of 1:

x2 � wx � x � w � x(x � w) � 1(x � w) Factor out x and 1.

� (x � 1)(x � w) Factor out x � w.

Check by using FOIL.

Now do Exercises 1–10

E X A M P L E 1

E X A M P L E 2 Factoring by Grouping with Rearranging Use grouping to factor each polynomial.

a) mn � 4m � m2 � 4n b) ax � b � bx � a

Solution a) We can factor out m from the first two terms to get m(n � 4), but we can’t get

another factor of n � 4 from the last two terms. By rearranging the terms we can factor by grouping:

mn � 4m � m2 � 4n � m2 � mn � 4m � 4n Rearrange terms. � m(m � n) � 4(m � n) Factor out m and 4. � (m � 4)(m � n) Factor out m � n.

For some four-term polynomials it is necessary to rearrange the terms before fac- toring out the common factors.

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332 Chapter 5 Factoring 5-12

Note that there are several rearrangements that will allow us to factor the polynomi- als in Example 2. For example, m2 � 4m � mn � 4n would also work for Example 2(a).

We saw in Section 5.1 that you could factor out a common factor with a positive sign or a negative sign. For example, we can factor �2x � 10 as 2(�x � 5) or �2(x � 5). We use this technique in Example 3.

U2V Factoring a Difference of Two Squares In Section 4.7, you learned that the product of a sum and a difference is a difference of two squares:

(a � b)(a � b) � a2 � ab � ab � b2 � a2 � b2

So a difference of two squares can be factored as a product of a sum and a difference, using the following rule.

Factoring a Difference of Two Squares

For any real numbers a and b,

a2 � b2 � (a � b)(a � b).

Factoring by Grouping with Negative Signs Use grouping to factor each polynomial.

a) 2x2 � 3x � 2x � 3 b) ax � 3y � 3x � ay

Solution a) We can factor out x from the first two terms and 1 from the last two terms:

2x2 � 3x � 2x � 3 � x(2x � 3) � 1(�2x � 3)

However, we didn’t get a common binomial. We can get a common binomial if we factor out �1 from the last two terms:

2x2 � 3x � 2x � 3 � x(2x � 3) � 1(2x � 3) Factor out x and �1. � (x � 1)(2x � 3) Factor out 2x � 3.

b) For this polynomial we have to rearrange the terms and factor out a common factor with a negative sign:

ax � 3y � 3x � ay � ax � 3x � ay � 3y Rearrange the terms. � x(a � 3) � y(a � 3) Factor out x and �y. � (x � y)(a � 3) Factor out a � 3.

Now do Exercises 19–28

E X A M P L E 3

b) ax � b � bx � a � ax � bx � a � b Rearrange terms. � x(a � b) � 1(a � b) Factor out x and 1. � (x � 1)(a � b) Factor out a � b.

Now do Exercises 11–18

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5-13 5.2 Special Products and Grouping 333

E X A M P L E 4 Factoring a difference of two squares Factor each polynomial.

a) y2 � 81 b) 9m2 � 16 c) 4x2 � 9y2

Solution a) Because 81 � 92, the binomial y2 � 81 is a difference of two squares:

y2 � 81 � y2 � 92 Rewrite as a difference of two squares.

� (y � 9)(y � 9) Factor.

Check by multiplying.

b) Because 9m2 � (3m)2 and 16 � 42, the binomial 9m2 � 16 is a difference of two squares:

9m2 � 16 � (3m)2 � 42 Rewrite as a difference of two squares.

� (3m � 4)(3m � 4) Factor.

Check by multiplying.

c) Because 4x2 � (2x)2 and 9y2 � (3y)2, the binomial 4x2 � 9y2 is a difference of two squares:

4x2 � 9y2 � (2x � 3y)(2x � 3y)

Now do Exercises 29–42

Don’t confuse a difference of two squares a2 � b2 with a sum of two squares a2 � b2. The sum a2 � b2 is not one of the special products and it can’t be factored.

U3V Factoring a Perfect Square Trinomial In Section 4.7 you learned how to square a binomial using the rule

(a � b)2 � a2 � 2ab � b2.

You can reverse this rule to factor a trinomial such as x2 � 6x � 9. Notice that

x2 � 6x � 9 � x2 � 2 � x � 3 � 32. ↑ ↑ a2 2ab b2

So if a � x and b � 3, then x2 � 6x � 9 fits the form a2 � 2ab � b2, and

x2 � 6x � 9 � (x � 3)2.

CAUTION

�

Note that the square of an integer is a perfect square. For example, 64 is a perfect square because 64 � 82. The square of a monomial in which the coefficient is an integer is also called a perfect square or simply a square. For example, 9m2 is a perfect square because 9m2 � (3m)2.

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334 Chapter 5 Factoring 5-14

Strategy for Identifying a Perfect Square Trinomial

A trinomial is a perfect square trinomial if

1. the first and last terms are of the form a2 and b2 (perfect squares), and

2. the middle term is 2ab or �2ab.

E X A M P L E 5 Identifying the special products Determine whether each binomial is a difference of two squares and whether each trinomial is a perfect square trinomial.

a) x2 � 14x � 49 b) 4x2 � 81

c) 4a2 � 24a � 25 d) 9y2 � 24y � 16

Solution a) The first term is x2, and the last term is 72. The middle term, �14x, is

�2 � x � 7. So this trinomial is a perfect square trinomial.

b) Both terms of 4x2 � 81 are perfect squares, (2x)2 and 92. So 4x2 � 81 is a difference of two squares.

c) The first term of 4a2 � 24a � 25 is (2a)2 and the last term is 52. However, 2 � 2a � 5 is 20a. Because the middle term is 24a, this trinomial is not a perfect square trinomial.

d) The first and last terms in a perfect square trinomial are both positive. Because the last term in 9y2 � 24y � 16 is negative, the trinomial is not a perfect square trinomial.

Now do Exercises 43–54

Note that the middle term in a perfect square trinomial may have a positive or a negative coefficient, while the first and last terms must be positive. Any perfect square trinomial can be factored as the square of a binomial by using the following rule.

Factoring Perfect Square Trinomials

For any real numbers a and b,

a2 � 2ab � b2 � (a � b)2

a2 � 2ab � b2 � (a � b)2.

A trinomial that is of the form a2 � 2ab � b2 or a2 � 2ab � b2 is called a perfect square trinomial. A perfect square trinomial is the square of a binomial. Perfect square trinomials will be used in solving quadratic equations by completing the square in Chapter 10. Perfect square trinomials can be identified using the following strategy.

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5-15 5.2 Special Products and Grouping 335

E X A M P L E 7 Factoring completely Factor each polynomial completely.

a) 2x3 � 50x b) 8x2y � 32xy � 32y c) 2x3 � 3x2 � 2x � 3

Solution a) The greatest common factor of 2x3 and 50x is 2x:

2x3 � 50x � 2x(x2 � 25) Check this step by multiplying. � 2x(x � 5)(x � 5) Difference of two squares

b) 8x2y � 32xy � 32y � 8y(x2 � 4x � 4) Check this step by multiplying. � 8y(x � 2)2 Perfect square trinomial

c) We can factor out x2 from the first two terms and 1 from the last two terms:

2x3 � 3x2 � 2x � 3 � x2(2x � 3) � 1(�2x � 3)

E X A M P L E 6 Factoring perfect square trinomials Factor.

a) x2 � 4x � 4 b) a2 � 16a � 64 c) 4x2 � 12x � 9

Solution a) The first term is x2, and the last term is 22. Because the middle term is �2 � 2 � x,

or �4x, this polynomial is a perfect square trinomial:

x2 � 4x � 4 � (x � 2)2

Check by expanding (x � 2)2.

b) a2 � 16a � 64 � (a � 8)2

Check by expanding (a � 8)2.

c) The first term is (2x)2, and the last term is 32. Because �2 � 2x � 3 � �12x, the polynomial is a perfect square trinomial. So

4x2 � 12x � 9 � (2x � 3)2.

Check by expanding (2x � 3)2.

Now do Exercises 55–72

U4V Factoring Completely To factor a polynomial means to write it as a product of simpler polynomials. A polynomial that can’t be factored using integers is called a prime or irreducible polynomial. The polynomials 3x, w � 1, and 4m � 5 are prime polynomials. Note that 4m � 5 � 4�m � �54��, but 4m � 5 is a prime polynomial because it can’t be factored using integers only.

A polynomial is factored completely when it is written as a product of prime polynomials. So (y � 8)(y � 1) is a complete factorization. When factoring polyno- mials, we usually do not factor integers that occur as common factors. So 6x(x � 7) is considered to be factored completely even though 6 could be factored.

Some polynomials have a factor common to all terms. To factor such polynomials completely, it is simpler to factor out the greatest common factor (GCF) and then factor the remaining polynomial. Example 7 illustrates factoring completely.

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U1V Factoring by Grouping

Factor by grouping. See Example 1.

1. bx � by � cx � cy 2. 3x � 3z � ax � az 3. ab � b2 � a � b 4. 2x2 � x � 2x � 1 5. wm � 3w � m � 3 6. ay � y � 3a � 3 7. 6x2 � 10x � 3xw � 5w 8. 5ax � 2ay � 5xy � 2y2

9. x2 � 3x � 4x � 12 10. y2 � 2y � 6y � 12

Factor by grouping. See Example 2.

11. mn � n � n2 � m 12. 2x3 � y � x � 2x2y 13. 10 � wm � 5m � 2w 14. 2a � 3b � 6 � ab 15. xa � ay � 3y � 3x 16. x3 � ax � 3a � 3x2

Exercises

U Study Tips V • As you study a chapter, make a list of topics and questions that you would put on the test, if you were to write it. • Write about what you read in the text. Sum things up in your own words.

5 .2

336 Chapter 5 Factoring 5-16

Warm-Ups ▼

Fill in the blank. 1. A is the square of an integer or an

algebraic expression. 2. A is the product of a sum and

a difference. 3. A trinomial of the form a2 � 2ab � b2 is a

trinomial. 4. A polynomial is one that can’t be factored. 5. A polynomial is when it is written

as a product of prime polynomials.

True or false? 6. We always factor out the GCF first. 7. The polynomial x2 � 16 is a difference of two squares.

8. The polynomial x2 � 8x � 16 is a perfect square trinomial.

9. The polynomial 9x2 � 21x � 49 is a perfect square trinomial.

10. The polynomial 16y � 1 is a prime polynomial. 11. The polynomial 4x2 � 4 is factored completely as

4(x2 � 1).

Remember that factoring reverses multiplication and every step of factoring can be checked by multiplication.

However, we didn’t get a common binomial. We can get a common binomial if we factor out �1 from the last two terms:

2x3 � 3x2 � 2x � 3 � x2(2x � 3) � 1(2x � 3) Factor out x2 and �1.

� (x2 � 1)(2x � 3) Factor out 2x � 3.

� (x � 1)(x � 1)(2x � 3) Difference of two squares

Now do Exercises 73–98

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5-17 5.2 Special Products and Grouping 337

17. a3 � w2 � aw � a2w 18. a4 � y � ay � a3

Factor by grouping. See Example 3.

19. w2 � w � bw � b 20. x2 � 2x � mx � 2m 21. w2 � aw � w � a 22. ap � 3a � p � 3 23. m2 � mx � x � m 24. �6n � 6b � b2 � bn 25. x2 � 7x � 5x � 35 26. y2 � 3y � 8y � 24 27. 2x2 � 14x � 5x � 35 28. 2y2 � 3y � 16y � 24

U2V Factoring a Difference of Two Squares

Factor each polynomial. See Example 4.

29. a2 � 4 30. h2 � 9 31. x2 � 49 32. y2 � 36 33. a2 � 121 34. w2 � 81 35. y2 � 9x2

36. 16x2 � y2

37. 25a2 � 49b2

38. 9a2 � 64b2

39. 121m2 � 1 40. 144n2 � 1 41. 9w2 � 25c2

42. 144w2 � 121a2

U3V Factoring a Perfect Square Trinomial

Determine whether each polynomial is a difference of two squares, a perfect square trinomial, or neither of these. See Example 5. See the Strategy for Identifying Perfect Square Trinomials box on page 334.

43. x2 � 20x � 100 44. x2 � 10x � 25 45. y2 � 40 46. a2 � 49 47. 4y2 � 12y � 9 48. 9a2 � 30a � 25 49. x2 � 8x � 64 50. x2 � 4x � 4 51. 9y2 � 25c2

52. 9x2 � 4 53. 9a2 � 6ab � b2

54. 4x2 � 4xy � y2

Factor each perfect square trinomial. See Example 6.

55. x2 � 2x � 1 56. y2 � 4y � 4

57. a2 � 6a � 9 58. w2 � 10w � 25

59. x2 � 12x � 36 60. y2 � 14y � 49

61. a2 � 4a � 4 62. b2 � 6b � 9

63. 4w2 � 4w � 1 64. 9m2 � 6m � 1

65. 16x2 � 8x � 1 66. 25y2 � 10y � 1

67. 4t2 � 20t � 25 68. 9y2 � 12y � 4

69. 9w2 � 42w � 49 70. 144x2 � 24x � 1

71. n2 � 2nt � t2 72. x2 � 2xy � y2

U4V Factoring Completely

Factor each polynomial completely. See Example 7.

73. 5x2 � 125 74. 3y2 � 27

75. �2x2 � 18 76. �5y2 � 20

77. a3 � ab2 78. x2y � y

79. 3x2 � 6x � 3 80. 12a2 � 36a � 27

81. �5y2 � 50y � 125 82. �2a2 � 16a � 32

83. x3 � 2x2y � xy2 84. x3y � 2x2y2 � xy3

85. �3x2 � 3y2

86. �8a2 � 8b2

87. 2ax2 � 98a

88. 32x2y � 2y3

89. w3 � w � w2 � 1

90. x3 � x2 � x � 1

91. x3 � x2 � 4x � 4

92. a2m � b2n � a2n � b2m

93. 3ab2 � 18ab � 27a

94. �2a2b � 8ab � 8b

95. �4m3 � 24m2n � 36mn2

96. 10a3 � 20a2b � 10ab2

97. x2a � b � bx2 � a

98. wx2 � 75 � 25w � 3x2

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Miscellaneous

Factor each polynomial completely.

99. 6a3y � 24a2y2 � 24ay3

100. 8b5c � 8b4c2 � 2b3c3

101. 24a3y � 6ay3

102. 27b3c � 12bc3

103. 2a3y2 � 6a2y 104. 9x3y � 18x2y2

105. ab � 2bw � 4aw � 8w2

106. 3am � 6n � an � 18m 107. (a � b) � b(a � b) 108. (a � b)w � (a � b) 109. (4x2 � 1)2x � (4x2 � 1) 110. (a2 � 9)a � 3(a2 � 9)

Applications

Use factoring to solve each problem.

111. Skydiving. The height in feet above the earth for a skydiver t seconds after jumping from an airplane at 6400 ft is approximated by the formula h(t) � �16t2 � 6400, provided t � 5.

a) Rewrite the formula with the right-hand side factored completely.

b) Use the result of part (a) to find h(2).

e) What is the approximate maximum revenue? f) Use the accompanying graph to estimate the price at

which the revenue is zero.

338 Chapter 5 Factoring 5-18

Figure for Exercise 111

h(t) � �16t 2 � 6400

112. Demand for pools. Tropical Pools sells an aboveground model for p dollars each. The monthly revenue for this model is given by the formula

R(p) � �0.08p2 � 300p. Revenue is the product of the price p and the demand (quantity sold).

a) Factor out the price on the right-hand side of the formula.

b) Write a formula D(p) for the monthly demand.

c) Find D(3000). d) Use the accompanying graph to estimate the price at

which the revenue is maximized. Approximately how many pools will be sold monthly at this price?

Figure for Exercise 112

300

200

100

0 0 2000 3000 4000

Price (dollars)

R ev

en ue

( th

ou sa

nd s

of d

ol la

rs )

1000

Figure for Exercise 113

x

Getting More Involved

114. Discussion

For what real number k does 3x2 � k factor as 3(x � 2)(x � 2)?

115. Writing

Explain in your own words how to factor a four-term polynomial by grouping.

116. Writing

Explain how you know that x2 � 1 is a prime polynomial.

113. Volume of a tank. The volume in cubic inches for a fish tank with a square base and height x is given by the formula

V(x) � x3 � 6x2 � 9x.

a) Rewrite the formula with the right-hand side factored completely.

b) Find an expression for the length of a side of the square base.

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5-19 5.3 Factoring the Trinomial ax2 � bx � c with a � 1 339

In This Section

U1V Factoring ax 2 � bx � c with

a � 1

U2V Factoring with Two Variables

U3V Factoring Completely

5.3 Factoring the Trinomial ax2 � bx � c with a � 1

In this section, we will factor the type of trinomials that result from multiplying two different binomials. We will do this only for trinomials in which the coefficient of x2, the leading coefficient, is 1. Factoring trinomials with a leading coefficient not equal to 1 will be done in Section 5.4.

U1V Factoring ax2 � bx � c with a � 1 To find the product of the binomials x � m and x � n, where x is the variable and m and n are constants, we use the distributive property as follows:

(x � m)(x � n) � (x � m)x � (x � m)n Distributive property

� x2 � mx � nx � mn Distributive property

� x2 � (m � n)x � mn Combine like terms.

Notice that in the trinomial the coefficient of x is the sum m � n and the constant term is the product mn. This observation is the key to factoring the trinomial ax2 � bx � c with a � 1. We first find two numbers that have a product of c (the constant term) and a sum of b (the coefficient of x). Then reverse the steps that we used in finding the product (x � m)(x � n). We summarize these ideas with the following strategy.

Strategy for Factoring x2 � bx � c by Grouping

To factor x2 � bx � c:

1. Find two integers that have a product of c and a sum equal to b.

2. Replace bx by the sum of two terms whose coefficients are the two numbers found in (1).

3. Factor the resulting four-term polynomial by grouping.

E X A M P L E 1 Factoring trinomials Factor.

a) x2 � 5x � 6 b) x2 � 8x � 12 c) a2 � 9a � 20

Solution a) To factor x2 � 5x � 6, we need two integers that have a product of 6 and a sum

of 5. If the product is positive and the sum is positive, then both integers must be positive. We can list all of the possibilities:

Product Sum

6 � 1 � 6 1 � 6 � 7 6 � 2 � 3 2 � 3 � 5

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340 Chapter 5 Factoring 5-20

We usually do not write out all of the steps shown in Example 1. We saw prior to Example 1 that

x2 � (m � n)x � mn � (x � m)(x � n).

So once you know m and n, you can simply write the factors, as shown in Example 2.

The only integers that have a product of 6 and a sum of 5 are 2 and 3. Now replace 5x with 2x � 3x and factor by grouping:

x2 � 5x � 6 � x2 � 2x � 3x � 6 Replace 5x by 2x � 3x.

� x(x � 2) � 3(x � 2) Factor out x and 3.

� (x � 3)(x � 2) Factor out x � 2.

Check by FOIL: (x � 3)(x � 2) � x2 � 5x � 6.

b) To factor x2 � 8x � 12, we need two integers that have a product of 12 and a sum of 8. Since the product and sum are both positive, both integers are positive.

Product Sum

12 � 1 � 12 1 � 12 � 13 12 � 2 � 6 2 � 6 � 8 12 � 3 � 4 3 � 4 � 7

The only integers that have a product of 12 and a sum of 8 are 2 and 6. Now replace 8x by 2x � 6x and factor by grouping:

x2 � 8x � 12 � x2 � 2x � 6x � 12 Replace 8x by 2x � 6x.

� x(x � 2) � 6(x � 2) Factor out x and 6.

� (x � 6)(x � 2) Factor out x � 2.

Check by FOIL: (x � 6)(x � 2) � x2 � 8x � 12.

c) To factor a2 � 9a � 20, we need two integers that have a product of 20 and a sum of �9. Since the product is positive and the sum is negative, both integers must be negative.

Product Sum

20 � (�1)(�20) �1 � (�20) � �21 20 � (�2)(�10) �2 � (�10) � �12 20 � (�4)(�5) �4 � (�5) � �9

Only �4 and �5 have a product of 20 and a sum of �9. Now replace �9a by �4a � (�5a) or �4a � 5a and factor by grouping:

a2 � 9a � 20 � a2 � 4a � 5a � 20 Replace �9a by �4a � 5a.

� a(a � 4) � 5(a � 4) Factor out a and �5.

� (a � 5)(a � 4) Factor out a � 4.

Check by FOIL: (a � 5)(a � 4) � a2 � 9a � 20.

Now do Exercises 1–14

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5-21 5.3 Factoring the Trinomial ax2 � bx � c with a � 1 341

E X A M P L E 3 Factoring trinomials Factor.

a) 2x � 8 � x2

b) �36 � t2 � 9t

Polynomials are easiest to factor when they are in the form ax2 � bx � c. So if a polynomial can be rewritten into that form, rewrite it before attempting to factor it. In Example 3, we factor polynomials that need to be rewritten.

Factoring trinomials more efficiently Factor.

a) x2 � 5x � 4

b) y2 � 6y � 16

c) w2 � 5w � 24

Solution a) To factor x2 � 5x � 4 we need two integers with a product of 4 and a sum of 5.

The only possibilities for a product of 4 are

(1)(4), (�1)(�4), (2)(2), and (�2)(�2).

Only 1 and 4 have a sum of 5. So,

x2 � 5x � 4 � (x � 1)(x � 4).

Check by using FOIL on (x � 1)(x � 4) to get x2 � 5x � 4.

b) To factor y2 � 6y � 16 we need two integers with a product of �16 and a sum of 6. The only possibilities for a product of �16 are

(�1)(16), (1)(�16), (�2)(8), (2)(�8), and (�4)(4).

Only �2 and 8 have a sum of 6. So,

y2 � 6y � 16 � (y � 8)(y � 2).

Check by using FOIL on (y � 8)(y � 2) to get y2 � 6y � 16.

c) To factor w2 � 5w � 24 we need two integers with a product of �24 and a sum of �5. The only possibilities for a product of �24 are

(�1)(24), (1)(�24), (�2)(12), (2)(�12), (�3)(8), (3)(�8), (�4)(6), and (4)(�6).

Only �8 and 3 have a sum of �5. So,

w2 � 5w � 24 � (w � 8)(w � 3).

Check by using FOIL on (w � 8)(w � 3) to get w2 � 5w � 24.

Now do Exercises 15–22

E X A M P L E 2

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342 Chapter 5 Factoring 5-22

Solution a) Before factoring, write the trinomial as x2 � 2x � 8. Now, to get a product of �8

and a sum of 2, use �2 and 4:

2x � 8 � x2 � x2 � 2x � 8 Write in ax2 � bx � c form.

� (x � 4)(x � 2) Factor and check by multiplying.

b) Before factoring, write the trinomial as t2 � 9t � 36. Now, to get a product of �36 and a sum of �9, use �12 and 3:

�36 � t2 � 9t � t2 � 9t � 36 Write in ax2 � bx � c form.

� (t � 12)(t � 3) Factor and check by multiplying.

Now do Exercises 23–24

To factor x2 � bx � c, we search through all pairs of integers that have a product of c until we find a pair that has a sum of b. If there is no such pair of integers, then the polynomial cannot be factored and it is a prime polynomial. Before you can con- clude that a polynomial is prime, be sure that you have tried all possibilities.

E X A M P L E 4 Prime polynomials Factor.

a) x2 � 7x � 6

b) x2 � 9

Solution a) Because the last term is �6, we want a positive integer and a negative integer

that have a product of �6 and a sum of 7. Check all possible pairs of integers:

Product Sum

�6 � (�1)(6) �1 � 6 � 5

�6 � (1)(�6) 1 � (�6) � �5

�6 � (2)(�3) 2 � (�3) � �1

�6 � (�2)(3) �2 � 3 � 1

None of these possible factors of �6 have a sum of 7, so we can be certain that x2 � 7x � 6 cannot be factored. It is a prime polynomial.

b) Because the x-term is missing in x2 � 9, its coefficient is 0. That is, x2 � 9 � x2 � 0x � 9. So we seek two positive integers or two negative integers that have a product of 9 and a sum of 0. Check all possibilities:

Product Sum

9 � (1)(9) 1 � 9 � 10

9 � (�1)(�9) �1 � (�9) � �10

9 � (3)(3) 3 � 3 � 6

9 � (�3)(�3) �3 � (�3) � �6

None of these pairs of integers have a sum of 0, so we can conclude that x2 � 9 is a prime polynomial. Note that x2 � 9 does not factor as (x � 3)2 because (x � 3)2 has a middle term: (x � 3)2 � x2 � 6x � 9.

Now do Exercises 25–52

U Helpful Hint V

Don’t confuse a2 � b2 with the differ- ence of two squares a2 � b2 which is not a prime polynomial:

a2 � b2 � (a � b)(a � b)

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5-23 5.3 Factoring the Trinomial ax2 � bx � c with a � 1 343

The prime polynomial x2 � 9 in Example 4(b) is a sum of two squares. There are many other sums of squares that are prime. For example,

x2 � 1, a2 � 4, b2 � 9, and 4y2 � 25

are prime. However, not every sum of two squares is prime. For example, 4x2 � 16 is a sum of two squares that is not prime because 4x2 � 16 � 4(x2 � 4).

U2V Factoring with Two Variables In Example 5, we factor polynomials that have two variables using the same technique that we used for one variable.

Sum of Two Squares

The sum of two squares a2 � b2 is prime, but not every sum of two squares is prime.

Polynomials with two variables Factor.

a) x2 � 2xy � 8y2 b) a2 � 7ab � 10b2 c) 1 � 2xy � 8x2y2

Solution a) To factor x2 � 2xy � 8y2 we need two integers with a product of �8 and a sum

of 2. The only possibilities for a product of �8 are

(�1)(8), (1)(�8), (�2)(4), and (2)(�4).

Only �2 and 4 have a sum of 2. Since (�2y)(4y) � �8y2, we have

x2 � 2xy � 8y2 � (x � 2y)(x � 4y).

Check by using FOIL on (x � 2y)(x � 4y) to get x2 � 2xy � 8y2.

b) To factor a2 � 7ab � 10b2 we need two integers with a product of 10 and a sum of �7. The only possibilities for a product of 10 are

(�1)(�10), (1)(10), (�2)(�5), and (2)(5).

Only �2 and �5 have a sum of �7. Since (�2b)(�5b) � 10b2, we have

a2 � 7ab � 10b2 � (a � 5b)(a � 2b).

Check by using FOIL on (a � 2b)(a � 5b) to get a2 � 7ab � 10b2.

c) As in part (a), we need two integers with a product of �8 and a sum of �2. The integers are �4 and 2. Since 1 factors as 1 � 1 and �8x2y2 � (�4xy)(2xy), we have

1 � 2xy � 8x2y2 � (1 � 2xy)(1 � 4xy).

Check by using FOIL.

Now do Exercises 53–64

E X A M P L E 5

U3V Factoring Completely In Section 5.2 you learned that binomials such as 3x � 5 (with no common factor) are prime polynomials. In Example 4 of this section we saw a trinomial that is a prime polynomial. There are infinitely many prime trinomials. When factoring a polynomial completely, we could have a factor that is a prime trinomial.

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U1V Factoring ax 2 � bx � c with a � 1

Factor each trinomial. Write out all of the steps as shown in Example 1. See the Strategy for Factoring x2 � bx � c by Grouping on page 339.

1. x2 � 4x � 3 2. y2 � 6y � 5

3. x2 � 9x � 18 4. w2 � 6w � 8

5. a2 � 7a � 10 6. b2 � 7b � 12

7. a2 � 7a � 12 8. m2 � 9m � 14

Exercises

U Study Tips V • Put important facts on note cards. Work on memorizing the note cards when you have a few spare minutes. • Post some note cards on your refrigerator door. Make this course a part of your life.

5 .3

344 Chapter 5 Factoring 5-24

Warm-Ups ▼

Fill in the blank. 1. If there are no two integers that have a of c and

a of b, then x2 � bx � c is prime. 2. We can check all factoring by the factors.

3. The sum of two squares a2 � b2 is .

4. Always factor out the first.

True or false? 5. x2 � 6x � 9 � (x � 3)2

6. x2 � 6x � 9 � (x � 3)2

7. x2 � 10x � 9 � (x � 9)(x � 1) 8. x2 � 8x � 9 � (x � 1)(x � 9) 9. x2 � 10xy � 9y2 � (x � y)(x � 9y)

10. x2 � 1 � (x � 1)(x � 1) 11. x2 � x �1 � (x � 1)(x � 1)

E X A M P L E 6 Factoring completely Factor each polynomial completely.

a) x3 � 6x2 � 16x b) 4x3 � 4x2 � 4x

Solution a) x3 � 6x2 � 16x � x(x2 � 6x � 16) Factor out the GCF.

� x (x � 8)(x � 2) Factor x2 � 6x � 16.

b) First factor out 4x, the greatest common factor:

4x3 � 4x2 � 4x � 4x(x2 � x � 1) To factor x2 � x � 1, we would need two integers with a product of 1 and a sum of 1. Because there are no such integers, x2 � x � 1 is prime, and the factorization is complete.

Now do Exercises 65–106

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5-25 5.3 Factoring the Trinomial ax2 � bx � c with a � 1 345

9. b2 � 5b � 6 10. a2 � 5a � 6

11. x2 � 3x � 10 12. x2 � x � 12

13. x2 � 5x � 24 14. a2 � 5a � 50

Factor each polynomial. If the polynomial is prime, say so. See Examples 2–4.

15. y2 � 7y � 10

16. x2 � 8x � 15

17. a2 � 6a � 8

18. b2 � 8b � 15

19. m2 � 10m � 16

20. m2 � 17m � 16

21. w2 � 9w � 10

22. m2 � 6m � 16

23. w2 � 8 � 2w

24. �16 � m2 � 6m

25. a2 � 2a � 12

26. x2 � 3x � 3

27. 15m � 16 � m2

28. 3y � y2 � 10

29. a2 � 4a � 12

30. y2 � 6y � 8

31. z2 � 25

32. p2 � 1

33. h2 � 49

34. q2 � 4

35. m2 � 12m � 20

36. m2 � 21m � 20

37. t2 � 3t � 10

38. x2 � 5x � 3

39. m2 � 18 � 17m

40. h2 � 36 � 5h

41. m2 � 23m � 24

42. m2 � 23m � 24

43. 5t � 24 � t2

44. t2 � 24 � 10t

45. t2 � 2t � 24

46. t2 � 14t � 24

47. t2 � 10t � 200

48. t2 � 30t � 200

49. x2 � 5x � 150

50. x2 � 25x � 150

51. 13y � 30 � y2

52. 18z � 45 � z2

U2V Factoring with Two Variables

Factor each polynomial. See Example 5.

53. x2 � 5ax � 6a2

54. a2 � 7ab � 10b2

55. x2 � 4xy � 12y2

56. y2 � yt � 12t2

57. x2 � 13xy � 12y2

58. h2 � 9hs � 9s2

59. x2 � 4xz � 33z2

60. x2 � 5xs � 24s2

61. 1 � 3ab � 28a2b2

62. 1 � xy � 20x2y2

63. 15a2b2 � 8ab � 1

64. 12m2n2 � 8mn � 1

U3V Factoring Completely

Factor each polynomial completely. Use the methods discussed in Sections 5.1 through 5.3. If the polynomial is prime say so. See Example 6.

65. 5x3 � 5x

66. b3 � 49b

67. w2 � 8w

68. x4 � x3

69. 2w2 � 162

70. 6w4 � 54w2

71. �2b2 � 98

72. �a3 � 100a

73. x3 � 2x2 � 9x � 18

74. x3 � 7x2 � x � 7

75. 4r2 � 9

76. t2 � 4z2

77. x2w2 � 9x2

78. a4b � a2b3

79. w2 � 18w � 81

80. w2 � 30w � 81

81. 6w2 � 12w � 18

82. 9w � w3

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83. 3y2 � 75

84. 5x2 � 500

85. ax � ay � cx � cy

86. y3 � y2 � 4y � 4

87. �2x2 � 10x � 12

88. �a3 � 2a2 � a

89. 32x2 � 2x4

90. 20w2 � 100w � 40

91. 3w2 � 27w � 54

92. w3 � 3w2 � 18w

93. 18w2 � w3 � 36w

94. 18a2 � 3a3 � 36a

95. 9y2 � 1 � 6y

96. 2a2 � 1 � 3a

97. 8vw2 � 32vw � 32v

98. 3h2t � 6ht � 3t

99. 6x3y � 30x2y2 � 36xy3

100. 3x3y2 � 3x2y2 � 3xy2

101. 5 � 8w � 3w2

102. �3 � 2y � 21y2

103. �3y3 � 6y2 � 3y

104. �4w3 � 16w2 � 20w

105. a3 � ab � 3b � 3a2

106. ac � xc � aw2 � xw2

Applications

Use factoring to solve each problem.

107. Area of a deck. The area in square feet for a rectangular deck is given by A(x) � x2 � 6x � 8.

a) Find A(6). b) If the width of the deck is x � 2 feet, then what is the

length?

109. Volume of a cube. Hector designed a cubic box with volume x3 cubic feet. After increasing the dimensions of the bottom, the box has a volume of x3 � 8x 2 � 15x cubic feet. If each of the dimensions of the bottom was increased by a whole number of feet, then how much was each increase?

110. Volume of a container. A cubic shipping container had a volume of a3 cubic meters. The height was decreased by a whole number of meters and the width was increased by a whole number of meters so that the volume of the container is now a3 � 2a2 � 3a cubic meters. By how many meters were the height and width changed?

Getting More Involved

111. Discussion

Which of the following products is not equivalent to the others? Explain your answer.

a) (2x � 4)(x � 3) b) (x � 2)(2x � 6) c) 2(x � 2)(x � 3) d) (2x � 4)(2x � 6)

112. Discussion

When asked to factor completely a certain polynomial, four students gave the following answers. Only one student gave the correct answer. Which one must it be? Explain your answer.

a) 3(x2 � 2x � 15) b) (3x � 5)(5x � 15) c) 3(x � 5)(x � 3) d) (3x � 15)(x � 3)

346 Chapter 5 Factoring 5-26

Figure for Exercise 107

Area � x2 � 6x � 8 ft2

L x � 2 ft

Figure for Exercise 108

Base

Area � x2 � 5x � 6 m2

x � 3 m

108. Area of a sail. The area in square meters for a triangular sail is given by A(x) � x2 � 5x � 6.

a) Find A(5). b) If the height of the sail is x � 3 meters, then what is

the length of the base of the sail?

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Mid-Chapter Quiz Sections 5.1 through 5.3 Chapter 5

Find the prime factorization of each integer.

1. 48 2. 140

Find the greatest common factor for each group of integers.

3. 36, 45 4. 60, 144, 240

Factor each expression by factoring out the greatest common factor.

5. 8w � 6y 6. 12x3 � 30x2

7. 15ab3 � 25a2b2 � 35a3b

Factor each expression.

8. (x � 3)x � (x � 3)5

9. m(m � 9) � 6(m � 9)

Factor completely.

10. 4y2 � 9w2

11. 4h2 � 12h � 9

12. w2 � 16w � 64

13. 10x3 � 250x

14. �6x2 � 36x � 54

15. aw � 3w � 6a � 18

16. bx � 5b � 6x � 30

17. ax2 � a � x2 � 1

18. x3 � 5x � 4x2

19. 2x3 � 18x

20. a2 � 12as � 32s2

In This Section

U1V The ac Method

U2V Trial and Error

U3V Factoring Completely

5.4 Factoring the Trinomial ax2 � bx � c with a � 1

In Section 5.3, we used grouping to factor trinomials with a leading coefficient of 1. In this section we will also use grouping to factor trinomials with a leading coefficient that is not equal to 1.

U1V The ac Method The first step in factoring ax2 � bx � c with a � 1 is to find two numbers with a prod- uct of c and a sum of b. If a 1, then the first step is to find two numbers with a product of ac and a sum of b. This method is called the ac method. The strategy for factoring by the ac method follows. Note that this strategy works whether or not the leading coefficient is 1.

Strategy for Factoring ax2 � bx � c by the ac Method

To factor the trinomial ax2 � bx � c:

1. Find two numbers that have a product equal to ac and a sum equal to b.

2. Replace bx by the sum of two terms whose coefficients are the two numbers found in (1).

3. Factor the resulting four-term polynomial by grouping.

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348 Chapter 5 Factoring 5-28

E X A M P L E 1 The ac method Factor each trinomial.