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

5

5.1

Factoring Out Common Factors Special Products and Grouping Factoring the Trinomial ax2 bx c with a 1 Factoring the Trinomial ax2 bx c with a 1 Difference and Sum of Cubes and a Strategy Solving Quadratic Equations by Factoring

5.2

Chapter 5.3 5.4 5.5 5.6

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.

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.

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5-2

5.1
In This Section U1V Prime Factorization of
Integers

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.

U2V Greatest Common Factor U3V Greatest Common Factor for
Monomials 4V Factoring Out the Greatest U Common Factor 5V Factoring Out the Opposite U of the GCF U6V Applications

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 22 3, because it expresses 12 as a product of

The one that is most useful to us is 12 prime numbers.

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

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.

E X A M P L E

1

Prime factorization
Find the prime factorization for 36.

Solution
U Helpful Hint V
The prime factorization of 36 can be found also with a factoring tree: 36 2 2 3 So 36 2 2 3 3. 18 9 3

We start by writing 36 as a product of two integers: 36 2 18 2 2 9 22 32 The prime factorization for 36 is 22 32.
Write 36 as 2 18. Replace 18 by 2 9. Use exponential notation.

2 2 3 3 Replace 9 by 3 3.

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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323

E X A M P L E

2

Factoring a large number
Find the prime factorization for 420.

Solution
U Helpful Hint V
The fact that every composite number has a unique prime factorization is known as the fundamental theorem of arithmetic.

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 20 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 3 ___ 105 5 __ 35 7
420 210 105 2 2 3 210 105 35

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

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

U2V Greatest Common Factor
The largest integer that is a factor of two or more integers is called the greatest common 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 factorizations 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.

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

324

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5-4

factors because 6 2 3 and 35 the GCF for 6 and 35 is 1.

5 7. However, because 6

1 6 and 35

1 35,

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 75 5 25 5 150 2 3 52 3 225 ___ 3 75 5 25 5 225 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.

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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) 15x 2, 9x 3 b) 12x 2y 2, 30x 2yz, 42x 3y

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 15x 2
2

5 3 x x

and

9x3

3 3 x x x,

we can see that 3x 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

U4V Factoring Out the Greatest Common Factor
In Chapter 4, we used the distributive property to multiply monomials and polynomials. For example, 6(5x If we start with 30x 18 and write 30x 18 6(5x 3), 3) 30x 18.

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.

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) 6x 4 12x 3 3x 2 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(5a
2

5a 8 8)

b) The GCF for 6, 12, and 3 is 3. We can factor x 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 2x 2 3x (2x
2 2

3x2 4x 4x 6x4 1) 12x3

3x2 1 3x2.

Check by multiplying: 3x 2(2x2

4x

1)

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5-6
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 x 2y5 x6y3 x 2y3 y2 x 2y3 x 4 x 2y3(y2 Check by multiplying. x4).

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.
CAUTION If the GCF is one of the terms of the polynomial, then you must remember 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.

E X A M P L E

6

A binomial factor
Factor out the greatest common factor. a) (a c) y(y b)w 3) (a (y b)6 3) b) x(x 2) 3(x 2)

Solution
a) The greatest common factor is a (a b)w (a b: b)6 2: 2) 3: 3) y(y (y 3) 1)(y 1(y 3) 3) (x 3)(x 2) (a b)(w 6)

b) The greatest common factor is x x(x 2) 3(x

c) The greatest common factor is y y(y 3) (y

Now do Exercises 63–70

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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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 c) x3 3y 2x2 8x b) a b

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 Check the answers by multiplying. b) a b 1(a b)
Factor out 1, the GCF of a and b.

3 is factored out.

1( a We can also write a c) x3 2x 2 8x

b) Factor out b x( x2 x (x 2 1(b 2x 2x

1, the opposite of the GCF.

a). 8) Factor out x. 8) Factor out
x.

Now do Exercises 71–86 CAUTION Be sure to change the sign of each term in parentheses when you factor out the opposite of the greatest common factor.

U6V Applications
E X A M P L E

8

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

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

Now do Exercises 87–90

Warm-Ups
Fill in the blank.

▼
True or false?
5. 6. 7. 8. 9. 10. There are only nine prime numbers. The prime factorization of 32 is 23 3. The integer 51 is a prime number. The GCF for 12 and 16 is 4. The GCF for x5y3 x4y7 is x4y3. We can factor out 2xy or 2xy from 2x2y

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.

6xy2.

5.1

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.

U1V Prime Factorization of Integers
Find the prime factorization of each integer. See Examples 1 and 2. 1. 18 3. 52 5. 98 7. 216 9. 460 11. 924 2. 20 4. 76 6. 100 8. 248 10. 345 12. 585

39. 40. 41. 42. 43. 44. 45. 46.

36y5 4y2( ) 42z4 3z2( ) 4 3 uv uv( ) x5y3 x2y( ) 4 3 4 14m n 2m ( ) 8y3z4 4z3( ) 33x4y3z2 3x3yz( 96a3b4c5 12ab3c3(

) )

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

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 15. 36, 60 17. 40, 48, 88 19. 76, 84, 100 21. 39, 68, 77 14. 18, 42 16. 42, 70 18. 15, 35, 45 20. 66, 72, 120 22. 81, 200, 539

47. 2w 48. 6y 49. 12x 50. 24a 51. x
3

4t 3 18y 36b 6x 30y2 5ay 15wa h3 y5 4k 3m6 3h3t 6 8x 24x
3

52. 10y4 53. 5ax 54. 6wz 55. h5 56. y6 57. 58.

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, 8x 3 25. 12x 3, 4x 2, 6x 27. 3x 2y, 2xy2 29. 24a2bc, 60ab2 31. 12u3v2, 25s2t4 33. 18a3b, 30a2b2, 54ab3 24. 12x 2, 4x 3 26. 3y5, 9y4, 15y 3 28. 7a2x 3, 5a3x 30. 30x2yz3, 75x 3yz6 32. 45m2n5, 56a4b8 34. 16x2z, 40xz2, 72z3

2k7m4 6h t
3 4 52

59. 2x 3 60. 6x 61. 12x t 62. 15x 2y2

6x 2 18x2

30x t 9xy2

24x 2t 2 6x2y

Factor out the GCF in each expression. See Example 6. 63. (x 64. (y 65. x(x 66. a(a 67. m(m 68. (x 69. a(y 3)a 4)3 1) 1) 9) 2)x 1)2 2)2 (x (y 5(x 3(a (m (x b(y 8(w 2) 1)2 2)2 3)b 4)z 1) 1) 9)

U4V Factoring Out the Greatest Common Factor
Complete the factoring of each monomial. 35. 27x 37. 24t2 9( ) 8t( ) 36. 51y 38. 18u2 3y( 3u( ) )

70. w(w

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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. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 8x 8y 2a 6b 4x 8x2 5x2 10x x 5 a 6 4 7a 7 5b 24a3 16a 2 30b 4 75b 3 12x 2 18x 20b 2 8b 2x3 6x2 14x 8x 4 6x 3 2x 2 6a2b 2 18u2v3 4ab3 15u4v5

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?

200 Surface area (in.2)

85. 4a3b 86. 12u5v6

100

0

1

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 accompanying figure.
?

2 Radius (inches)

3

Figure for Exercise 89

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 x cm

Is the greatest common factor of or negative? Explain.

6x2

3x positive

92. Writing
Area x2 50x cm2

Figure for Exercise 88

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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5-10

Math at Work

Kayak Design
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 original 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 BW 2 SW 2
2

(SL)2

can be used to calculate the appropriate paddle length. BL is the length of the paddle’s blade. BS is a boating style factor, which is 1.2 for touring, 1.0 for river running, 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 contact 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 designing, building, and even fitting a kayak is a blend of art and science.

5.2
In This Section U1V Factoring by Grouping U2V Factoring a Difference of Two
Squares

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.

U3V Factoring a Perfect Square
Trinomial 4V Factoring Completely U

U1V Factoring by Grouping
The product of two binomials may have four terms. For example, (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) (x a)(x 3)

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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.

E X A M P L E

1

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 (y 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 (x Check by using FOIL. w) 1)(x 1(x w) w) Factor out x and 1. Factor out x w. 2) 5)(x 5(x 2) 2) Factor out y and 5. Factor out x 2.

Now do Exercises 1–10

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

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 m(m n) 4(m n) (m 4)(m n) Rearrange terms. Factor out m and 4. Factor out m n.

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Chapter 5 Factoring

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

Now do Exercises 11–18

Note that there are several rearrangements that will allow us to factor the polynomials 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.

E X A M P L E

3

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(2 x 3) 1( 2 x 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(2 x 3) (x 1)(2 x 1(2 x 3) 3) Factor out x and 1. 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 x(a 3) y(a 3) (x y)(a 3) Rearrange the terms. Factor out x and y. Factor out a 3.

Now do Exercises 19–28

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).

5-13

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Special Products and Grouping

333

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.

E X A M P L E

4

Factoring a difference of two squares
Factor each polynomial. a) y 2 81 b) 9m 2 16 c) 4x 2 9y 2

Solution
a) Because 81 y2 92, the binomial y2 81 y2 (y Check by multiplying. b) Because 9m2 squares: 9m2 16 (3m)2 and 16 42, the binomial 9m2 16 is a difference of two 92 9)(y 9) 81 is a difference of two squares:
Rewrite as a difference of two squares. Factor.

(3m)2 (3m

42 4)(3m 4)

Rewrite as a difference of two squares. Factor.

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

4x2

9y2

(2x

3y)(2x

3y)

Now do Exercises 29–42 CAUTION Don’t confuse a difference of two squares a2 b2 with a sum of two 2 2 2 2 b . The sum a b is not one of the special products and squares a 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. 6x 32.
↑ b2

You can reverse this rule to factor a trinomial such as x 2 x2 6x 9 x2
↑ a2

9. Notice that

2 x 3
2ab

So if a

x and b

3, then x 2 x2

6x 6x

9 fits the form a 2 9 (x 3)2.

2ab

b2, and

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5-14

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.

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 c) 4a
2

14x 24a

49 25

b) 4x2 d) 9y
2

81 24y 16

Solution
a) The first term is x 2, and the last term is 72. The middle term, 2 x 7. So this trinomial is a perfect square trinomial. 14x, is 81 is a

b) Both terms of 4x2 81 are perfect squares, (2x)2 and 9 2. So 4x 2 difference of two squares.

c) The first term of 4a2 24a 25 is (2a)2 and the last term is 5 2. 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 a2 2ab 2ab b2 b2 (a (a b)2 b)2.

5-15

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Special Products and Grouping

335

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 or 4x, this polynomial is a perfect square trinomial: x2 Check by expanding (x b) a2 16a 64 (a Check by expanding (a 2)2. 8)2 8)2. 2 2x 3 12x, the 4x 4 (x 2)2 2 2 x,

c) The first term is (2x)2, and the last term is 32. Because polynomial is a perfect square trinomial. So 4x 2 Check by expanding (2x 3)2. 12x 9 (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 5 , but 4m 5 is a prime polynomial because it can’t be factored 4 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 polynomials, 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.

E X A M P L E

7

Factoring completely
Factor each polynomial completely. a) 2x 3 50x b) 8x 2y 32xy 32y c) 2x 3 3x2 2x 3

Solution
a) The greatest common factor of 2x3 and 50x is 2x: 2x3 b) 8x y
2

50x 32y

2x(x2 25) 2x(x 5)(x 8y(x 8y(x 3x2
2

Check this step by multiplying.

5)

Difference of two squares Perfect square trinomial

32xy

4x 2)2

4) Check this step by multiplying.

c) We can factor out x2 from the first two terms and 1 from the last two terms: 2x3 2x 3 x2(2x 3) 1( 2x 3)

336

Chapter 5 Factoring

5-16
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 (x
2

3) 1)(2x 1)(x

1(2x 3) 1)(2x

3) 3)

Factor out x2 and Factor out 2x 3.

1.

(x

Difference of two squares

Now do Exercises 73–98

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

Warm-Ups
Fill in the blank.

▼
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).

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.

5.2

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.

U1V Factoring by Grouping
Factor by grouping. See Example 1. 1. 2. 3. 4. 5. 6. 7. 8. bx 3x ab 2x2 wm ay 6x2 5ax by cx cy 3z ax az b2 a b x 2x 1 3w m 3 y 3a 3 10x 3xw 5w 2ay 5xy 2y2

9. x2 10. y2

3x 2y

4x 6y

12 12

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

5-17
17. a3 18. a4 w2 aw a2w y ay a3

5.2

Special Products and Grouping

337

Factor each perfect square trinomial. See Example 6. 55. x2 57. a2 59. x2 61. a2 63. 4w2 65. 16x2 67. 4t2 69. 9w2 71. n2 2x 6a 12x 4a 4w 8x 20t 42w 2nt t2 1 9 36 4 1 1 25 49 56. y2 58. w2 60. y2 62. b2 64. 9m2 66. 25y2 68. 9y2 70. 144x2 72. x2 2xy 4y 10w 14y 6b 6m 10y 12y 24x y2 4 25 49 9 1 1 4 1

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

U2V Factoring a Difference of Two Squares
Factor each polynomial. See Example 4. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. a2 4 h2 9 x 2 49 y2 36 a2 121 w 2 81 y2 9x2 16x 2 y2 25a2 49b2 9a2 64b2 121m2 1 144n2 1 9w2 25c2 144w2 121a2

U4V Factoring Completely
Factor each polynomial completely. See Example 7. 73. 5x2 75. 2x2 125 18 ab2 6x 3 125 xy2 74. 3y2 76. 5y2 y 36a 16a 2x2y2 27 32 xy3 27 20

77. a3 79. 3x2 81. 83. x3 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 5y2

78. x2y 80. 12a2 82. 2a2

50y 2x2y

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. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. x2 x2 y2 a2 4y2 9a2 x2 x2 9y2 9x2 9a2 4x2 20x 100 10x 25 40 49 12y 9 30a 25 8x 64 4x 4 25c2 4 6ab b2 4xy y2

84. x3y

3x2 3y2 8a2 8b2 2ax2 98a 32x2y 2y3 w3 w w2 1 x3 x2 x 1 x3 x2 4x 4 a2m b2n a2n b2m 3ab2 18ab 27a 2a2b 8ab 8b 4m3 24m2n 36mn2 10a3 20a2b 10ab2 x2a b bx2 a wx2 75 25w 3x2

338

Chapter 5 Factoring

5-18
e) What is the approximate maximum revenue? f) Use the accompanying graph to estimate the price at which the revenue is zero.

Miscellaneous
Factor each polynomial completely. 99. 100. 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 6a3y 24a2y2 24ay3 8b5c 8b4c2 2b3c3 24a3y 6ay3 27b3c 12bc3 2a3y2 6a2y 9x3y 18x2y2 ab 2bw 4aw 8w2 3am 6n an 18m (a b) b(a b) (a b)w (a b) (4x2 1)2x (4x2 1) (a2 9)a 3(a2 9)

300 Revenue (thousands of dollars) 200 100 0

0

1000 2000 3000 4000 Price (dollars)

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).

Figure for Exercise 112

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.

h(t)

16t 2

6400

x

Figure for Exercise 111

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 113

Getting More Involved
114. Discussion For what real number k does 3x2 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 polynomial. 1 is a prime k factor as

5-19

5.3

Factoring the Trinomial ax2

bx

c with a

1

339

5.3
In This Section U1V Factoring ax2 U2V U3V bx c with a 1 Factoring with Two Variables Factoring Completely

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 x 2, 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 x2 x2 m)x (x mx nx (m n)x m)n Distributive property mn Distributive property 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 ax 2 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 x 2
To factor x2 bx c:

bx

c by Grouping

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) x 2 5x 6 b) x 2 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 6 6 1 6 2 3 Sum 1 2 6 3 7 5

340

Chapter 5 Factoring

5-20
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 x(x (x Check by FOIL: (x
2

2x 2) 3)(x 2)

3x 3(x 2) x2

6 2)

Replace 5x by 2x Factor out x and 3. Factor out x 2.

3x.

3)(x

5x

6.

b) To factor x 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 12 12 12 1 12 2 6 3 4 Sum 1 2 3 12 13 6 8 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 x(x (x Check by FOIL: (x
2

2x 2) 6)(x 2)

6x 6(x 2) x2

12

Replace 8x by 2x
Factor out x 2.

6x.

2) Factor out x and 6.

6)(x

8x

12.

c) To factor a 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 20 20 20 Only 4a ( 1)( 20) ( 2)( 10) ( 4)( 5) Sum 1 2 4 ( 20) ( 10) ( 5) 21 12 9 9a by

4 and 5 have a product of 20 and a sum of ( 5a) or 4a 5a and factor by grouping: a2 9a 20 a2 a(a (a 4a 4) 5)(a 4) 5a 5(a 4) a2 9a 20

9. Now replace

Replace

9a by

4a 5.

5a.

4) Factor out a and
Factor out a 4.

Check by FOIL: (a

5)(a

20.

Now do Exercises 1–14

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.

5-21

5.3

Factoring the Trinomial ax2

bx

c with a

1

341

E X A M P L E

2

Factoring trinomials more efficiently
Factor. a) x 2 b) y2 c) w2 5x 6y 5w 4 16 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 Check by using FOIL on (x
2

5x

4

(x

1)(x

4). 5x 4. 16 and a sum of 6.

1)(x

4) to get x2

b) To factor y 6y 16 we need two integers with a product of 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 Check by using FOIL on ( y
2

6y

16 8)(y

(y

8)(y

2). 6y 16. 24 and a sum

2) to get y 2

c) To factor w 5w 24 we need two integers with a product of 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 w2 5w 5. So, 24 8)(w (w 8)(w 3). 5w 24.

Check by using FOIL on (w

3) to get w2

Now do Exercises 15–22

Polynomials are easiest to factor when they are in the form ax 2 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.

E X A M P L E

3

Factoring trinomials
Factor. a) 2x b) 36 8 t2 x2 9t

342

Chapter 5 Factoring

5-22

Solution
a) Before factoring, write the trinomial as x 2 and a sum of 2, use 2 and 4: 2x 8 x2 x2 (x 2x 4)(x 8 2x 8. Now, to get a product of bx c form.

8

Write in ax2

2) Factor and check by multiplying. 9t 36. Now, to get a product of bx c form.

b) Before factoring, write the trinomial as t 2 and a sum of 9, use 12 and 3: 36 t2 9t t2 (t 9t 12)(t 36

36

Write in ax2

3) Factor and check by multiplying.

Now do Exercises 23–24

To factor x 2 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 conclude 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) x 2 b) x
2

7x 9

6

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 6 6 6 6 ( 1)(6) (1)( 6) (2)( 3) ( 2)(3) Sum 1 1 2 2 6 5 5 1 ( 6) ( 3) 3 1

U Helpful Hint V
Don’t confuse a2 b2 with the difference of two squares a2 b2 which is not a prime polynomial: a2 b2 (a b)(a b)

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 9 9 9 9 (1)(9) ( 1)( 9) (3)(3) ( 3)( 3) Sum 1 1 3 3 9 3 10 10 6 9 is 6 ( 9) ( 3)

None of these pairs of integers have a sum of 0, so we can conclude that x 2 a prime polynomial. Note that x 2 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

5-23

5.3

Factoring the Trinomial ax2

bx

c with a

1

343

The prime polynomial x 2 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 16 is are prime. However, not every sum of two squares is prime. For example, 4x2 a sum of two squares that is not prime because 4x2 16 4(x2 4). Sum of Two Squares The sum of two squares a2

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

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.

E X A M P L E

5

Polynomials with two variables
Factor. a) x 2 2xy 8y 2 b) a 2 7ab 10b2 c) 1 2xy 8x2y2

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

8 and a sum

8y 2, we have 4y).
2

2xy

8y

2

(x

2y)(x

Check by using FOIL on (x

2y)(x

4y) to get x

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 a2 7ab 7. Since ( 2b)( 5b) 10b2 2b)(a (a 5b)(a 2b). 7ab 10b2. 5b) to get a2 10b2, we have

Check by using FOIL on (a

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 Check by using FOIL. 2xy 8x2y2 (1 2xy)(1 4xy).

Now do Exercises 53–64

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.

344

Chapter 5 Factoring

5-24

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 6x 2 16x x (x 2 x (x 4x3
2

6x 8)(x 4x2

16) Factor out the GCF. 2) 4x
Factor x2 6x 16.

b) First factor out 4x, the greatest common factor: 4x (x2 x 1) To factor x 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

Warm-Ups
Fill in the blank.

▼
True or false?
5. 6. 7. 8. 9. 10. 11. x2 x2 x2 x2 x2 x2 x2 6x 9 (x 3)2 6x 9 (x 3)2 10x 9 (x 9)(x 1) 8x 9 (x 1)(x 9) 10xy 9y2 (x y)(x 9y) 1 (x 1)(x 1) x 1 (x 1)(x 1)

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. 2 2 3. The sum of two squares a b is . 4. Always factor out the first.

5.3

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.

U1V Factoring ax2

bx

c with a

1

3. x 2 5. a2 7. a2

9x 7a 7a

18 10 12

4. w 2 6. b2 8. m 2

6w 7b 9m

8 12 14

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. x 2 4x 3 2. y 2 6y 5

5-25
9. b2 11. x2 13. x2 5b 3x 5x 6 10 24 10. a 2 12. x2 14. a2 5a x 5a 6 12 50

5.3

Factoring the Trinomial ax2

bx

c with a

1

345

49. x2 50. x
2

5x 25x 30 45

150 150 y2 z2

51. 13y 52. 18z

U2V Factoring with Two Variables
Factor each polynomial. If the polynomial is prime, say so. See Examples 2– 4. 15. y 2 16. x 2 17. a
2

Factor each polynomial. See Example 5. 53. x2 54. a2 55. x2 56. y
2

5ax 7ab 4xy yt 13xy 9hs 4xz 5xs 3ab xy
2 2

6a2 10b2 12y 2 12t 2 12y2 9s 2 33z2 24s2 28a2b2 20x2y2 8ab 8mn 1 1

7y 8x 6a 8b 10m 17m 9w 6m 8 m 2a 3x 16 y2 4a 6y 25 1 49 4 12m 21m 3t 5x 18 36 23m 23m 24 24 2t 14t 10t 30t
2

10 15 8 15 16 16 10 16 2w 6m 12 3 m2 10 12 8

18. b2 19. m2 20. m 2 21. w
2

57. x 2 58. h2 59. x 2 60. x
2

22. m 2 23. w 2 24. 25. a 26. x 16
2 2

61. 1 62. 1

63. 15a2b2 64. 12m n

27. 15m 28. 3y 29. a 2 30. y 2 31. z
2

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 66. b 68. x
3 2

5x 49b 8w x3

32. p2 33. h2 34. q
2

67. w

4 2

35. m2 36. m2 37. t2 38. x
2 2

20 20 3 17m 5h 24 24 t2 10t

69. 2w 70. 6w 71. 72. 73. x3 74. x 76. t
3

162 54w2 98 100a 2x2 7x
2

4

10

2b2 a
3

39. m

9x x 7

18

40. h2 41. m 42. m 44. t2 45. t2 46. t 48. t
2 2 2

75. 4r
2 2

2

9 4z2
2

43. 5t

77. x w 78. a4b 79. w2 80. w
2

9x2 a2b3 18w 30w 12w w
3

24 24 200 200

81 81 18

47. t2
2

81. 6w2 82. 9w

346
83. 3y2 84. 5x 85. ax 86. y3 87. 88. 2x a3
2 2

Chapter 5 Factoring

5-26
108. Area of a sail. The area in square meters for a triangular sail is given by A(x) x2 5x 6.

75 500 ay y2 cx 4y cy 4

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?

10x 12 2a2 a 2x4 100w 27w 3w
2

89. 32x2 90. 20w 2 91. 3w 92. w
3 2

40 54 x 3m

18w 36w 36a 6y 3a
Base Area Figure for Exercise 108 x2 5x 6 m2
3

93. 18w2 94. 18a 95. 9y
2 2

w3 3a 1 1

96. 2a2 97. 8vw
3 2

32vw 6ht 30x y 3w 6y2 16w2 3b aw2 ab xc 3t
2 2

32v 36xy3 3xy 2 109. Volume of a cube. Hector designed a cubic box with volume x 3 cubic feet. After increasing the dimensions of the bottom, the box has a volume of x 3 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?

98. 3h2t 99. 6x y 100. 3x 3y 2 101. 5 102. 103. 104. 105. a 3 3y3 4w
3 3

3x 2y 2
2

8w 2y

21y2 3y 20w 3a2 xw2

106. ac

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) x 2 6x 8. a) Find A(6). b) If the width of the deck is x length? 2 feet, then what is the

Getting More Involved
111. Discussion Which of the following products is not equivalent to the others? Explain your answer. a) (2x c) 2(x 4)(x 2)(x 3) 3) b) (x 2)(2x 6) 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(x 2 2x c) 3(x 5)(x 15) 3) b) (3x d) (3x 5)(5x 15)(x 15) 3)

L Area x2 6x 8 ft 2

x

2 ft

Figure for Exercise 107

5-27

5.4

Factoring the Trinomial ax2

bx

c with a

1

347

Mid-Chapter Quiz
Find the prime factorization of each integer. 1. 48 2. 140

Sections 5.1 through 5.3

Chapter 5

Factor completely. 10. 4y2 11. 4h2 12. w
2

9w2 12h 16w 250x 36x 3w 5b a 5x 18x 12as 32s2 6a 6x x
2

Find the greatest common factor for each group of integers. 3. 36, 45 4. 60, 144, 240

9 64

Factor each expression by factoring out the greatest common factor. 5. 8w 7. 15ab
3

13. 10x3 14. 6x
2

54 18 30 1

6y 25a b
2 2

6. 12x3 35a b
3

30x2

15. aw 16. bx 17. ax 18. x3
2

Factor each expression. 8. (x 9. m(m 3)x 9) (x 3)5 9)

4x2

6(m

19. 2x3 20. a
2

5.4
In This Section U1V The ac Method U2V Trial and Error U3V Factoring Completely

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 product 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 ax 2
To factor the trinomial ax2 bx

bx c: c by the ac Method

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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E X A M P L E

1

The ac method
Factor each trinomial. a) 2x2 b) 2x2 c) 10x2 7x x 13x 6 6 3

Solution
a) In 2x2 7x 6 we have a 2, b ac 7, and c 2 6 12. 6. So,

Now we need two integers with a product of 12 and a sum of 7. The pairs of integers with a product of 12 are 1 and 12, 2 and 6, and 3 and 4. Only 3 and 4 have a sum of 7. Replace 7x by 3x 4x and factor by grouping: 2x2 7x 6 2x2 (2x (2x Check by FOIL. b) In 2x2 x 6 we have a 2, b ac 1, and c 2( 6) 12. 6. So, 3x 3)x 3)(x 4x (2x 2) 6
Replace 7x by 3x Factor out 2x 3. 4x.

3)2 Factor out the common factors.

Now we need two integers with a product of 12 and a sum of 1. We can list the possible pairs of integers with a product of 12 as follows: 1 and 12 2 and 2 and 6 3x 6 3 and 3 and 4 4x and factor by grouping:
3x 3. 4x.

4

1 and 12 Only

3 and 4 have a sum of 1. Replace x by 2x2 x 6 2x2 (2x (2x 3x 3)x 3)(x 4x (2x 2) 6

Replace x by Factor out 2x

3)2 Factor out the common factors.

Check by FOIL. c) Because ac 10( 3) 30, we need two integers with a product of 30 and a sum of 13. The product is negative, so the integers must have opposite signs. We can list all pairs of factors of 30 as follows: 1 and 30 2 and 15 3 and 10 5 and 5 and 6 6

1 and 30

2 and 15

3 and 10 2 and 15: 3

The only pair that has a sum of 13 is 10x2 13x 3 10x2 (5x (5x Check by FOIL. 2x 1)2x 1)(2x

15x (5x 3)

Replace 13x by Factor out 5x

2x 1.

15x.

1)3 Factor out the common factors.

Now do Exercises 1–38

5-29

5.4

Factoring the Trinomial ax2

bx

c with a

1

349

E X A M P L E

2

Factoring a trinomial in two variables by the ac method
Factor 8x2 14xy 3y2

Solution
Since a 8, b 24 and a sum of follow: 14, and c 3, we have ac 24. Two numbers with a product of 14 must both be negative. The possible pairs with a product of 24

1 and 2 and Only 2 and grouping: 12 have a sum of 8x2

24 12 14. Replace

3 and 4 and

8 6 2xy 3y2 y)( 3y) 12xy and factor by

14xy by

14xy

3y2

8x2 (4x (4x

2xy y)2x y)(2x

12xy (4x 3y)

Check by FOIL.

Now do Exercises 39–44

U2V Trial and Error
After you have gained some experience at factoring by the ac method, you can often find the factors without going through the steps of grouping. For example, consider the polynomial 3x2
2

7x

6.

The factors of 3x can only be 3x and x. The factors of 6 could be 2 and 3 or 1 and 6. We can list all of the possibilities that give the correct first and last terms, without regard to the signs: (3x 3)(x 2) (3x 2)(x 3) (3x 6)(x 1) (3x 1)(x 6) Because the factors of 6 have unlike signs, one binomial factor is a sum and the other binomial is a difference. Now we try some products to see if we get a middle term of 7x: (3x (3x
U Helpful Hint V
If the trinomial has no common factor, then neither binomial factor can have a common factor.

3)(x 3)(x

2) 2)

3x2 3x2

3x 3x

6 6

Incorrect Incorrect

Actually, there is no need to try (3x 3)(x 2) or (3x 6)(x 1) because each contains a binomial with a common factor. A common factor in the binomial causes a common factor in the product. But 3x2 7x 6 has no common factor. So the factors must come from either (3x 2)(x 3) or (3x 1)(x 6). So we try again: (3x (3x 2)(x 2)(x 3) 3) 3x2 3x2 7x 7x 6 Incorrect 6 Correct

Even though there may be many possibilities in some factoring problems, it is often possible to find the correct factors without writing down every possibility. We can use a bit of guesswork in factoring trinomials. Try whichever possibility you think might work. Check it by multiplying. If it is not right, then try again. That is why this method is called trial and error.

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E X A M P L E

3

Trial and error
Factor each trinomial using trial and error. a) 2x2 5x 3 b) 3x2 11x 6

Solution
a) Because 2x2 factors only as 2x x and 3 factors only as 1 3, there are only two possible ways to get the correct first and last terms, without regard to the signs:

U Helpful Hint V
The ac method is more systematic than trial and error. However, trial U Helpful Hint Vfaster and easier, and error can be especially if your is moresecond trial The ac method first or systematic is correct. and error. However, trial than trial and error can be faster and easier, especially if your first or second trial is correct.

(2x 1)(x 3)

and

(2x 3)(x 1)

Because the last term of the trinomial is negative, one of the missing signs must be , and the other must be . The trinomial is factored correctly as 2x2 Check by using FOIL. b) There are four possible ways to factor 3x2 (3x 1)(x 6) (3x 6)(x 1) 11x 6: 5x 3 (2x 1)(x 3).

(3x 2)(x 3) (3x 3)(x 2)

The first binomials of (3x 6)(x 1) and (3x 3)(x 2) have a common factor of 3. Since there is no common factor in 3x2 11x 6, we can rule out both of these possibilities. Since the last term in 3x2 11x 6 is positive and the middle term is negative, both signs in the factors must be negative. So the correct factorization is either (3x 1)(x 6) or (3x 2)(x 3). By using FOIL we can verify that (3x 2)(x 3) 3x2 11x 6. So the polynomial is factored correctly as 3x2 11x 6 (3x 2)(x 3).

Now do Exercises 45–64

Factoring by trial and error is not just guessing. In fact, if the trinomial has a positive leading coefficient, we can determine in advance whether its factors are sums or differences.

Using Signs in Trial and Error 1. If the signs of the terms of a trinomial are , then both factors are sums: x2 5x 6 (x 2)(x 3). 2. If the signs are , then both factors are differences: x2 5x 6 (x 2)(x 3). 3. If the signs are or , then one factor is a sum and the other is a difference: x2 x 6 (x 3)(x 2) and x2 x 6 (x 3)(x 2). In Example 4 we factor a trinomial that has two variables.

5-31

5.4

Factoring the Trinomial ax2

bx

c with a

1

351

E X A M P L E

4

Factoring a trinomial with two variables by trial and error
Factor 6x2 7xy 2y2.

Solution
We list the possible ways to factor the trinomial: (3x 2y)(2x y) (3x y)(2x 2y) (6x 2y)(x y) (6x y)(x 2y)

Note that there is a common factor 2 in (2x 2y) and in (6x 2y). Since there is no common factor of 2 in the original trinomial, the second and third possibilities will not work. Because the last term of the trinomial is positive and the middle term is negative, both factors must contain subtraction symbols. Only the first possibility will give a middle term of 7xy when subtraction symbols are used in both factors. So, 6x2 7xy 2y2 (3x 2y)(2x y).

Now do Exercises 65–74

U3V Factoring Completely
You can use the latest factoring technique along with the techniques that you learned earlier to factor polynomials completely. Remember always to first factor out the greatest common factor (if it is not 1).

E X A M P L E

5

Factoring completely
Factor each polynomial completely. a) 4x3 b) 12x2y 14x2 6xy 6x 6y

Solution
a) 4x3 14x2 6x 2x(2x2 2x(2x Check by multiplying. b) 12x 2y 6xy
2

7x 1)(x

3) 3)

Factor out the GCF, 2x. Factor 2x2 7x 3.

6y

6y(2x2

x

1) Factor out the GCF, 6y.

To factor 2x x 1 by the ac method, we need two numbers with a product of 2 and a sum of 1. Because there are no such numbers, 2x2 x 1 is prime and the factorization is complete.

Now do Exercises 75–84

Our first step in factoring is to factor out the greatest common factor (if it is not 1). If the first term of a polynomial has a negative coefficient, then it is better to factor out the opposite of the GCF so that the resulting polynomial will have a positive leading coefficient.

352

Chapter 5 Factoring

5-32

E X A M P L E

6

Factoring out the opposite of the GCF
Factor each polynomial completely. a) b) 18x3 3a2 51x2 2a 21 15x

Solution
a) The GCF is 3x. Because the first term has a negative coefficient, we factor out 3x: 18x3 51x2
2

15x

3x(6x2 3x(3x

17x 1)(2x

5)

Factor out

3x. 17x 5.

5) Factor 6x2

b) The GCF for 3a 2a cient, factor out 1: 3a2 2a 21

21 is 1. Because the first term has a negative coeffi1(3a2 1(3a 2a 7)(a 21) Factor out
2

1.

3) Factor 3a 2a 21. Now do Exercises 85–100

Warm-Ups
Fill in the blank.

▼
True or false?
3. 4. 5. 6. 7. 8. 2x2 2x2 3x2 2x2 2x2 12x2 3x 5x 10x 7x 16x 13x 1 3 (2x (2x 3 (3x 9 (2x 9 (2x 3 (3x 1)(x 1) 1)(x 3) 1)(x 3) 9)(x 1) 9)(2x 1) 1)(4x 3)

1. If there are no two integers that have a of ac and a of b, then ax2 bx c is prime. 2. In the method we make educated guesses at the factors and then check by FOIL.

5.4

Exercises
U Study Tips V
• Pay particular attention to the examples that your instructor works in class or presents to you online. • The examples and homework assignments should give you a good idea of what your instructor expects from you.

U1V The ac Method
Find the following. See Example 1. 1. Two integers that have a product of 12 and a sum of 7 2. Two integers that have a product of 20 and a sum of 12

3. Two integers that have a product of 30 and a sum of 4. Two integers that have a product of 36 and a sum of 5. Two integers that have a product of 12 and a sum of

17 20 4

5-33
6. Two integers that have a product of of 7 8 and a sum

5.4

Factoring the Trinomial ax2

bx

c with a

1

353
1 7 5 2 5 2 5 10 10 y2 5b2

47. 6x2 49. 5a2 51. 4w2 53. 15x2 55. 8x2 57. 15x2 59. 4x2 61. 2x2 63. 3x2 65. 10x2 67. 42a2

5x 11a 8w x 6x 31x 4x 18x x

1 2 3 2 1 2 3 90 10

48. 15y2 50. 3y2 52. 6z2 54. 15x2 56. 8x2 58. 15x2 60. 4x2 62. 3x2 64. 3x2 y2 b2 66. 8x2 68. 10a2

8y 10y 13z 13x 22x 31x 12x 11x 17x 2xy 27ab

Each of the following trinomials is in the form ax2 bx c. For each trinomial, find two integers that have a product of ac and a sum of b. Do not factor the trinomials. See Example 1. 7. 6x2 9. 6y
2

7x 11y w

2 3 1

8. 5x2 10. 6z
2

17x 19z 17t

6 10 4

11. 12w2

12. 15t2

Factor each trinomial using the ac method. See Example 1. See the Strategy for Factoring ax2 bx c by the ac Method box on page 347. 14. 2x2 11x 5 13. 2x2 3x 1 15. 2x2 17. 3t 2 19. 2x2 21. 6x
2

9x 7t 5x 7x 5x 7x 13b 11y 2x 2x 9t 13x

4 2 3 3 4 6 6 3 1 1 2 2

16. 2h2 18. 3t2 20. 3x2 22. 21x 24. 6x2 26. 3a2 28. 7y2 30. 35x 32. 6x2 34. 8x2 36. 9t2 38. 15x2
2 2

7h 8t x 2x

3 5 2 3

3xy 13ab

Complete the factoring. 69. 70. 71. 72. 73. 74. 3x2 2x2 5x2 4x2 6a2 4b2 7x 2 (x 2)( x 15 (x 3)( 11x 2 (5x 1)( 19x 5 (4x 1)( 17a 5 (3a 1)( 16b 15 (2b 5)( ) ) ) ) ) )

23. 3x2 25. 2x2 27. 5b2 29. 4y
2

5x 3 14a 15 16y 2x 15 1

U3V Factoring Completely
Factor each polynomial completely. See Examples 5 and 6. 75. 81w3 77. 4w2 w 2w 12x2 11w 3zx 21x2 2ab xy2 t2 2t3 30 36x 35 18z 18x 15b2 3y2 2t 4t 2 76. 81w3 78. 2x2 80. 24y 82. 8y2 84. a2b 86. 8x3 w2 28x 12y2 14y 2ab 4x2 2a2b 6x 6t 14t2 6 12 4t 98 12 15 15b 2x 15a2

31. 3x2 33. 8x2 35. 9t2 37. 15x2

4x 5 10x 3 5t 7x 4 2

79. 27 81. 6w2

Use the ac method to factor each trinomial. See Example 2. 39. 4a2 41. 6m2 43. 3x2 16ab 7mn 8xy 15b2 5n2 5y2 40. 10x2 42. 3a2 44. 3m2 17xy 2ab 13mn 3y2 21b2 12n2 83. 3x2z 85. 9x3 87. a2 89. 2x2y2

88. a2b2 90. 18x2 92. 36t2

U2V Trial and Error
Factor each trinomial using trial and error. See Examples 3 and 4. 45. 5a2 6a 1 46. 7b2 8b 1

91.

6t3

93. 12t4

94. 12t3

354

Chapter 5 Factoring

5-34
a) Rewrite the formula by factoring the right-hand side completely. b) Use the factored version of the formula to find N(3). c) Use the accompanying graph to estimate the time at which the workers are most efficient. d) Use the accompanying graph to estimate the maximum number of components assembled per hour during an 8-hour shift.

95. 4x2y 8xy2 3y3 96. 9x2 24xy 9y2 97. 4w2 7w 3 98. 30w2 w 1 99. 12a3 22a 2b 6ab2 100. 36a2b 21ab2 3b3

Applications
Solve each problem. 101. Height of a ball. If a ball is thrown straight upward at 40 feet per second from a rooftop 24 feet above the ground, then its height in feet above the ground t seconds after it is thrown is given by h(t) 16t2 40t 24. a) Find h(0), h(1), h(2), and h(3). b) Rewrite the formula with the polynomial factored completely. c) Find h(3) using the result of part (b).
40 ft/sec

Getting More Involved
103. Exploration Find all positive and negative integers b for which each polynomial can be factored. a) x2 bx 3 c) 2x2 bx 15 104. Exploration Find two integers c (positive or negative) for which each polynomial can be factored. Many answers are possible. b) 3x2 bx 5

h(t)

16 t 2

40 t

24

a) x 2 x c b) x2 2x c c) 2x 2 3x c 105. Cooperative learning

Figure for Exercise 101

102. Worker efficiency. In a study of worker efficiency at Wong Laboratories it was found that the number of components assembled per hour by the average worker t hours after starting work could be modeled by the formula N(t) 3t3 23t2 8t.

Working in groups, cut two large squares, three rectangles, and one small square out of paper that are exactly the same size as shown in the accompanying figure. Then try to place the six figures next to one another so that they form a large rectangle. Do not overlap the pieces or leave any gaps. Explain how factoring 2x2 3x 1 can help you solve this puzzle. x x x x 1 x 1 x 1 1 x 1

300 Number of components

200 Figure for Exercise 105 100

106. Cooperative learning Working in groups, cut four squares and eight rectangles out of paper as in Exercise 105 to illustrate the trinomial 4x2 7x 3. Select one group to demonstrate how to arrange the 12 pieces to form a large rectangle. Have another group explain how factoring the trinomial can help you solve this puzzle.

0

0 1 2 3 4 5 6 7 8 Time (hours)

Figure for Exercise 102

5-35

5.5

Difference and Sum of Cubes and a Strategy

355

5.5
In This Section U1V Factoring a Difference or Sum of Two Cubes

Difference and Sum of Cubes and a Strategy

In Sections 5.1 to 5.4, we established the general idea of factoring and some special cases. In this section we will see two more special cases. We will then summarize all of the factoring that we have done with a factoring strategy.

U2V Factoring a Difference of Two
Fourth Powers 3V The Factoring Strategy U

U1V Factoring a Difference or Sum of Two Cubes
We can use division to discover that a b is a factor of a3 b3 (a difference of two cubes) and a b is a factor of a3 b3 (a sum of two cubes): a a2 b a3 a3 ab b2 0a2b 0ab2 a2b a2b 0ab2 a2b ab2 ab2 ab2 b
3

a

a2 b a3 a3

b3 b3 0

ab b2 0a2b 0ab2 a2b a2b 0ab2 a2b ab2 ab2 ab2

b3

b3 b3 0

In each division the remainder is 0. So in each case the dividend is equal to the divisor times the quotient. These results give us two new factoring rules. Factoring a Difference or Sum of Two Cubes a3 b3 (a b)(a2 ab a3 b3 (a b)(a2 ab

b2) b2)

Use the following strategy to factor a difference or sum of two cubes.

Strategy for Factoring a3

b3 or a3

b3

1. The first factor is the original polynomial without the exponents, and the middle term in the second factor has the opposite sign from the first factor:

a3 – b3

(a – b)(a2 ab ↑ ↑ opposite signs

b2)

a3

b3

(a

opposite signs

b)(a2 – ab ↑ ↑

b2)

2. Recall the two perfect square trinomials a2 2ab b2 and a2 – 2ab b2. The second factor is almost a perfect square trinomial. Just delete the 2.

It is helpful also to compare the differences and sums of squares and cubes: a2 a3 b2 b3 (a (a b)(a b) b)(a2 ab b
2

)

a2 a3

b2 Prime b3 (a b)(a2

ab

b2)

The factors a2 ab b2 and a2 ab b2 are prime. They can’t be factored. The 2ab b2, which are almost the perfect square trinomials a2 2ab b2 and a2 same, are not prime. They can be factored: a2 2ab b2 (a b)2 and a2 2ab b2 (a b)2.

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Chapter 5 Factoring

5-36

E X A M P L E

1

Factoring a difference or sum of two cubes
Factor each polynomial. a) w3 8 b) x3 1 c) 8y3 27

Solution
a) Because 8 23, w3 8 is a difference of two cubes. To factor w3 a w and b 2 in the formula a3 b3 (a b)(a2 ab b2): w3 8 (w 2)(w2 2w 4) x and b 1 in 8, let

b) Because 1 13, the binomial x3 1 is a sum of two cubes. Let a the formula a3 b3 (a b)(a2 ab b2): x3 c) 8y3 27 (2y)3 (2y 33 3)(4y 2 6y 1 (x 1)(x2 x 1)

This is a difference of two cubes.

9) Let a

2y and b

3 in the formula.

Now do Exercises 1–16

In Example 1, we used the first three perfect cubes, 1, 8, and 27. You should verify that 1, 8, 27, 64, 125, 216, 343, 512, 729, and 1000 are the first 10 perfect cubes.
CAUTION The polynomial (a and b 1, then

b)3 is not equivalent to a3 (a b)3 (2 1)3 13 1

b3 because if a

2

and a3 Likewise, (a b3 23 13 8 1 b3. 7.

b)3 is not equivalent to a3

U2V Factoring a Difference of Two Fourth Powers
A difference of two fourth powers of the form a4 b4 is also a difference of two 2 2 2 2 (b ) . It can be factored by the rule for factoring a difference of two squares, (a ) squares: a4 b4

(a2)2 (a2
(a

(b2)2 b2)(a2
b)(a

Write as a difference of two squares.

b) b)(a2
2

Difference of two squares

b

2

)

Factor completely.

Note that the sum of two squares a2

b2 is prime and cannot be factored.

E X A M P L E

2

Factoring a difference of two fourth powers
Factor each polynomial completely. a) x4 16 b) 81m4 n4

5-37

5.5

Difference and Sum of Cubes and a Strategy

357

Solution
a) x4 16

(x2)2 42 (x2 4)(x2
(x 2)(x n4

Write as a difference of two squares.

4) 2)(x
2

Difference of two squares

4)

Factor completely. Write as a difference of two squares.

b) 81m4

(9m2)2 (n2)2 (9m2 n2)(9m2
(3m n)(3m

n

2

)
2

Factor.

n)(9m

n

2

)

Factor completely.

Now do Exercises 17–24 CAUTION A difference of two squares or cubes can be factored, and a sum of two cubes can be factored. But the sums of two squares x2 4 and 9m2 n2 in Example 2 are prime.

U3V The Factoring Strategy
The following is a summary of the ideas that we use to factor a polynomial completely.

Strategy for Factoring Polynomials Completely
1. Factor out the GCF (with a negative coefficient if necessary). 2. When factoring a binomial, check to see whether it is a difference of two squares, a difference of two cubes, or a sum of two cubes. A sum of two squares does not factor. 3. When factoring a trinomial, check to see whether it is a perfect square trinomial. 4. If the polynomial has four terms, try factoring by grouping. 5. When factoring a trinomial that is not a perfect square, use the ac method or the trial-and-error method. 6. Check to see whether any of the factors can be factored again.

We will use the factoring strategy in Example 3.

E X A M P L E

3

Factoring polynomials
Factor each polynomial completely. a) 2a2b c) 3x4 24ab 15x3 72b 72x2 b) 3x3 d) 60y3 6x2 85y2 75x 25y 150

Solution
a) 2a2b 24ab 72b 2b(a2 12a 2b(a 6)2 150 36)
First factor out the GCF, 2b. Factor the perfect square trinomial.

b) 3x3

6x2

75x

3[x3 2x 2 25x 50] Factor out the GCF, 3. 3[x 2(x 2) 25(x 2)] Factor out common factors. 3(x 2 25)(x 2) Factor by grouping. 3(x 5)(x 5)(x 2) Factor the difference of two squares.

358

Chapter 5 Factoring

5-38
c) Factor out 3x2 to get 3x4 15x3 72x2 3x2(x2 5x 24). To factor the trinomial, find two numbers with a product of 24 and a sum of 5. For a product of 24 we have 1 24, 2 12, 3 8, and 4 6. To get a sum of 5 and a product of 24 choose 8 and 3: 3x4 15x3 72x2 3x2(x2 5x 3x2(x 3)(x 24) 8)

d) Factor out 5y to get 60y3 85y2 25y 5y(12y2 17y 5). By the ac method we need two numbers that have a product of 60 (ac) and a sum of 17. The numbers are 20 and 3. Now factor by grouping: 60y3 85y2 25y 5y(12y2 17y 5y(12y2 20y 5y[4y(3y 5) 5y(3y 5)(4y 5) Factor out 5y. 3y 5) 17y 20y 3y 1(3y 5)] Factor by grouping. 1) Factor out 3y 5.

Now do Exercises 25–92

Warm-Ups
Fill in the blank.

▼
True or false?
5. For any real number x, x2 4 (x 2)2. 6. The trinomial 4x2 6x 9 is a perfect square trinomial. 7. The binomial 4y2 25 is prime. 8. If the GCF is not 1, then you should factor it out first. 9. You can factor y2 5y my 5m by grouping. 10. You can factor x2 ax 3x 3a by grouping.

1. If there is no , then the dividend is the divisor times the quotient. 2. The binomial a3 b3 is a of two cubes. 3. The binomial a3 b3 is a of two cubes. 4. If a3 b3 is divided by a b, then the remainder is .

5.5

Exercises
U Study Tips V
• If you have a choice, sit at the front of the class. It is easier to stay alert when you are at the front. • If you miss what is going on in class, you miss what your instructor feels is important and most likely to appear on tests and quizzes.

U1V Factoring a Difference or Sum of Two Cubes
Factor each difference or sum of cubes. See Example 1. 1. m3 1 2. z3 27

3. 4. 5. 6.

x3 y3 a3 b3

8 27 125 216

5-39
7. c3 8. d
3

5.5

Difference and Sum of Cubes and a Strategy

359

343 1000 1 1

37. 9x 2 38. 9x
2 2

6x 6x 1 25 z4 1 xy xy2 10y 20x
2

1 3

9. 8w3 10. 125m3 11. 8t3 12. 125n3 13. x3 14. m3 15. 8t3 16. u
3

39. 9m 40. 4b 42. y4 41. w4 8

2

27 y3 n3 y3 125v
3

43. 6x 2y 44. 5x 2y 2 45. y 2 46. x
2

2y 6y2 25 25 3 18 2 6

47. 48a

24a 24b 4m 4a 16t q4 24a 18x
4

U2V Factoring a Difference of Two Fourth Powers
Factor each polynomial completely. See Example 2. 17. x 4 18. m4 19. x4 20. a4 21. 16b4 22. 625b4 23. a4 24. 16a4 y4 n4 1 81 1 1 m4 81b4

48. 8b

2 2

49. 16m 50. 32a 51. s
4

2

52. 81 53. 9a2 54. 3x 56. 4x
2 2

16 48 6 12

55. 24x
2

26x 6x 27 20b2 27a 25a

57. 3m 58. 5a

2

2

U3V The Factoring Strategy
Factor each polynomial completely. If a polynomial is prime, say so. See Example 3. See the Strategy for Factoring Polynomials Completely box on page 357. 25. 2x 26. 3x 27. a2 28. x
2 2 2 2 3

59. 3a2 60. a 61. 8 62. x 3 63. w
2 2 2

2x 2 6x 2 4t
2

9x

18 12x 4 y
2

64. 9x 66. x 60 27 4x 6a
3 3

4y2 5x 2 2x
2

65. 6x 3 67. a b 68. 2m2 69. x 3 70. 71. 72. x3 73. 2x
3

12x x 2

29. 4x
3 3

8x 18x 4x
2 2

4ab 1800 2x 2 4x 28mn3 x 2ma2 1 a3 8 50x

30. 3x 31. x

32. a

5a

2x 3 7m3n x2

33. 5max 2 34. 3bmw2 35. 2x 36. 3x
2 2

20ma 12bm 1 5

3x 8x

16

74. m a

2

360
75. 2w4 76. m n 77. 3a w 78. 8a 79. 5x
3 2 2 4

Chapter 5 Factoring

5-40

16w mn 4a 500 16y 2 2n 5b 3x 81a2 4w 8w 7a 3 3y 4 4 5 30a 2 20y 3 9aw 14bn wm bw wn 5a
4

Extra Factoring Exercises
Factor each polynomial completely. 27w 97. 3w 2 98. 4z2 99. 81 100. 9 101. w2 102. 6z
2

18aw

30w 16z b2 4p2 8w 12z 6x 36m 5a 3w 7x 6x
2 2

75 16

80. 25x 2 81. 2m 82. aw 83. 3x
4

103. 3x 2 104. 6m 105. ax 106. w
2 2

105 96 20 cw 3c 12 27 6 40 4x

84. 3a5 85. 4w 2 86. 4w 2 87. a 4 88. 2y 5 89. 4aw3 90. 9bn 3 91. t 2 92. t 3

107. 12x2 108. 8x2 109. 110. 111. w 112. y3 113. y3 114. a
3 4

12aw2 15bn2 6t 12t 2 9 36t

9x 8x
3

15x 4x 27 1 y2 2a
2

y 4a

1 8

Getting More Involved
93. Discussion Are there any values for a and b for which (a b)3 a3 b3? Find a pair of values for a and b for which (a b)3 a3 b3. Is (a b)3 equivalent to a3 b3? Explain your answers. 94. Writing ab Explain why a prime polynomials. 95. Discussion The polynomial a6 1 is a sum of two squares and a sum of two cubes. You can’t factor it as a sum of two squares, but you can factor any sum of two cubes. Factor a6 1. 96. Discussion Factor a
6 2

115. m

81 256 2 ab xy 8b2 12y2 9my3 25w2a3
2

116. t 4 117. a2 118. x2 119. m3y 120. w4a 121. x b and a
2 2 4 5

6m2y2 10w 3a2 2x 6y y3 16a 2 18x 17m
3 2 3

4x

ab

b are

2

122. y

4

9y3

123. y 7 124. a6 125. x 2 126. m 127. 128. 129. x
4 2

72 72
2

6a

5a

4a 18

12 x

15x

8x ab3 24tx 30yz 9x2 25z2

b and a

6

6

b completely.

6

130. a4 131. 16t 2 132. 9y2

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Solving Quadratic Equations by Factoring

361

5.6
In This Section U1V The Zero Factor Property U2V Fractions and Decimals U3V Applications

Solving Quadratic Equations by Factoring

The techniques of factoring can be used to solve equations involving polynomials. These equations cannot be solved by the other methods that you have learned. After you learn to solve equations by factoring, you will use this technique to solve some new types of problems.

U1V The Zero Factor Property
In this chapter you learned to factor polynomials such as x 2 x 2 x 6 0 is called a quadratic equation. Quadratic Equation If a, b, and c are real numbers with a 0, then ax 2 bx c 0 is called a quadratic equation. A quadratic equation always has a second-degree term because it is specified in the definition that a is not zero. The main idea used to solve quadratic equations, the zero factor property, is simply a fact about multiplication by zero. The Zero Factor Property The equation a b 0 is equivalent to a 0 or b 0. x 6. The equation

We will use the zero factor property most often to solve quadratic equations that have two factors, as shown in Example 1. However, this property holds for more than two factors as well. If a product of any number of factors is zero, then at least one of the factors is zero. The following strategy gives the steps to follow when solving a quadratic equation by factoring. Of course, this method applies only to quadratic equations in which the quadratic polynomial can be factored. Methods that can be used for solving all quadratic equations are presented in Chapter 10.

Strategy for Solving an Equation by Factoring
1. 2. 3. 4. 5. 6.

Rewrite the equation with 0 on one side. Factor the other side completely. Use the zero factor property to get simple linear equations. Solve the linear equations. Check the answer in the original equation. State the solution(s) to the original equation.

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5-42

E X A M P L E

1

Using the zero factor property
Solve x 2 x 6 0.

Solution
First factor the polynomial on the left-hand side:

U Helpful Hint V
Some students grow up believing that the only way to solve an equation is to “do the same thing to each side.” Then along come quadratic equations and the zero factor property. For a quadratic equation, we write an equivalent compound equation that is not obtained by “doing the same thing to each side.”

x We now check that For x x2 x

3 x

0 3

x2 x 6 (x 3)(x 2) or x 2 or x

0 0 Factor the left-hand side. 0 Zero factor property 2 Solve each equation. For x 2: x2 x 6

3 and 2 satisfy the original equation. 3: 6 ( 3)2 9 3 0 6 2) ( 3) 6 6 (2)2 (2) 4 2 6 0 6

The solutions to x 2 x the equation (x 3)(x

0 are 3 and 2. Checking 3 and 2 in the factored form of 0 will help you understand the zero factor property: 3)( 3 3)(2 2) 2) (0)( 5) (5)(0) 0 0 (2

( 3

For each solution to the equation, one of the factors is zero and the other is not zero. All it takes to get a product of zero is one of the factors being zero.

Now do Exercises 1–12

A sentence such as x 3 or x 2, which is made up of two or more equations connected with the word “or,” is called a compound equation. In Example 2, we again solve a quadratic equation by using the zero factor property to write a compound equation.

E X A M P L E

2

Using the zero factor property
Solve the equation 3x 2 3x.

Solution
First rewrite the equation with 0 on the right-hand side: 3x 2 3x 3x 3x(x 1) or x 1 or x
2

3x 0 0 0
Add 3x to each side. Factor the left-hand side. Zero factor property

3x x Check 0 and

0 0

1 Solve each equation. 3x. For x 3( 1)2 3 1: 3( 1) 3 1.

1 in the original equation 3x 2 For x 3(0)2 0 0: 3(0) 0

There are two solutions to the original equation, 0 and

Now do Exercises 13–20

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363

CAUTION If in Example 2 you divide each side of 3x 2 3x by 3x, you would get x 1 but not the solution x 0. For this reason we usually do not divide each side of an equation by a variable.

E X A M P L E

3

Using the zero factor property
Solve (2x 1)(x 1) 14.

Solution
To write the equation with 0 on the right-hand side, multiply the binomials on the left and then subtract 14 from each side: (2x 2x 2x (2x 2x 5 2x x Check
5 2
2

1)(x
2

1) 1 15 3) 3 x x

14 Original equation 14 Multiply the binomials. 0 0 0 3 3
Subtract 14 from each side. Factor. Zero factor property

x x

5)(x or 5 5 2 x or or

0

and 3 in the original equation: 2 5 2 1 5 2 1 ( 5 ( 4) 14 (2 3 1)(3 1) (7)(2) 14 1) 7 2 5 2 2 2

So the solutions are

5 2

and 3.

Now do Exercises 21–26

CAUTION In Example 3, we started with a product equal to 14. Because 1 14 14, 1 1 2 7 14, 14, and so on, we cannot make any 28 14, 3 42 2 conclusion about the factors that have a product of 14. If the product of two factors is zero, then we can conclude that one or the other factor is zero.

If a perfect square trinomial occurs in a quadratic equation with 0 on one side, then there are two identical factors of the trinomial. In this case it is not necessary to set both factors equal to zero. The solution can be found from one factor.

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E X A M P L E

4

An equation with a repeated factor
Solve 5x2 30x 45 0.

Solution
Notice that the trinomial on the left-hand side has a common factor: 5x 2 5(x
2

30x

45

0 0 0 0 0 3
Factor out the GCF. Factor the perfect square trinomial. Divide each side by 5. Zero factor property

6x 9) 5(x 3)2 (x 3)2 x 3 x

Even though x x 3 0. If x

3 occurs twice as a factor, it is not necessary to write x 3 in 5x2 30x 45 0, we get 5 32 30 3 45 0,

3

0 or

which is correct. So the only solution to the equation is 3.

Now do Exercises 27–30 CAUTION Do not include 5 in the solution to Example 4. Dividing by 5 eliminates it. Instead of dividing by 5 we could have applied the zero factor property to 5(x 3)2 0. Since 5 is not 0, we must have (x 3)2 0 or x 3 0.

If the left-hand side of the equation has more than two factors, we can write an equivalent equation by setting each factor equal to zero.

E X A M P L E

5

An equation with three solutions
Solve 2x 3 x2 8x 4 0.

Solution
We can factor the four-term polynomial by grouping:

U Helpful Hint V
Compare the number of solutions in Examples 1 through 5 to the degree of the polynomial. The number of real solutions to any polynomial equation is less than or equal to the degree of the polynomial. This fact is known as the fundamental theorem of algebra.

2x 3 x 2 8x 4 x (2x 1) 4(2x 1) (x 2 4)(2x 1) (x 2)(x 2)(2x 1) x 2 0 or x 2
2

x To check let x

0 0 Factor out the common factors. 0 Factor out 2x 1. 0 Difference of two squares 0 or 2x 1 0 Zero factor property 1 2 or x 2 or x Solve each equation. 2 2, 1, and 2 in 2x3 x2 8x 4 0:
2

2( 2)3 1 2 2
3

( 2)2 1 2
2

8( 2) 1 8 2 8(2)

4 4 4 2,

0 0 0
1 , 2

2(2)3

22

Since all of these equations are correct, the solutions are

and 2.

Now do Exercises 31–38

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Solving Quadratic Equations by Factoring

365

U2V Fractions and Decimals
If the coefficients in an equation are not integers, we might be able to convert them into integers. Fractions can be eliminated by multiplying each side of the equation by the least common denominator (LCD). To eliminate decimals multiply each side by the smallest power of 10 that will eliminate all of the decimals.

E X A M P L E

6

Converting to Integers
Solve. a) 1 2 x 12 1 x 6 2 0 b) 0.02x2 0.19x 0.1

Solution
a) The LCD for 6 and 12 is 12. So multiply each side of the equation by 12: 1 2 1 x x 2 12 6 1 2 1 x x 2 12 12 6 x 2 2x 24 (x 6)(x 4) 6 0 or x 4 x 6 or x 0 12(0) 0 0 0 4
Original equation Multiply each side by 12. Simplify. Factor. Zero factor property

x Check:

1 1 ( 6)2 ( 6) 12 6 1 1 (4)2 (4) 2 12 6 The solutions are 6 and 4.

2 4 3

3 2 3

1 2

2 0

0

b) Multiply each side by 100 to eliminate the decimals: 0.02x 2 0.19x 100(0.02x 2 0.19x) 2x2 19x 2 2x 19x 10 (2x 1)(x 10) 1 0 or x 10 1 x or x 2
1 2

2x

0.1 Original equation 100(0.1) Multiply each side by 100. 10 Simplify. 0 Get 0 on one side. 0 Factor. 0 Zero factor property 10

The solutions are

and 10. You might want to use a calculator to check.

Now do Exercises 39-46 CAUTION You can multiply each side of the equation in Example 6(a) by 12 to clear the fractions and get an equivalent equation, but multiplying the polynomial 1 2 1 x x 2 by 12 to clear the fractions is not allowed. It would result 12 6 in an expression that is not equivalent to the original.

366

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5-46

Note that all of the equations in this section can be solved by factoring. However, we can have equations involving prime polynomials. Such equations cannot be solved by factoring but can be solved by the methods in Chapter 10.

U3V Applications
There are many problems that can be solved by equations like those we have just discussed.

E X A M P L E

7

Area of a garden
Merida’s garden has a rectangular shape with a length that is 1 foot longer than twice the width. If the area of the garden is 55 square feet, then what are the dimensions of the garden?

Solution
If x represents the width of the garden, then 2x 1 represents the length. See Fig. 5.1. Because the area of a rectangle is the length times the width, we can write the equation x(2x
2x Figure 5.1 1 ft x ft

1)

55.

We must have zero on the right-hand side of the equation to use the zero factor property. So we rewrite the equation and then factor: 2x 2 (2x 2x 11 x The width is certainly not
11 . 2

x

55 5) 5 x

0 0 Factor. 0 Zero factor property 5

11)(x or x

U Helpful Hint V
To prove the Pythagorean theorem start with two identical squares with sides of length a b, and partition them as shown. b c a

0 11 2

or

So we use x 2x 1

5 to get the length: 1 11

2(5)

b

b2

b

We check by multiplying 11 feet and 5 feet to get the area of 55 square feet. So the width is 5 ft, and the length is 11 ft.

Now do Exercises 65–66 a c b b a c c2 c b a c b a c a2 a a b a

The Pythagorean theorem was one of the earliest theorems known to ancient civilizations. It is named for the Greek mathematician and philosopher Pythagoras. Builders from ancient to modern times have used the theorem to guarantee they had right angles when laying out foundations. The Pythagorean theorem says that in any right triangle the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. The Pythagorean Theorem The triangle shown in Fig. 5.2 is a right triangle if and only if a2 b2 c2. a Figure 5.2

Hypotenuse c b

There are eight identical triangles in the diagram. Erasing four of them from each original square will leave smaller squares with areas a2, b2, and c2. Since the original squares had equal areas, the remaining areas must be equal. So a2 b2 c2.

Legs

5-47

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Solving Quadratic Equations by Factoring

367

If you do an Internet search, you can find sites that have many different proofs to this theorem. One proof is shown in the Helpful Hint in the margin.

E X A M P L E

8

Using the Pythagorean theorem
The length of a rectangle is 1 meter longer than the width, and the diagonal measures 5 meters. What are the length and width?

Solution
If x represents the width of the rectangle, then x 1 represents the length. Because the two sides are the legs of a right triangle, we can use the Pythagorean theorem to get a relationship between the length, width, and diagonal. See Fig. 5.3. x2
5

(x
2

1) 2 1 24 12 4) 4 x x

52 25 0 0 0 0 4

Pythagorean theorem Simplify. Divide each side by 2. Zero factor property The length cannot be negative.

x x 2

x 2x 2 x

2x 2x x 3)(x or

2

x Figure 5.3

1

(x x x 3 x 1 3 4

0 or

To check this answer, we compute 32 3 meters by 4 meters.

42

52, or 9

16

25. So the rectangle is

Now do Exercises 67–68

CAUTION The hypotenuse is the longest side of a right triangle. So if the lengths of the sides of a right triangle are 5 meters, 12 meters, and 13 meters, then the length of the hypotenuse is 13 meters, and 52 122 132.

Warm-Ups
Fill in the blank.

▼
True or false?
7. The equation x(x 2) 3 is equivalent to x 3 or x 2 3. 8. Equations solved by factoring always have two different solutions. 9. The equation ad 0 is equivalent to a 0 or d 0. 10. The solution set to (x 1)(x 4) 0 is {1, 4}.

1. A equation has the form ax2 bx c 0 where a 0. 2. A equation is two equations connected with the word “or.” 3. The property says that if ab 0, then a 0 or b 0. 4. Some quadratic equations can be solved by . 5. We do not usually each side of an equation by a variable. 6. The theorem says that a triangle is a right triangle if and only if the sum of the squares of the legs is equal to the square of the hypotenuse.

11. If a, b, and c are the sides of any triangle, then a2 b2 c2. 12. The solution set to 3(x 4)(x 5) 0 is {3, 4, 5}.

5.6

Exercises
U Study Tips V
• Avoid cramming. When you have limited time to study for a test, start with class notes and homework assignments. Work one or two problems of each type. • Don’t get discouraged if you cannot work the hardest problems. Instructors often ask some relatively easy questions to see if you understand the basics.

U1V The Zero Factor Property
Solve by factoring. See Example 1. See the Strategy for Solving an Equation by Factoring box on page 361. 1. (x 2. (a 3. (2x 4. (3k 5. x2 6. x
2

24. (b 25. 2(4

3)(3b 5h) 1)

4) 3h2 1

10

5)(x 6)(a 5)(3x 8)(4k 3x 7x 9w 6t 2y 3q m h 2

4) 5) 4) 3) 0

0 0 0 0 26. 2w(4w

Solve each equation. See Examples 4 and 5. 27. 2x 2 28. 3x 2 29. 4m2 50 48 12m 20y 9x x3 4w2 2a2 3n2 w2 y2 20x 24x 9 4 0 0

12 14 27 24 18 1 3

0 0 0 0 0 0 0

7. w 2 8. t 2 9. y 2 10. q
2

30. 25y2 31. 32. 33. 34. 35. 36. x3 25x w3 a3 n3 w3

11. 2m 2 12. 2h 2

0 0 4w 16 a 2 3 n 25w 25 2y 0 3m 0

37. 6y 3 38. 12m 3

Solve each equation. See Examples 2 and 3. 13. 14. 15. 16. 17. 18. x2 w2 m2 h2 a2 p2 x 2w 7m 5h a 20 p 42 5x 10x 2)(x 2)(x 3)(2a 6) 6) 1) 3 7 12 20 15

13m 2

U2V Fractions and Decimals
Solve each equation. See Example 6. 1 5 39. x 2 x 1 0 6 6 1 2 3 40. x x 1 0 10 10 1 2 41. x2 x 3 0 9 3 1 2 3 42. x x 5 0 10 2 43. 0.01x 2 0.08x 0.2 44. 0.01x 2 45. 0.3x 2 0.07x 0.7x 0.7x 2 1 0.1 0 0

19. 2x2 20. 3x2 21. (x 22. (x 23. (a

46. 0.1x 2

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Solving Quadratic Equations by Factoring

369

Miscellaneous
Solve each equation. 47. x 2 48. x 2 49. 4x 2 50. 25x 2 51. a 52. x 54.
3

16 36 9 1 a 4x
2

0 0

67. Violla’s bathroom. The length of Violla’s bathroom is 2 feet longer than twice the width. If the diagonal measures 13 feet, then what are the length and width? 68. Rectangular stage. One side of a rectangular stage is 2 meters longer than the other. If the diagonal is 10 meters, then what are the lengths of the sides?

3

10 m

53. 3x

15x

18

0

x

2m xm

2x 2 2x 24 0 11 z 6 55. z2 2 8 m 1 56. m2 3 57. (t 3)(t 5) 9 58. 3x(2x 59. (x 2)2
2

Figure for Exercise 68

1) x2

18 10
2

69. Consecutive integers. The sum of the squares of two consecutive integers is 13. Find the integers. 70. Consecutive integers. The sum of the squares of two consecutive even integers is 52. Find the integers. 71. Two numbers. The sum of two numbers is 11, and their product is 30. Find the numbers. 72. Missing ages. Molly’s age is twice Anita’s. If the sum of the squares of their ages is 80, then what are their ages? 73. Three even integers. The sum of the squares of three consecutive even integers is 116. Find the integers. 74. Two odd integers. The product of two consecutive odd integers is 63. Find the integers. 75. Consecutive integers. The product of two consecutive integers is 5 more than their sum. Find the integers. 76. Consecutive even integers. If the product of two consecutive even integers is 34 larger than their sum, then what are the integers? 77. Two integers. Two integers differ by 5. If the sum of their squares is 53, then what are the integers? 78. Two negative integers. Two negative integers have a sum of 10. If the sum of their squares is 68, then what are the integers? 79. Lucy’s kids. The sum of the squares of the ages of Lucy’s two kids is 100. If the boy is two years older than the girl, then what are their ages?

60. (x 3) (x 2) 17 1 2 1 1 61. x x 8 2 16 1 2 1 62. h h 1 0 2 18 63. a3 3a2 25a 75 64. m4 m3 100m2 100m

U3V Applications
Solve each problem. See Examples 7 and 8. 65. Dimensions of a rectangle. The perimeter of a rectangle is 34 feet, and the diagonal is 13 feet long. What are the length and width of the rectangle? 66. Address book. The perimeter of the cover of an address book is 14 inches, and the diagonal measures 5 inches. What are the length and width of the cover?

ADDRESS BOOK
5 in.

Figure for Exercise 66

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5-50
87. Throwing a sandbag. A balloonist throws a sandbag downward at 24 feet per second from an altitude of 720 feet. Its height (in feet) above the ground after t seconds is given by S(t) 16t2 24t 720. a) Find S(1). b) What is the height of the sandbag 2 seconds after it is thrown? c) How long does it take for the sandbag to reach the ground? [On the ground, S(t) 0.] 88. Throwing a wrench. An angry construction worker throws his wrench downward from a height of 128 feet with an initial velocity of 32 feet per second. The height of the wrench above the ground after t seconds is given by S(t) 16t2 32t 128. a) What is the height of the wrench after 1 second? b) How long does it take for the wrench to reach the ground? 89. Glass prism. One end of a glass prism is in the shape of a triangle with a height that is 1 inch longer than twice the base. If the area of the triangle is 39 square inches, then how long are the base and height?

80. Sheri’s kids. The sum of the squares of the ages of Sheri’s three kids is 114. If the twin girls are three years younger than the boy, then what are their ages? 81. Area of a rectangle. The area of a rectangle is 72 square feet. If the length is 6 feet longer than the width, then what are the length and the width? 82. Area of a triangle. The base of a triangle is 4 inches longer than the height. If its area is 70 square inches, then what are the base and the height? 83. Legs of a right triangle. The hypotenuse of a right triangle is 15 meters. If one leg is 3 meters longer than the other, then what are the lengths of the legs? 84. Legs of a right triangle. If the longer leg of a right triangle is 1 cm longer than the shorter leg and the hypotenuse is 5 cm, then what are the lengths of the legs? 85. Skydiving. If there were no air resistance, then the height (in feet) above the earth for a skydiver t seconds after jumping from an airplane at 10,000 feet would be given by h(t) 16t2 10,000.

a) Find the time that it would take to fall to earth with no air resistance; that is, find t for which h(t) 0. A skydiver actually gets about twice as much free fall time due to air resistance. b) Use the accompanying graph to determine whether the skydiver (with no air resistance) falls farther in the first 5 seconds or the last 5 seconds of the fall. c) Is the skydiver’s velocity increasing or decreasing as she falls?

2x

1 in.

10 9 8 7 6 5 4 3 2 1 0

Height (thousands of feet)

x in. Figure for Exercise 89

90. Areas of two circles. The radius of a circle is 1 meter longer than the radius of another circle. If their areas differ by 5 square meters, then what is the radius of each?
0 5 10 15 20 Time (seconds) 25

Figure for Exercise 85

91. Changing area. Last year Otto’s garden was square. This year he plans to make it smaller by shortening one side 5 feet and the other 8 feet. If the area of the smaller garden will be 180 square feet, then what was the size of Otto’s garden last year? 92. Dimensions of a box. Rosita’s Christmas present from Carlos is in a box that has a width that is 3 inches shorter than the height. The length of the base is 5 inches longer than the height. If the area of the base is 84 square inches, then what is the height of the package?

86. Skydiving. If a skydiver jumps from an airplane at a height of 8256 feet, then for the first five seconds, her height above the earth is approximated by the formula h(t) 16t2 8256. How many seconds does it take her to reach 8000 feet?

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371

x in.

MIAMI

B

x

3 in. 13 mi

x

5 in.

Figure for Exercise 92

N
A Figure for Exercise 97

93. Flying a kite. Imelda and Gordon have designed a new kite. While Imelda is flying the kite, Gordon is standing directly below it. The kite is designed so that its altitude is always 20 feet larger than the distance between Imelda and Gordon. What is the altitude of the kite when it is 100 feet from Imelda? 94. Avoiding a collision. A car is traveling on a road that is perpendicular to a railroad track. When the car is 30 meters from the crossing, the car’s new collision detector warns the driver that there is a train 50 meters from the car and heading toward the same crossing. How far is the train from the crossing? 95. Carpeting two rooms. Virginia is buying carpet for two square rooms. One room is 3 yards wider than the other. If she needs 45 square yards of carpet, then what are the dimensions of each room? 96. Winter wheat. While finding the amount of seed needed to plant his three square wheat fields, Hank observed that the side of one field was 1 kilometer longer than the side of the smallest field and that the side of the largest field was 3 kilometers longer than the side of the smallest field. If the total area of the three fields is 38 square kilometers, then what is the area of each field? 97. Sailing to Miami. At point A the captain of a ship determined that the distance to Miami was 13 miles. If she sailed north to point B and then west to Miami, the distance would be 17 miles. If the distance from point A to point B is greater than the distance from point B to Miami, then how far is it from point A to point B? 98. Buried treasure. Ahmed has half of a treasure map, which indicates that the treasure is buried in the desert 2x 6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to

the north, and then walk 2x 4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x? 99. Broken Bamboo I. A 10 chi high bamboo stalk is broken by the wind. The top touches the ground 3 chi from its base as shown in the accompanying figure. At what height did the stalk break? This problem appeared in a book by Chinese mathematician Yang Hui in 1261.

3 chi Figure for Exercise 99

100. Broken Bamboo II. A section of bamboo that is 5 chi in length is broken from a stalk of bamboo of unknown height. If the broken section touches the ground 3 chi from the base as in Exercise 99, then what was the original height of the bamboo stalk? 101. Emerging markets. Catarina’s investment of $16,000 in an emerging market fund grew to $25,000 in two years. Find the average annual rate of return by solving the equation 16,000(1 r)2 25,000. 102. Venture capital. Henry invested $12,000 in a new restaurant. When the restaurant was sold two years later, he received $27,000. Find his average annual return by solving the equation 12,000(1 r) 2 27,000.

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5-52

5

Chapter

Wrap-Up

Summary
Examples A positive integer larger than 1 that has no integral factors other than 1 and itself A polynomial that cannot be factored is prime. 1. Find the GCF for the coefficients of the monomials. 2. Form the product of the GCF of the coefficients and each variable that is common to all of the monomials, where the exponent on each variable equals the smallest power of that variable in any of the monomials. Use the distributive property to factor out the GCF from all terms of a polynomial. 2, 3, 5, 7, 11 x2 3 and x 2

Factoring Prime number

Prime polynomial Strategy for finding the GCF for monomials

x

5 are prime.

12x 3yz, 8x 2y3 GCF 4x 2y

Factoring out the GCF Special Cases Difference of two squares Perfect square trinomial Difference or sum of two cubes Factoring Polynomials Factoring by grouping

2x 3

4x

2x(x 2

2)

Examples a
2

b

2

(a b2 b2 (a (a

b)(a

b) b)2 b)2 ab ab b2) b2)

m2

9

(m

3)(m

3)

a2 a2 a3 a3

2ab 2ab b3 b3

(a (a b)(a2 b)(a2

x 2 6x 9 (x 3)2 4h2 12h 9 (2h 3)2 t3 p3 8 1 (t 2)(t 2 2t 4) (p 1)( p2 p 1)

Examples Factor out common factors from groups of terms. 6x 6w ax 6(x w) (6 a)(x aw a(x w) w)

Strategy for factoring ax 2 bx c by the ac method

1. Find two numbers that have a product equal to ac and a sum equal to b. 2. Replace bx by two terms using the two new numbers as coefficients. 3. Factor the resulting four-term polynomial by grouping.

6x 2

17x 12 6x 2 9x 8x 12 (2x 3)3x (2x 3)4 (2x 3)(3x 4)

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Chapter 5 Enriching Your Mathematical Word Power

373

Factoring by trial and error Strategy for factoring polynomials completely

Try possible factors of the trinomial and check by using FOIL. If incorrect, try again. 1. First factor out the greatest common factor. 2. When factoring a binomial, check to see whether it is a difference of two squares, a difference of two cubes, or a sum of two cubes. The sum of two squares (with no common factor) is prime. 3. When factoring a trinomial, check to see whether it is a perfect square trinomial. 4. If the polynomial has four terms, try factoring by grouping. 5. When factoring a trinomial that is not a perfect square, use the ac method or trial and error. 6. Check to see whether any factors can be factored again.

2x 2

5x

12

(2x

3)(x

4)

x4 x2 x3 x3 x2 x2 x2 x2 x2 x4

4x 2 x 2(x 2 4) 4 (x 2)(x 2) 8 (x 2)(x 2 2x 8 (x 2)(x 2 2x 4 is prime. 6x 6x bx 7x 4x2 9 9 2x

4) 4)

(x 3)2 (x 3)2 2b x(x b) 2(x (x 2)(x b) 12 (x 3)(x 4) x2(x2 4) x2(x 2)(x

b)

2)

Solving Equations Zero factor property The equation a b 0 is equivalent to a 0 or b 0.

Examples x(x 1) 0 x 0 or x 1 0

Strategy for solving an equation by factoring

1. Rewrite the equation with 0 on the rightx 2 3x 18 hand side. x 2 3x 18 0 2. Factor the left-hand side completely. (x 6)(x 3) 0 3. Set each factor equal to zero to get linear x 6 0 or x 3 0 equations. x 6 or x 3 4. Solve the linear equations. 5. Check the answers in the original equation. ( 6)2 3( 6) 18, 32 3(3) 6. State the solution(s) to the original equation. The solutions are 6 and 3.

18

Enriching Your Mathematical Word Power
Fill in the blank. 1. A number is an integer greater than 1 that has no integral factors other than itself and 1. 2. An integer larger than 1 that is not prime is 3. A polynomial that has no factors is a 4. Writing a polynomial as a product is 5. Writing a polynomial as a product of primes is factoring . 6. The largest integer that is a factor of two or more integers is the common factor. 7. The square of a monomial in which the coefficient is an integer is a square. . . polynomial. 8. The trinomial a2 trinomial. 9. The polynomial a3 10. The polynomial a 11. A ax2 bx
3

2ab

b2 is a perfect of two cubes. of two cubes.

b3 is a b is a
3

equation is an equation of the form c 0. factor property, if ab 0 then

12. According to the a 0 or b 0.

13. The theorem indicates that a triangle is a right triangle if and only if the sum of the squares of the legs is equal to the square of the hypotenuse.

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Chapter 5 Factoring

5-54

Review Exercises
5.1 Factoring Out Common Factors Find the prime factorization for each integer. 1. 144 3. 58 5. 150 2. 121 4. 76 6. 200 43. r 2 45. y2 47. u2 4r 6y 26u 60 55 120 44. x 2 46. a2 48. v2 13x 6a 22v 40 40 75

Find the greatest common factor for each group. 7. 36, 90 9. 8x, 12x 2 8. 30, 42, 78 10. 6a 2b, 9ab 2, 15a 2b 2

Factor completely. 49. 3t 3 51. 5w 3 12t 2 25w2 3a2b2 xy2 25w ab3 50. 52. 4m4 3t 3 36m2 3t 2 xy3 100h2 c with a 22x x 2p 40q ab 40ab 10 5 50 2b2 50b2 5 1 6t y4

Complete the factorization of each binomial. 11. 3x 13. 2a 6 3( ) 20 2( ) 12. 7x 2 x 14. a2 a x( a( ) )

53. 2a3b 55. 9x 3

54. 6x2y2 56. h4

Factor each polynomial by factoring out the GCF. 15. 2a a 2 17. 6x 2y 2 9x 5y 19. 3x 2y 12xy 9y2 16. 9 3b 18. a 3b 5 a 3b 2 20. 2a2 4ab2 ab 5.4 Factoring the Trinomial ax2 bx Factor each polynomial completely. 57. 14t 2 59. 6x 2 5.2 Special Products and Grouping Factor each polynomial completely. 21. 22. 23. 24. 25. 26. 27. y ac w2 a2 abc mnx y2
2

t 19x 5p

3 7 4 8p2 8p 5y2 2y2

58. 15x2 60. 2x2 62. 3p2 64. 6q2

61. 6p2 63. 30p3

y by b mc aw2 mw2 2a 2w aw 3x ax 3a 3 c 3ab 5 5nx m 400 8w 20y 4r 24t 2 12xy 5x xy 4 18t 36y
2

65. 6 x 2 67. 32x 2 28. 4m 30. t 2 32. 2a2 34. 3m2 36. t 2 38. 9y 40. x 2 bx
2 2

29xy 16xy

66. 10a2 68. 8a2

9 20t 4a 75 9w 2 12xy xy ax 4x
2

29. w2 31. 4y2 33. r 2 35. 8t 3 37. x
2

16 25

100 2

5.5 Difference and Sum of Cubes and a Strategy Factor completely. 69. 5x 3 71. 9x 2 73. n2 75. x 3 77. x 2y 79. w2 81. a2 83. x 3 40x 3x 2 64 2x 2 x 2 70. w2 72. ax 3 74. 4t2 76. 16x2 78. 3x 2 6w ax h2 2x 27 3n 1 18 3 9

16xy 2 4w 2a x2 5 1 x 1

39. x 2

5y

ay 1

80. 2n2 82. 2w2

5.3 Factoring the Trinomial ax2 Factor each polynomial. 41. b2 5b 24

c with a 2a 35

12w 9y 2

42. a2

84. 9x 2y 2

5-55
85. a2 87. 2x 2 ab 2a 16x 1000 q4 3a2 a 3 2b 24 86. 4m2 88. 6x 2 90. 8p3 92. z4 94. y3 20m 21x 1 81 5y2 8y 40 25 45

Chapter 5 Review Exercises

375

89. m3 91. p4 93. a3

114. Racquetball. The volume of rubber (in cubic centimeters) in a hollow rubber ball used in racquetball is given by 4 4 3 V R3 r , 3 3 where the inside radius is r centimeters and the outside radius is R centimeters. a) Rewrite the formula by factoring the right-hand side completely. b) The accompanying graph shows the relationship between r and V when R 3. Use the graph to estimate the value of r for which V 100 cm3.

5.6 Solving Quadratic Equations by Factoring Solve each equation. 95. x 3 97. (a 99. 2m2 101. m3 103. (x 105. p2 5x 2 2)(a 9m 4m2 3)2 1 p 4 3) 5 9m x2 1 8 0.01 5 0 6 0 36 96. 2m2 98. (w 100. 12x2 102. w3 104. (h 106. t 2 10m 2)(w 5x 5w2 2)2 1 3) 3 w (h 13 t 6 0.03y 0.02 12

Volume (cm3)

50 0 5 1)2 9

150 100 50 0 0 1 2 3 Inside radius (centimeters)

Figure for Exercise 114

107. 0.1x2

0.07x

108. 0.2y2

Applications Solve each problem. 109. Positive numbers. Two positive numbers differ by 6, and their squares differ by 96. Find the numbers. 110. Consecutive integers. Find three consecutive integers such that the sum of their squares is 77.

115. Leaning ladder. A 10-foot ladder is placed against a building so that the distance from the bottom of the ladder to the building is 2 feet less than the distance from the top of the ladder to the ground. What is the distance from the bottom of the ladder to the building?

10 ft

111. Dimensions of a notebook. The perimeter of a notebook is 28 inches, and the diagonal measures 10 inches. What are the length and width of the notebook? 112. Two numbers. The sum of two numbers is 8.5, and their product is 18. Find the numbers. 113. Poiseuille’s law. According to the nineteenth-century physician Poiseuille, the velocity (in centimeters per second) of blood r centimeters from the center of an artery of radius R centimeters is given by v kR2 kr2, where k is a constant. Rewrite the formula by factoring the right-hand side completely. x 2

x

Figure for Exercise 115

116. Towering antenna. A guy wire of length 50 feet is attached to the ground and to the top of an antenna. The height of the antenna is 10 feet larger than the distance from the base of the antenna to the point where the guy wire is attached to the ground. What is the height of the antenna?

376

Chapter 5 Factoring

5-56

Chapter 5 Test
Give the prime factorization for each integer. 1. 66 2. 336 Solve each equation. 21. x 2
6x 9 0

22. 2x2

5x

12

0

Find the greatest common factor (GCF) for each group. 3. 48, 80 5. 6y2, 15y3 4. 42, 66, 78 6. 12a2b, 18ab2, 24a3b3 23. 3x3
1 2 x 8 12x 3 x 4

24. (2x

1)(3x

5)

5

Factor each polynomial completely. 7. 5x 2 9. 3a3b 11. 4b 13. ax 15. 6b2 17. 2a2 19. x3
2

25.
12xy 2 24 12y 2

1

0

26. 0.3x 2

1.7x

1

0

10x 3ab3 28b ay 7b 13a 125 49 bx 5 15 by

8. 6x 2y 2 10. a2 12. 3m 14. ax 16. m2 18. z3 20. a4
3

2a

Write a complete solution to each problem. 27. If the length of a rectangle is 3 feet longer than the width and the diagonal is 15 feet, then what are the length and width?

27m 2a 4mn 9z2 ab3 5x 4n2 18z 10

28. The sum of two numbers is 4, and their product is the numbers.

32. Find

29. A ball is dropped from a height of 64 feet. Its height above the earth in feet is given by h(t) 16t2 number of seconds after it is dropped. a) Find h (1). b) How long does it take the ball to fall to the earth? 64, where t is the

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Chapter 5 Making Connections

377

MakingConnections
Simplify each expression. 1. 2.
91 17 4 17 91

A Review of Chapters 1–5
Simplify each expression. Write answers without negative exponents. All variables represent nonzero real numbers. 33. t 8 35. t 2 t2 t8 34. t 8 t 2 36. (t 8)2 38.
6 8

18 6 1

3. 5 4. 32 5. 2
5

2(7 24

3)

37. 39.

4(6)( 2) 0.07(63)

8t 8 2t 2 6x 15x

3y 5 9y2 4w 3 24w 6 x 2 3y3 40
2

40. 42.
2

6. 0.07(37)

2

Perform the indicated operations. 7. x 2x 9.
6 2 2x

41. ( 2 x 3y2)3 2x 43. 32 1 2

8. x 10.

6 2x 2

44.

1 3

3

11. 2 3y 4z 13. 2 15. 17. (5x 19. (3 2(3x 4z) 4) 2)(3x 4)

12. 2(3y 14. (x 16. 5x 18. (5x 20. 9x 3

4z) 3) 2(3x 2)(3x 6
2

2(5 4) 4x

x) 1)

Solve each inequality. State the solution set in interval notation and sketch its graph. 45. 2x 5 3x 4

9x2 6x 3x

7x 7

14

46. 4

5x

11

Find the solution set to each equation. 21. 2x 22. 2x 23. (x 24. (2x 25. 3x(x 26. x
2

3 1 3)(x

0 0 5) 1) 0 0 0

47.

2 x 3

3

5

48. 0.05(x

120)

24

0

3)(2x 3) x 3x 3x x 0 1

Factor each expression completely. 49. 4p3 50. 3m 14.9 0.5x 40) 2 51. 52.
4 2

27. 3x 28. 3x 29. 0.01x 30. 0.05x 31. 2x2 32. 2x2

12p2 12m
2 3

32p 9m2 3 q 3an 3bn

12a 2b

12a 8 a 7 54

0.04(x 18 7x 15

53. ab 54. 2am 55. 7x3

qb

2bm

0

56. 2a3

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Chapter 5 Factoring

5-58

Solve each problem. 57. The area of a rectangular garden is 750 square yards and the length is 30 yards. What is the width? 58. The perimeter of a rectangular canvas is 66 inches and its length is 19 inches. Find the width. 59. The area of a rectangular balcony is 66 square feet. If the length is 5 feet more than the width, then what is the length? 60. A craft shop charges five cents per square inch for a rectangular piece of copper. If the width is 3 inches less than the length and the charge is $5.40, then what is the width? 61. The diagonal measure of a small television screen is 1 inch greater than the length and 2 inches greater than the width. Find the length and width. 62. Another ace. Professional tennis players can serve a tennis ball at speeds over 120 mph into a rectangular region that has a perimeter of 69 feet and an area of 283.5 square feet. Find the length and width of the service region.
Photo for Exercise 62

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Chapter 5 Critical Thinking

379

Critical Thinking

For Individual or Group Work

Chapter 5

These exercises can be solved by a variety of techniques, which may or may not require algebra. So be creative and think critically. Explain all answers. Answers are in the Instructor’s Edition of this text. 1. Counting cubes. What is the total number of cubes that are in each of the following diagrams? a) b) 4. Chess board. There are 64 squares on a square chess board. How many squares are neither diagonal squares nor edge squares?

c)

d)

2. More cubes. Imagine a large cube that is made up of 125 small cubes like those in Exercise 1. What is the total number of cubes that could be found in this arrangement? 3. Timely coincidence. Starting at 8 A.M. determine the number of times in the next 24 hours for which the hour and minute hands on a clock coincide.

Photo for Exercise 4

5. Last digit. Find the last digit in 39999. 6. Reconciling remainders. Find a positive integer smaller than 500 that has a remainder of 3 when divided by 5, a remainder of 6 when divided by 9, and a remainder of 8 when divided by 11. 7. Exact sum. Find this sum exactly: 1 2 1 22 1 23 1 24 1 219

Photo for Exercise 3

8. Ten-digit number. Find a 10-digit number whose first digit is the number of 1’s in the 10-digit number, whose second digit is the number of 2’s in the 10-digit number, whose third digit is the number of 3’s in the 10-digit number, and so on. The ninth digit must be the number of nines in the 10-digit number and the tenth digit must be the number of zeros in the 10-digit number.