factoring, solvingequations, and problem solving 5 · pdf file5.2 factoring the difference of...

241

■ Algebraic equations can be used to solve a large variety of problems involving geometric relationships.

5.1 Factoring by Using theDistributive Property

5.2 Factoring the Difference of Two Squares

5.3 Factoring Trinomials of the Form x2 � bx � c

5.4 Factoring Trinomials of the Form ax2 � bx � c

5.5 Factoring, SolvingEquations, and Problem Solving

5Factoring, SolvingEquations, andProblem Solving

flower garden is in the shape of a right triangle with one leg 7 meters longer

than the other leg and the hypotenuse 1 meter longer than the longer leg.

Find the lengths of all three sides of the right triangle. A popular geometric formula,

called the Pythagorean theorem, serves as a guideline for setting up an equation to

solve this problem. We can use the equation to determine

that the sides of the right triangle are 5 meters, 12 meters, and 13 meters long.

The distributive property has allowed us to combine similar terms and multiply

polynomials. In this chapter, we will see yet another use of the distributive property

as we learn how to factor polynomials. Factoring polynomials will allow us to solve

other kinds of equations, which will, in turn, help us to solve a greater variety of word

problems.

x2 � 1x � 7 22 � 1x � 8 22

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242

I N T E R N E T P R O J E C TPythagoras is widely known for the Pythagorean theorem pertaining to right triangles. Do anInternet search to determine at least two other fields where Pythagoras made significantcontributions. Pythagoras also founded a school. While conducting your search, find the namegiven to the students attending Pythagoras' school and some of the school rules for students. Canyou think of any modern-day schools that might have the same requirements?

5.1 Factoring by Using the Distributive PropertyO B J E C T I V E S

1 Find the Greatest Common Factor

2 Factor Out the Greatest Common Factor

3 Factor by Grouping

4 Solve Equations by Factoring

5 Solve Word Problems Using Factoring

1 Find the Greatest Common FactorIn Chapter 1, we found the greatest common factor of two or more whole numbers byinspection or by using the prime factored form of the numbers. For example, by in-spection we see that the greatest common factor of 8 and 12 is 4. This means that 4 isthe largest whole number that is a factor of both 8 and 12. If it is difficult to determinethe greatest common factor by inspection, then we can use the prime factorizationtechnique as follows:

We see that is the greatest common factor of 42 and 70.It is meaningful to extend the concept of greatest common factor to monomials.

Consider the next example.

Find the greatest common factor of and .

Solution

Therefore, the greatest common factor is .

▼ PRACTICE YOUR SKILL

Find the greatest common factor of 14a2 and 7a5. 7a2 ■

By the greatest common factor of two or more monomials we mean the mono-mial with the largest numerical coefficient and highest power of the variables that is afactor of the given monomials.

2 # 2 # x # x � 4x2

12x3 � 2 # 2 # 3 # x # x # x

8x2 � 2 # 2 # 2 # x # x

12x38x2

2 # 7 � 14

70 � 2 # 5 # 7

42 � 2 # 3 # 7

E X A M P L E 1

05-W4801-AM1.qxd 8/19/08 8:45 PM 42REVISED PAGES

Find the greatest common factor of , , and .

Solution

Therefore, the greatest common factor is .


Find the greatest common factor of 18m2n4, 4m3n5, and 10m4n3. 2m2n3 ■

2 Factor Out the Greatest Common FactorWe have used the distributive property to multiply a polynomial by a monomial; forexample,

Suppose we start with and want to express it in factored form. We use thedistributive property in the form ab � ac � a(b � c).

3x is the greatest common factor of 3x2 and 6x

Use the distributive property

The next four examples further illustrate this process of factoring out the great-est common monomial factor.

Factor .

Solution

ab � ac � a(b � c)


Factor 15a2 � 21a6. 3a2 (5 � 7a4) ■

Factor .

Solution


Factor 8m3n2 � 2m6n. 2m3n(4n � m3) ■

� 6xy12x � 3y 2 12x2y � 18xy2 � 6xy12x 2 � 6xy13y 2

12x2y � 18xy2

� 4x213x � 2 2 12x3 � 8x2 � 4x213x 2 � 4x212 2

12x3 � 8x2

� 3x1x � 2 2 3x2 � 6x � 3x1x 2 � 3x12 2

3x2 � 6x

3x 1x � 2 2 � 3x2 � 6x

2 # 2 # 2 # x # y � 8xy

32xy � 2 # 2 # 2 # 2 # 2 # x # y

24x3y2 � 2 # 2 # 2 # 3 # x # x # x # y # y

16x2y � 2 # 2 # 2 # 2 # x # x # y

32xy24x3y216x2y

5.1 Factoring by Using the Distributive Property 243

E X A M P L E 2

E X A M P L E 3

E X A M P L E 4


Factor .

Solution


Factor 48y8 � 16y6 � 24y4. 8y4(6y4 � 2y2 � 3) ■

Factor .

Solution


Factor 8b3 � 8b2. 8b2(b � 1) ■

We want to emphasize the point made just before Example 3. It is important torealize that we are factoring out the greatest common monomial factor. We could factor an expression such as in Example 6 as , ,

, or even , but it is the form that we want. We

can accomplish this by factoring out the greatest common monomial factor; we some-times refer to this process as factoring completely. A polynomial with integralcoefficients is in completely factored form if these conditions are met:

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

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

Thus , , and are not completely factored be-

cause they violate condition 2. The form violates both conditions

1 and 2.

3 Factor by GroupingSometimes there may be a common binomial factor rather than a common monomialfactor. For example, each of the two terms of x(y � 2) � z(y � 2) has a commonbinomial factor of (y � 2). Thus we can factor (y � 2) from each term and get

x(y � 2) � z(y � 2) � (y � 2)(x � z)

Consider a few more examples involving a common binomial factor:

x1x � 5 2 � 41x � 5 2 � 1x � 5 2 1x � 4 2 x1x � 2 2 � 31x � 2 2 � 1x � 2 2 1x � 3 2 a1b � c 2 � d1b � c 2 � 1b � c 2 1a � d 2

12

118x2 � 18x 23x 13x � 2 23 13x2 � 3x 29 1x2 � x 2

9x 1x � 1 212

118x2 � 18x 23x 13x � 3 23 13x2 � 3x 29 1x2 � x 29x2 � 9x

� 9x1x � 1 2 9x2 � 9x � 9x1x 2 � 9x11 2

9x2 � 9x

� 6x314 � 5x � 7x2 2 24x3 � 30x4 � 42x5 � 6x314 2 � 6x315x 2 � 6x317x2 2

24x3 � 30x4 � 42x5

244 Chapter 5 Factoring, Solving Equations, and Problem Solving

E X A M P L E 5

E X A M P L E 6


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

ab � 3a � bc � 3c

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

ab � 3a � bc � 3c � a(b � 3) � c(b � 3)

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

a(b � 3) � c(b � 3) � (b � 3)(a � c)

This factoring process is called factoring by grouping. Let’s consider two more exam-ples of factoring by grouping.

Factor each polynomial completely.

(a) x2 � x � 5x � 5 (b) 6x2 � 4x � 3x � 2

Solution

(a) Factor x from first twoterms and 5 from last twoterms

Factor common binomialfactor of (x � 1) from bothterms

(b) Factor 2x from first twoterms and �1 from last twoterms

Factor common binomialfactor of (3x � 2) from bothterms


Factor each polynomial completely.

(a) ab � 5a � 3b � 15 (b) xy � 2x � 4y � 8 ■

(a) (a � 3)(b � 5) (b) (x � 4)(y � 2)

4 Solve Equations by FactoringSuppose we are told that the product of two numbers is 0. What do we know about thenumbers? Do you agree we can conclude that at least one of the numbers must be 0?The next property formalizes this idea.

� 13x � 2 2 12x � 1 2

6x2 � 4x � 3x � 2 � 2x13x � 2 2 � 113x � 2 2

� 1x � 1 2 1x � 5 2

x2 � x � 5x � 5 � x1x � 1 2 � 51x � 1 2


E X A M P L E 7

Property 5.1

For all real numbers a and b,

ab � 0 if and only if a � 0 or b � 0

Property 5.1 provides us with another technique for solving equations.


Solve .

Solution

To solve equations by applying Property 5.1, one side of the equation must be a prod-uct, and the other side of the equation must be zero. This equation already has zeroon the right-hand side of the equation, but the left-hand side of this equation is a sum.We will factor the left-hand side, x2 � 6x, to change the sum into a product.


The solution set is {�6, 0}. (Be sure to check both values in the original equation.)


Solve y2 � 7y � 0. {0, 7} ■

Solve .

Solution

In order to solve this equation by Property 5.1, we will first get zero on the right-handside of the equation by adding �12x to each side. Then we factor the expression onthe left-hand side of the equation.

Added �12x to both sides

x � 0 or x � 12 � 0 ab � 0 if and only if a � 0 or b � 0

x � 0 or x � 12

The solution set is {0, 12}.


Solve a2 � 15a. {0,15} ■

Remark: Note in Example 9 that we did not divide both sides of the original equationby x. Doing so would cause us to lose the solution of 0.

Solve .

Solution


The solution set is .e 0, 34f

x � 0 or x �34

x � 0 or 4x � 3

x � 0 or 4x � 3 � 0

x14x � 3 2 � 0

4x2 � 3x � 0

4x2 � 3x � 0

x1x � 12 2 � 0

x2 � 12x � 0

x2 � 12x

x2 � 12x

x � 0 or x � �6

x � 0 or x � 6 � 0

x1x � 6 2 � 0

x2 � 6x � 0

x2 � 6x � 0


E X A M P L E 8

E X A M P L E 9

E X A M P L E 1 0



Solve 5y2 � 2y � 0. ■

Solve .

Solution

In order to solve this equation by Property 5.1, we will factor the left-hand side of theequation. The greatest common factor of the terms is (x � 2).


The solution set is .


Solve m(m � 2) � 5(m � 2) � 0. {�5, 2} ■

5 Solve Word Problems Using FactoringEach time we expand our equation-solving capabilities, we gain more techniquesfor solving word problems. Let’s solve a geometric problem with the ideas we learnedin this section.

Apply Your SkillThe area of a square is numerically equal to twice its perimeter. Find the length of aside of the square.

Solution

Sketch a square and let s represent the length of each side (see Figure 5.1). Then thearea is represented by and the perimeter by 4s. Thus

Because 0 is not a reasonable answer to the problem, the solution is 8. (Be sure tocheck this solution in the original statement of the example!)


The area of a square is numerically equal to three times its perimeter. Find the lengthof a side of the square. The length is 12 ■

s � 0 or s � 8

s � 0 or s � 8 � 0

s1s � 8 2 � 0

s2 � 8s � 0

s2 � 8s

s2 � 214s 2s2

5�3, �26 x � �2 or x � �3

x � 2 � 0 or x � 3 � 0

1x � 2 2 1x � 3 2 � 0

x1x � 2 2 � 31x � 2 2 � 0

x 1x � 2 2 � 3 1x � 2 2 � 0

e�25

, 0 f


s s

s

s

Figure 5.1

E X A M P L E 1 1

E X A M P L E 1 2


For Problems 1–10, answer true or false.

1. The greatest common factor of 6x2y3 � 12x3y2 � 18x4y is 2x2y.2. If the factored form of a polynomial can be factored further, then it has not met

the conditions for being considered “factored completely.”3. Common factors are always monomials.4. If the product of x and y is zero, then x is zero or y is zero.5. The factored form 3a(2a2 � 4) is factored completely.6. The solutions for the equation x(x � 2) � 7 are 7 and 5.7. The solution set for x2 � 7x is {7}.8. The solution set for x(x � 2) � 3(x � 2) � 0 is {2, 3}.9. The solution set for �3x � x2 is {�3, 0}.

10. The solution set for x(x � 6) � 2(x � 6) is {�6}.


Problem Set 5.1

C O N C E P T Q U I Z

1 Find the Greatest Common Factor

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

1. 6y 2. 8x

3. 12xy 4.

5. 6.

7. 2x 8.

9.

10.

2 Factor Out the Greatest Common Factor

For Problems 11– 46, factor each polynomial completely.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.4x4y(13y � 15x2)

29. 30.

31. 3x14 � 5y � 7x 212x � 15xy � 21x212xy317x � 1 28x2y15y � 1 284x2y3 � 12xy340x2y2 � 8x2y14x5y215y � 3x3 270x5y3 � 42x8y252x4y2 � 60x6y7a2b51a � 6b 29a2b1b3 � 3 27a3b5 � 42a2b69a2b4 � 27a2b915xy � 8zw 2818ab � 9cd 245xy � 72zw64ab � 72cdx2y112y � 29 2xy

2116y � 25x 212x2y2 � 29x2y16xy3 � 25x2y25ab2113b � 9a 212a2b13 � 5ab3 265ab3 � 45a2b236a2b � 60a3b47x2y14y � 7 26xy12y � 5x 228x2y2 � 49x2y12xy2 � 30x2y

4x13 � 7x2 212x � 28x39x12x � 5 218x2 � 45x

8x13 � 5y 224x � 40xy7y12x � 3 214xy � 21y

613x � 4y 218x � 24y412x � 3y 28x � 12y

7ab370a3b3, 42a2b4, and 49ab5

8a2b216a2b2, 40a2b3, and 56a3b4

72xy, 36x2y, and 84xy26x3, 8x, and 24x2

48ab248a2b2 and 96ab414ab242ab3 and 70a2b2

9x272x3 and 63x260x2y and 84xy2

32x and 40xy24y and 30xy

32.

33. 34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

3 Factor by Grouping

For Problems 47– 60, use the process of factoring by groupingto factor each polynomial.

47.

48.

49.

50.

51. 1a � b 2 1c � 1 2ac � bc � a � b

1x � y 2 12 � a 22x � 2y � ax � ay

1x � y 2 1b � c 2bx � by � cx � cy

1x � y 2 17 � z 27x � 7y � zx � zy

1x � y 2 15 � b 25x � 5y � bx � by

1x � 8 2 14x � 5 24x1x � 8 2 � 51x � 8 21x � 1 2 12x � 3 22x1x � 1 2 � 31x � 1 21x � 7 2 1x � 9 2x1x � 7 2 � 91x � 7 21x � 3 2 1x � 6 2x1x � 3 2 � 61x � 3 21y � 6 2 1x � 3 2x1y � 6 2 � 31y � 6 21b � 4 2 1a � c 2a1b � 4 2 � c1b � 4 21c � d 2 1a � 2 2a1c � d 2 � 21c � d 21y � 1 2 1x � z 2x1y � 1 2 � z1y � 1 2

12a2b212a � 3b2 � 5a2b 224a3b2 � 36a2b4 � 60a4b3

7ab12ab2 � 5b � 7a2 214a2b3 � 35ab2 � 49a3b

2a17 � 9a2 � 13a4 214a � 18a3 � 26a5

4y 2111y

3 � 6y � 5 244y5 � 24y3 � 20y2x21x2 � x � 1 2x12x2 � 3x � 4 2 x4 � x3 � x22x3 � 3x2 � 4x

5y16x2 � 8x � 11 230x2y � 40xy � 55y

12xy

Blue problem numbers indicate Enhanced WebAssign Problems.


82.

83. {4, 6}

84. {�9, 2}


For Problems 85–91, set up an equation and solve eachproblem.

85. The square of a number equals nine times that number.Find the number. 0 or 9

86. Suppose that four times the square of a number equals20 times that number. What is the number? 0 or 5

87. The area of a square is numerically equal to five times itsperimeter. Find the length of a side of the square.20 units

88. The area of a square is 14 times as large as the area of atriangle. One side of the triangle is 7 inches long, and thealtitude to that side is the same length as a side of thesquare. Find the length of a side of the square. Also findthe areas of both figures, and be sure that your answerchecks. 49 inches, 2401 square inches, square inches

89. Suppose that the area of a circle is numerically equal tothe perimeter of a square whose length of a side is thesame as the length of a radius of the circle. Find the length of a side of the square. Express your answer interms of p.

90. One side of a parallelogram, an altitude to that side, and one side of a rectangle all have the same measure. If anadjacent side of the rectangle is 20 centimeters long, and thearea of the rectangle is twice the area of the parallelogram,find the areas of both figures. See answer below

91. The area of a rectangle is twice the area of a square. If therectangle is 6 inches long, and the width of the rectangleis the same as the length of a side of the square, find thedimensions of both the rectangle and the square.

4p

3432

x1x � 9 2 � 21x � 9 241x � 6 2 � x1x � 6 2 � 0

e 23

, 7 fx13x � 2 2 � 713x � 2 2 � 052.

53.

54.

55.

56.

57.

58.

59.

60.

4 Solve Equations by Factoring

For Problems 61–84, solve each equation.

61. {0, 8} 62. {0, 12}

63. 64.

65. 66.

67. 68.

69. 70.

71. 72.

73. 74.

75. 76.

77. 78.

79. 80.

81. 5�5, 46x1x � 5 2 � 41x � 5 2 � 0

e�75

, 0 f7x � �5x2e�52

, 0 f5x � �2x2

5�15, 0615x � �x250, 13613x � x2

50, 969x � x2 � 050, 767x � x2 � 0

e0, 23f12x2 � 8xe0,

32f4x2 � 6x

50, 466n2 � 24n � 05�5, 063n2 � 15n � 0

e�25

, 0 f5x2 � �2xe�37

, 0 f7x2 � �3x

e0, 74f4y2 � 7y � 0e0,

32f2y2 � 3y � 0

5�2, 06n2 � �2n50, 56n2 � 5n

5�7, 06x2 � 7x � 05�1, 06x2 � x � 0

x2 � 12x � 0x2 � 8x � 0

15n � 2 2 14n � 3 220n2 � 8n � 15n � 6

12n � 1 2 13n � 4 26n2 � 3n � 8n � 4

13x � 2 2 1x � 6 23x2 � 2x � 18x � 12

12x � 1 2 1x � 5 22x2 � x � 10x � 5

1x � 4 2 1x � 9 2x2 � 4x � 9x � 36

1x � 2 2 1x � 8 2x2 � 2x � 8x � 16

1x � 3 2 1x � 7 2x2 � 3x � 7x � 21

1x � 5 2 1x � 12 2x2 � 5x � 12x � 60

1x � y 2 11 � a 2x � y � ax � ay


THOUGHTS INTO WORDS

92. Suppose that your friend factors like this:

Is this correct? Would you suggest any changes?

� 12xy12x � 3 2 � 14xy 2 13 2 12x � 3 2

24x2y � 36xy � 4xy16x � 9 224x2y � 36xy 93. The following solution is given for the equation

.

The solution set is {0, 10}. Is this solution correct? Wouldyou suggest any changes?

x � 0 or x � 10

x � 0 or x � 10 � 0

x1x � 10 2 � 0

x2 � 10x � 0

x1x � 10 2 � 0

x1x � 10 2 � 0

90. The area of the parallelogram is 10(10) � 100 square centimeters and the area of the rectangle is 10(20) � 200 square centimeters

The square is 3 inches by 3 inches and the rectangle is 3 inchesby 6 inches


94. The total surface area of a right circular cylinder is givenby the formula A � � , where r represents theradius of a base, and h represents the height of the cylin-der. For computational purposes, it may be more conve-nient to change the form of the right side of the formulaby factoring it.

Use to find the total surface area of

each of the following cylinders. Use as an approxi-mation for .

a. r � 7 centimeters and h � 12 centimeters836 square centimeters

b. r � 14 meters and h � 20 meters2992 square meters

c. r � 3 feet and h � 4 feet132 square feet

d. r � 5 yards and h � 9 yards440 square yards

95. The formula A � P � Prt yields the total amount of money accumulated (A) when P dollars are invested at r percent simple interest for t years. For computationalpurposes it may be convenient to change the right side ofthe formula by factoring.

p

227

A � 2pr1r � h 2 � 2pr1r � h 2

A � 2pr2 � 2prh

2prh2pr2

Use A � P(1 � rt) to find the total amount of moneyaccumulated for each of the following investments.

a. $100 at 8% for 2 years $116

b. $200 at 9% for 3 years $254

c. $500 at 10% for 5 years $750

d. $1000 at 10% for 10 years $2000

For Problems 96 –99, solve each equation for the indicatedvariable.

96. for x

97. for x

98. for y

99. for y y �c

1 � a � by � ay � by � c � 0

y � 0 or y �b5a

5ay2 � by

x � 0 or x �c

b2b2x2 � cx � 0

x �c

a � bax � bx � c

� P11 � rt 2 A � P � Prt


FURTHER INVESTIGATIONS

Answers to the Example Practice Skills1. 7a2 2. 2m2n3 3. 3a2(5 � 7a4) 4. 2m3n(4n � m3) 5. 8y4(6y4 � 2y2 � 3) 6. 8b2(b � 1)

7. (a) (a � 3)(b � 5) (b) (x � 4)(y � 2) 8. {0, 7} 9. {0, 15} 10. 11. {�5, 2}12. The length is 12.

Answers to the Concept Quiz1. False 2. True 3. False 4. True 5. False 6. False 7. False 8. True 9. True 10. False

e�25

, 0 f

5.2 Factoring the Difference of Two SquaresO B J E C T I V E S

1 Factor the Difference of Two Squares

2 Solve Equations by Factoring the Difference of Two Squares


1 Factor the Difference of Two SquaresIn Section 4.3, we noted some special multiplication patterns. One of these pat-terns was

1a � b 2 1a � b 2 � a2 � b2


To apply the difference-of-two-squares pattern is a fairly simple process, as these nextexamples illustrate. The steps inside the box are often performed mentally.

Because multiplication is commutative, the order of writing the factors is not impor-tant. For example, (x � 6)(x � 6) can also be written as (x � 6)(x � 6).

Remark: You must be careful not to assume an analogous factoring pattern for thesum of two squares; it does not exist. For example, ≠ (x � 2)(x � 2) because (x � 2)(x � 2) � � 4x � 4. We say that the sum of two squares is not factorable us-ing integers. The phrase “using integers” is necessary because � 4 could be

written as , but such factoring is of no help. Furthermore, we do not

consider as factoring .It is possible that both the technique of factoring out a common monomial fac-

tor and the difference of two squares pattern can be applied to the same polynomial.In general, it is best to look for a common monomial factor first.

Factor .

Solution

Common factor of 2

Difference of squares


Factor 3y2 � 12. 3(y � 2)(y � 2) ■

In Example 1, by expressing as 2(x � 5)(x � 5), we say that it has been fac-tored completely. That means the factors 2, x � 5, and x � 5 cannot be factored anyfurther using integers.

Factor completely .

Solution

Common factor of 2y

Difference of squares


Factor completely 32x3 � 2x. 2x(4x � 1)(4x � 1) ■

Sometimes it is possible to apply the difference-of-two-squares pattern morethan once. Consider the next example.

� 2y13y � 2 2 13y � 2 2 18y3 � 8y � 2y19y2 � 4 2

18y3 � 8y

2x2 � 50

� 21x � 5 2 1x � 5 2 2x2 � 50 � 21x2 � 25 2

2x2 � 50

x2 � 411 2 1x2 � 4 21212x2 � 8 2

x2x2

x2 � 4

� 18 � y 2 18 � y 2 18 2 2 � 1y 2 2 64 � y2 �

� 13x � 4y 2 13x � 4y 2 13x 2 2 � 14y 2 2 9x2 � 16y2 �

� 12x � 5 2 12x � 5 2 12x 2 2 � 15 2 2 4x2 � 25 �

� 1x � 6 2 1x � 6 2 1x 2 2 � 16 2 2 x2 � 36 �

5.2 Factoring the Difference of Two Squares 251

Difference of Two Squares

a2 � b2 � (a � b)(a � b)

Here is another version of that pattern:

E X A M P L E 1

E X A M P L E 2


Factor completely .

Solution


Factor completely a4 � 81. (a2 � 9)(a � 3)(a � 3) ■

The following examples should help you to summarize the factoring ideaspresented thus far.

is not factorable using integers

is not factorable using integers


Each time we learn a new factoring technique, we also develop more power for solv-ing equations. Let’s consider how we can use the difference-of-squares factoring pat-tern to help solve certain kinds of equations.

Solve .

Solution

The solution set is {�5, 5}. Check these answers!


Solve x2 � 64. {�8, 8} ■

Solve .

Solution

9x2 � 25 � 0

9x2 � 25

9x2 � 25

x � �5 or x � 5

x � 5 � 0 or x � 5 � 0

1x � 5 2 1x � 5 2 � 0

x2 � 25 � 0

x2 � 25

x2 � 25

9x � 17y

a2 � 9

36x2 � 49y2 � 16x � 7y 2 16x � 7y 23 � 3x2 � 3 11 � x2 2 � 3 11 � x 2 11 � x 225 � y2 � 15 � y 2 15 � y 25x2 � 20 � 5 1x2 � 4 2

� 1x2 � 4 2 1x � 2 2 1x � 2 2 x4 � 16 � 1x2 � 4 2 1x2 � 4 2

x4 � 16


E X A M P L E 3

E X A M P L E 4

E X A M P L E 5

Remember: ab � 0 if and only if a � 0 or b � 0




Solve 4x2 � 9. ■

Solve .

Solution

Divide both sides by 5

The solution set is {�2, 2}. Check it!


Solve 3x2 � 75. {�5, 5} ■

Solve .

Solution

The solution set is {�3, 0, 3}.


Solve x3 � 49x � 0. {�7, 0, 7} ■

The more we know about solving equations, the more easily we can solve wordproblems.

x � 0 or x � 3 or x � �3

x � 0 or x � 3 � 0 or x � 3 � 0

x1x � 3 2 1x � 3 2 � 0

x1x2 � 9 2 � 0

x3 � 9x � 0

x3 � 9x � 0

y � �2 or y � 2

y � 2 � 0 or y � 2 � 0

1y � 2 2 1y � 2 2 � 0

y2 � 4 � 0

y2 � 4

5y2

5�

205

5y2 � 20

5y2 � 20

e�32

, 32f

e�53

, 53f

x � �53

or x �53

3x � �5 or 3x � 5

3x � 5 � 0 or 3x � 5 � 0

13x � 5 2 13x � 5 2 � 0


E X A M P L E 6

E X A M P L E 7



Apply Your SkillThe combined area of two squares is 20 square centimeters. Each side of onesquare is twice as long as a side of the other square. Find the lengths of the sides ofeach square.

Solution

We can sketch two squares and label the sides of the smaller square s (see Figure 5.2).Then the sides of the larger square are 2s. The sum of the areas of the two squares is20 square centimeters, so we set up and solve the following equation:

Because s represents the length of a side of a square, we must disregard the solu-tion �2. Thus one square has sides of length 2 centimeters, and the other square hassides of length 2(2) � 4 centimeters.


The combined area of two squares is 250 square feet. Each side of one square is threetimes as long as a side of the other square. Find the lengths of the sides of each square.5 feet and 15 feet ■


1. A binomial that has two perfect square terms that are subtracted is called the dif-ference of two squares.

2. The sum of two squares is factorable using integers.3. When factoring it is usually best to look for a common factor first.4. The polynomial 4x2 � y2 factors into (2x � y)(2x � y).5. The completely factored form of y4 � 81 is (y2 � 9)(y2 � 9).6. The solution set for x2 � �16 is {�4}.7. The solution set for 5x3 � 5x � 0 is {�1, 0, 1}.8. The solution set for x4 � 9x2 � 0 is {�3, 0, 3}.

s � �2 or s � 2

s � 2 � 0 or s � 2 � 0

1s � 2 2 1s � 2 2 � 0

s2 � 4 � 0

s2 � 4

5s2 � 20

s2 � 4s2 � 20

s2 � 12s 2 2 � 20


E X A M P L E 8

Problem Set 5.2

1 Factor the Difference of Two Squares

For Problems 1–12, use the difference-of-squares pattern tofactor each polynomial.

1. 2.

3. 4. 1x � 11 2 1x � 11 21x � 10 2 1x � 10 2 x2 � 121x2 � 100

1x � 5 2 1x � 5 2x2 � 251x � 1 2 1x � 1 2x2 � 1

5. 6.

7. 8.

9. 10.

11. 12. 12 � 3n 2 12 � 3n 211 � 2n 2 11 � 2n 2 4 � 9n21 � 4n2

12a � 9b 2 12a � 9b 216a � 5b 2 16a � 5b 2 4a2 � 81b236a2 � 25b2

17y � 8x 2 17y � 8x 213x � y 2 13x � y 2 49y2 � 64x29x2 � y2

1x � 6y 2 1x � 6y 21x � 2y 2 1x � 2y 2 x2 � 36y2x2 � 4y2


s

s

2s

2s

Figure 5.2



61. 62.

63. 64.{�10, 0, 10}

65. 66.

67. 68.


For Problems 69–80 set up an equation and solve the problem.

69. Forty-nine less than the square of a number equals zero.Find the number. �7 or 7

70. The cube of a number equals nine times the number.Find the number. �3, 0, or 3

71. Suppose that five times the cube of a number equals80 times the number. Find the number. �4, 0, or 4

72. Ten times the square of a number equals 40. Find thenumber. �2 or 2

73. The sum of the areas of two squares is 234 square inches.Each side of the larger square is five times the length of aside of the smaller square. Find the length of a side ofeach square. 3 inches and 15 inches

74. The difference of the areas of two squares is 75 squarefeet. Each side of the larger square is twice the length ofa side of the smaller square. Find the length of a side ofeach square. 5 feet and 10 feet

75. Suppose that the length of a certain rectangle is

times its width, and the area of that same rectangle is 160 square centimeters. Find the length and width of therectangle. 20 centimeters and 8 centimeters

76. Suppose that the width of a certain rectangle is three-fourths of its length and that the area of this same rect-angle is 108 square meters. Find the length and width ofthe rectangle. 12 meters and 9 meters

77. The sum of the areas of two circles is 80p square meters.Find the length of a radius of each circle if one of them istwice as long as the other. 4 meters and 8 meters

78. The area of a triangle is 98 square feet. If one side of thetriangle and the altitude to that side are of equal length,find the length. 14 feet

79. The total surface area of a right circular cylinder is square centimeters. If a radius of the base and the

altitude of the cylinder are the same length, find thelength of a radius. 5 centimeters

80. The total surface area of a right circular cone is square feet. If the slant height of the cone is equal inlength to a diameter of the base, find the length of a radius. 8 feet

192p

100p

212

e�

14

, 0, 14f64x3 � 4xe�

12

, 0, 12f36x3 � 9x

e�59

, 59f81x2 � 25e�9

8,

98f64x2 � 81

5�7, 0, 762x3 � 98x � 04x3 � 400x � 0

e�

12

, 12f3 � 12x2 � 0e�1

3,

13f5 � 45x2 � 0For Problems 13– 44, factor each polynomial completely. Indi-

cate any that are not factorable using integers. Don’t forget tolook for a common monomial factor first.

13. 14.

15. 16.

17. 18.

19. 20.

21. Not factorable 22.

23. 24.Not factorable

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

39. 40.

41. 16x4 � 81y4 42. x4 � 1

43. 81 � x4 44. 81x4 � 16y4


For Problems 45– 68, solve each equation.

45. 46.

47. 48.

49. 50.

51. 52.

53. 54.

55. 56.

57. 58.

59. 60. 5�2, 0, 262n3 � 8n5�4, 0, 46n3 � 16n

5�1, 0, 16x3 � x � 05�4, 0, 463x3 � 48x � 0

5�2, 267x2 � 285�5, 563x2 � 75

e�67

, 67f49x2 � 36e�

25

, 25f25x2 � 4

5�9, 96n2 � 81 � 05�11, 116n2 � 121 � 0

e�32

, 32f4x2 � 9e�4

3,

43f9x2 � 16

5�12, 126144 � n25�2, 264 � n2

5�1, 16x2 � 15�3, 36x2 � 9

13x � 2y 2 13x � 2y 2 19x2 � 4y 2 213 � x 2 13 � x 2 19 � x

2 2

1x � 1 2 1x � 1 2 1x2 � 1 212xy � 3y 2 12x � 3y 2 14x2 � 9y2 22xy14x � 3y 2 14x � 3y 23xy15x � 2y 2 15x � 2y 232x3y � 18xy375x3y � 12xy3

91x � 1 2 1x � 1 29x2 � 941x � 4 2 1x � 4 24x2 � 64

4x11 � 3x 2 11 � 3x 25x11 � 2x 2 11 � 2x 2 4x � 36x35x � 20x3

6x1x2 � 4 26x3 � 24x3x1x2 � 16 23x3 � 48x

x31x2 � 2 2x5 � 2x3x21x2 � 1 2x4 � x2

12 � x 2 12 � x 2 14 � x2 21x � 3 2 1x � 3 2 1x2 � 9 2 16 � x4x4 � 81

9a21a2 � 9 29a4 � 81a24a21a2 � 4 24a4 � 16a2

315 � x 2 15 � x 275 � 3x2413 � x 2 13 � x 236 � 4x2

16x2 � 25y29x15x � 4y 245x2 � 36xy

613x � 7y 218x � 42yx2 � 9y2

2x3 � 2xx1x � 5 2 1x � 5 2x3 � 25x

81x � 2y 2 1x � 2y 221x � 3y 2 1x � 3y 2 8x2 � 32y22x2 � 18y2

121x2 � 5 212x2 � 6081x2 � 4 28x2 � 32

71x � 1 2 1x � 1 27x2 � 751x � 2 2 1x � 2 25x2 � 20


2x1x � 1 2 1x � 1 2


The following patterns can be used to factor the sum of twocubes and the difference of two cubes, respectively.

Consider these examples:

Use the sum-of-two-cubes and the difference-of-two-cubespatterns to factor each polynomial.

84. 85.1x � 2 2 1x2 � 2x � 4 21x � 1 2 1x2 � x � 1 2x3 � 8x3 � 1

x3 � 1 � 1x 23 � 11 23 � 1x � 1 2 1x2 � x � 1 2x3 � 8 � 1x 23 � 12 23 � 1x � 2 2 1x2 � 2x � 4 2

a3 � b3 � 1a � b 2 1a2 � ab � b2 2a3 � b3 � 1a � b 2 1a2 � ab � b2 2

86. 87.

88. 89.

90. 91.

92. 93.

94. 95.

96. 97.

98. 99.14 � x 2 116 � 4x � x2 213n � 5 2 19n2 � 15n � 25 2 64 � x327n3 � 125

12 � n 2 14 � 2n � n2 2 125x3 � 8y38 � n313x � 2y 2 19x2 � 6xy � 4y2 21ab � 1 2 1a2b2 � ab � 1 2 27x3 � 8y3a3b3 � 112x � y 2 14x2 � 2xy � y2 21x � 2y 2 1x2 � 2xy � 4y2 2 8x3 � y3x3 � 8y311 � 3a 2 11 � 3a � 9a2 211 � 2x 2 11 � 2x � 4x2 2 1 � 27a31 � 8x3

12x � 3y 2 14x2 � 6xy � 9y2 2 27a3 � 64b38x3 � 27y31n � 4 2 1n2 � 4n � 16 21n � 3 2 1n2 � 3n � 9 2 n3 � 64n3 � 27


THOUGHTS INTO WORDS

81. How do we know that the equation x2 � 1 � 0 has nosolutions in the set of real numbers?

82. Why is the following factoring process incomplete?

How should the factoring be done?

16x2 � 64 � 14x � 8 2 14x � 8 2

83. Consider the following solution:

The solution set is {�3, 3}. Is this a correct solution? Doyou have any suggestion to offer the person who workedon this problem?

4 � 0 or x � �3 or x � 3

4 � 0 or x � 3 � 0 or x � 3 � 0

41x � 3 2 1x � 3 2 � 0

41x2 � 9 2 � 0

4x2 � 36 � 0


Answers to the Example Practice Skills1. 3(y � 2)(y � 2) 2. 2x(4x � 1)(4x � 1) 3. (a2 � 9)(a � 3)(a � 3) 4. {�8, 8} 5. 6. {�5, 5}7. {�7, 0, 7} 8. 5 feet and 15 feet

Answers to the Concept Quiz1. True 2. False 3. True 4. False 5. False 6. False 7. True 8. True

e�32

, 32f

5.3 Factoring Trinomials of the Form x2 � bx � cO B J E C T I V E S

1 Factor Trinomials of the Form x2 � bx � c

2 Use Factoring of Trinomials to Solve Equations

3 Solve Word Problems Including Consecutive Number Problems

4 Use the Pythagorean Theorem to Solve Problems

1 Factor Trinomials of the Form x2 � bx � cOne of the most common types of factoring used in algebra is to express a trinomial asthe product of two binomials. In this section, we will consider trinomials where thecoefficient of the squared term is 1, that is, trinomials of the form .x2 � bx � c

13a � 4b 2 19a2 � 12ab � 16b2 2

15x � 2y 2 125x2 � 10xy � 4y2 2


Again, to develop a factoring technique we first look at some multiplicationideas. Consider the product (x � r)(x � s), and use the distributive property to showhow each term of the resulting trinomial is formed.

14243

x2 � (s � r)x � rs

Note that the coefficient of the middle term is the sum of r and s and that the last termis the product of r and s. These two relationships are used in the next examples.

Factor .

Solution

We need to fill in the blanks with two numbers whose sum is 7 and whose prod-uct is 12.

� (x � _____)(x � _____)

This can be done by setting up a table showing possible numbers.

x2 � 7x � 12

x2 � 7x � 12

1x � r 2 1x � s 2 � x1x 2 � x1s 2 � r1x 2 � r1s 2

5.3 Factoring Trinomials of the Form x2 � bx � c 257

Product Sum

1(12) � 12 1 � 12 � 13

2(6) � 12 2 � 6 � 8

3(4) � 12 3 � 4 � 7

The bottom line contains the numbers that we need. Thus


Factor y2 � 9y � 20. (y � 5)(y � 4) ■

Factor .

Solution

We need two numbers whose product is 24 and whose sum is �11.

x2 � 11x � 24

x2 � 7x � 12 � 1x � 3 2 1x � 4 2

Product Sum

1�4 2 1�6 2 � 24 �4 � 1�6 2 � �10

1�3 2 1�8 2 � 24 �3 � 1�8 2 � �11

1�2 2 1�12 2 � 24 �2 � 1�12 2 � �14

1�1 2 1�24 2 � 24 �1 � 1�24 2 � �25

The third line contains the numbers that we want. Thus


Factor m2 � 8m � 15. (m � 3)(m � 5) ■

x2 � 11x � 24 � 1x � 3 2 1x � 8 2

E X A M P L E 1

E X A M P L E 2


The bottom line is the key line. Thus


Factor y2 � 5y � 24. (y � 8)(y � 3) ■

Factor .

Solution

We need two numbers whose product is �8 and whose sum is �2.

x2 � 2x � 8

x2 � 3x � 10 � 1x � 5 2 1x � 2 2


Product Sum

�215 2 � �10 �2 � 5 � 3

21�5 2 � �10 2 � 1�5 2 � �3

�1110 2 � �10 �1 � 10 � 9

11�10 2 � �10 1 � 1�10 2 � �9

Product Sum

�214 2 � �8 �2 � 4 � 2

21�4 2 � �8 2 � 1�4 2 � �2

�118 2 � �8 �1 � 8 � 7

11�8 2 � �8 1 � 1�8 2 � �7

The third line has the information we want.


Factor y2 � 4y � 12. (y � 2)(y � 6) ■

The tables in the last four examples illustrate one way of organizing yourthoughts for such problems. We showed complete tables; that is, for Example 4, we in-cluded the last line even though the desired numbers were obtained in the third line.If you use such tables, keep in mind that as soon as you get the desired numbers, thetable need not be continued beyond that point. Furthermore, there will be times thatyou will be able to find the numbers without using a table. The key ideas are the prod-uct and sum relationships.

Factor .

Solution

x2 � 13x � 12

x2 � 2x � 8 � 1x � 4 2 1x � 2 2

Factor .

Solution

We need two numbers whose product is �10 and whose sum is 3.

x2 � 3x � 10E X A M P L E 3

E X A M P L E 4

E X A M P L E 5

Product Sum

(�1)(�12) � 12 (�1) � (�12) � �13


We need not complete the table.


Factor y2 � 7y � 6. (y � 1)(y � 6) ■

In the next example, we refer to the concept of absolute value. Recall that theabsolute value is the number without regard for the sign. For example,

Factor .

Solution

Note that the coefficient of the middle term is �1. Therefore, we are looking for twonumbers whose product is �56; their sum is �1, so the absolute value of the negativenumber must be 1 larger than the absolute value of the positive number. The numbersare �8 and 7, and we have


Factor a2 � a � 12. (a � 4)(a � 3) ■

Factor .

Solution

x2 � 10x � 12

x2 � x � 56 � 1x � 8 2 1x � 7 2

x2 � x � 56

0 4 0 � 4 and 0�4 0 � 4

x2 � 13x � 12 � 1x � 1 2 1x � 12 2


Product Sum

314 2 � 12 3 � 4 � 7

216 2 � 12 2 � 6 � 8

1112 2 � 12 1 � 12 � 13

Because the table is complete and no two factors of 12 produce a sum of 10, we con-clude that

is not factorable using integers.


Factor y2 � 7y � 18. Not factorable ■

In a problem such as Example 7, we need to be sure that we have tried all pos-sibilities before we conclude that the trinomial is not factorable.

2 Use Factoring of Trinomials to Solve EquationsThe property ab � 0 if and only if a � 0 or b � 0 continues to play an important roleas we solve equations that involve the factoring ideas of this section. Consider the fol-lowing examples.

x2 � 10x � 12

E X A M P L E 6

E X A M P L E 7


Solve .

Solution

Factor the left side

Use ab � 0 if and only if a � 0 or b � 0

The solution set is {�5, �3}.


Solve y2 � 14y � 24 � 0. {2, 12} ■

Solve .

Solution

The solution set is {�6, 1}.


Solve a2 � 3a � 28 � 0. {�7, 4} ■

Solve .

Solution

The solution set is {�5, 9}.


Solve x2 � 18x � 40. {�2, 20} ■

Don’t forget that we can always check to be absolutely sure of our solutions.Let’s check the solutions for Example 10. If y � 9, then y2 � 4y � 45 becomes

45 � 45

81 � 36 � 45

92 � 419 2 � 45

y � 9 or y � �5

y � 9 � 0 or y � 5 � 0

1y � 9 2 1y � 5 2 � 0

y2 � 4y � 45 � 0

y2 � 4y � 45

y2 � 4y � 45

x � �6 or x � 1

x � 6 � 0 or x � 1 � 0

1x � 6 2 1x � 1 2 � 0

x2 � 5x � 6 � 0

x2 � 5x � 6 � 0

x � �3 or x � �5

x � 3 � 0 or x � 5 � 0

1x � 3 2 1x � 5 2 � 0

x2 � 8x � 15 � 0

x2 � 8x � 15 � 0


E X A M P L E 8

E X A M P L E 9

E X A M P L E 1 0


If y � �5, then becomes

3 Solve Word Problems Including Consecutive Number Problems

The more we know about factoring and solving equations, the more easily we cansolve word problems.

Apply Your SkillFind two consecutive integers whose product is 72.

Solution

Let n represent one integer. Then n � 1 represents the next integer.

The product of the two integers is 72

If n � �9, then n � 1 � �9 � 1 � �8. If n � 8, then n � 1 � 8 � 1 � 9. Thus theconsecutive integers are �9 and �8 or 8 and 9.


Find two consecutive integers whose product is 110. �11, �10 or 10, 11 ■

Apply Your SkillA rectangular plot is 6 meters longer than it is wide. The area of the plot is 16 squaremeters. Find the length and width of the plot.

Solution

We let w represent the width of the plot, and then w � 6 represents the length (seeFigure 5.3).

Figure 5.3

w

w + 6

n � �9 or n � 8

n � 9 � 0 or n � 8 � 0

1n � 9 2 1n � 8 2 � 0

n2 � n � 72 � 0

n2 � n � 72

n1n � 1 2 � 72

45 � 45

25 � 20 � 45

1�5 2 2 � 41�5 2 � 45

y2 � 4y � 45


E X A M P L E 1 1

E X A M P L E 1 2


Using the area formula A � lw, we obtain

The solution �8 is not possible for the width of a rectangle, so the plot is 2 meterswide, and its length (w � 6) is 8 meters.


A rectangular plot is 2 yards longer than it is wide. The area of the plot is 80 squareyards. Find the width and length of the plot. 8 yards, 10 yards ■


The Pythagorean theorem, an important theorem pertaining to right triangles, canalso serve as a guideline for solving certain types of problems. The Pythagorean the-orem states that in any right triangle, the square of the longest side (called thehypotenuse) is equal to the sum of the squares of the other two sides (called legs); seeFigure 5.4. We can use this theorem to help solve a problem.

Figure 5.4

Apply Your SkillSuppose that the lengths of the three sides of a right triangle are consecutive wholenumbers. Find the lengths of the three sides.

Solution

Let s represent the length of the shortest leg. Then s � 1 represents the length of theother leg, and s � 2 represents the length of the hypotenuse. Using the Pythagoreantheorem as a guideline, we obtain the following equation:

Sum of squares of two legs � Square of hypotenuse6447448 64748

�

Solving this equation yields

2s2 � 2s � 1 � s2 � 4s � 4

s2 � s2 � 2s � 1 � s2 � 4s � 4

1s � 2 2 2s2 � 1s � 1 2 2

a2 + b2 = c2

a

bc

w � �8 or w � 2

w � 8 � 0 or w � 2 � 0

1w � 8 2 1w � 2 2 � 0

w2 � 6w � 16 � 0

w2 � 6w � 16

w1w � 6 2 � 16


E X A M P L E 1 3


19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

2 Use Factoring of Trinomials to Solve Equations


31. 32.

33. 34. x2 � 9x � 8 � 0x2 � 9x � 18 � 0

x2 � 9x � 20 � 0x2 � 10x � 21 � 0

a2 � 3ab � 54b2a2 � 4ab � 32b2

x2 � 4xy � 12y2x2 � 3xy � 10y2

x2 � 12x � 64x2 � 10x � 48

x2 � 8x � 36x2 � 6x � 72

x2 � 21x � 90x2 � 21x � 80

y2 � y � 30y2 � y � 721 Factor Trinomials of the Form x2 � bx � c

For Problems 1–30, factor each trinomial completely. Indicateany that are not factorable using integers.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18. x2 � 11x � 42x2 � 5x � 66

x2 � 14x � 32x2 � 18x � 72

t2 � 20t � 96t2 � 12t � 24

n2 � 4n � 45n2 � 6n � 40

n2 � 3n � 18n2 � 6n � 27

n2 � 7n � 10n2 � 11n � 28

x2 � 5x � 4x2 � 11x � 18

x2 � 11x � 24x2 � 13x � 40

x2 � 9x � 14x2 � 10x � 24


Add �s2 to both sides

Add �4s to both sides

Add �4 to both sides

The solution of �1 is not possible for the length of a side, so the shortest side (s) is oflength 3. The other two sides (s � 1 and s � 2) have lengths of 4 and 5.


The length of one leg of a right triangle is 7 inches more then the length of theother leg. The length of the hypotenuse is 13 inches. Find the length of the two legs.5 inches and 12 inches ■


1. Any trinomial of the form x2 � bx � c can be factored (using integers) into theproduct of two binomials.

2. To factor x2 � 4x � 60 we look for two numbers whose product is �60 andwhose sum is �4.

3. A trinomial of the form x2 � bx � c will never have a common factor other than 1.

4. If n represents an odd integer, then n � 1 represents the next consecutive oddinteger.

5. The Pythagorean theorem only applies to right triangles.6. In a right triangle the longest side is called the hypotenuse.7. The polynomial x2 � 25x � 72 is not factorable.8. The polynomial x2 � 27x � 72 is not factorable.9. The solution set of the equation x2 � 2x � 63 � 0 is {�9, 7}.

10. The solution set of the equation x2 � 5x � 66 � 0 is {�11, �6}.

s � 3 or s � �1

s � 3 � 0 or s � 1 � 0

1s � 3 2 1s � 1 2 � 0

s2 � 2s � 3 � 0

s2 � 2s � 1 � 4

s2 � 2s � 1 � 4s � 4

Problem Set 5.3

Not factorable

Not factorable

1x � 3 2 1x � 14 21x � 11 2 1x � 6 21x � 6 2 1x � 12 2

1t � 8 2 1t � 12 21n � 5 2 1n � 9 21n � 10 2 1n � 4 21n � 3 2 1n � 6 21n � 9 2 1n � 3 21n � 2 2 1n � 5 21n � 7 2 1n � 4 21x � 1 2 1x � 4 21x � 2 2 1x � 9 21x � 3 2 1x � 8 21x � 5 2 1x � 8 21x � 2 2 1x � 7 21x � 4 2 1x � 6 2

Not factorable

Not factorable

51, 8653, 665�5, �465�7, �36

1a � 6b 2 1a � 9b 21a � 8b 2 1a � 4b 21x � 2y 2 1x � 6y 21x � 2y 2 1x � 5y 21x � 4 2 1x � 16 2

1x � 12 2 1x � 6 21x � 6 2 1x � 15 21x � 5 2 1x � 16 21y � 5 2 1y � 6 21y � 9 2 1y � 8 2




35. 36.

37. 38.

39. 40.

41. 42.

43. {2, 14}

44. {3, 15}

45. 46.

47. 48.

49. {�6, 4}

50. {�2, 8}

3 Solve Word Problems IncludingConsecutive Number Problems

For Problems 51– 64, set up an equation and solve eachproblem.

51. Find two consecutive integers whose product is 56.�8 and �7 or 7 and 8

52. Find two consecutive odd whole numbers whose productis 63. 7 and 9

53. Find two consecutive even whole numbers whose productis 168. 12 and 14

54. One number is 2 larger than another number. The sum oftheir squares is 100. Find the numbers.�8 and �6, or 6 and 8

55. Find four consecutive integers such that the product ofthe two larger integers is 22 less than twice the product ofthe two smaller integers. �4, �3, �2, �1 or 7, 8, 9, 10

56. Find three consecutive integers such that the product ofthe two smaller integers is 2 more than 10 times thelargest integer. 11, 12, and 13 or �2, �1, and 0

57. One number is 3 smaller than another number. Thesquare of the larger number is 9 larger than 10 times thesmaller number. Find the numbers. 4 and 7 or 0 and 3

58. The area of the floor of a rectangular room is 84 squarefeet. The length of the room is 5 feet more than its width.Find the width and length of the room.7 feet and 12 feet

�x2 � 6x � 16 � 0

�x2 � 2x � 24 � 055, 7652, 86 x1x � 12 2 � �35x1x � 10 2 � �16

5�10, 26x2 � 8x � 205�12, 16x2 � 11x � 12

x2 � 18x � 45 � 0

x2 � 16x � 28 � 0

t2 � t � 72 � 0t2 � t � 56 � 0

n2 � 8n � 48 � 0n2 � 6n � 40 � 0

n2 � 3n � 18 � 0n2 � 5n � 36 � 0

x2 � x � 12 � 0x2 � 3x � 10 � 0 59. Suppose that the width of a certain rectangle is 3 inchesless than its length. The area is numerically 6 less thantwice the perimeter. Find the length and width of therectangle. 9 inches by 6 inches

60. The sum of the areas of a square and a rectangle is 64 square centimeters. The length of the rectangle is 4 centimeters more than a side of the square, and the width of the rectangle is 2 centimeters more than a side ofthe square. Find the dimensions of the rectangle and the square.

8 centimeters by 6 centimeters and 4 centimeters by 4 centimeters61. The perimeter of a rectangle is 30 centimeters, and the

area is 54 square centimeters. Find the width and length of the rectangle. [Hint: Let w represent the width;then 15 � w represents the length.]6 centimeters by 9 centimeters

62. The perimeter of a rectangle is 44 inches, and its area is 120 square inches. Find the width and length of the rectangle. 10 inches by 12 inches

63. An apple orchard contains 84 trees. The number of treesper row is 5 more than the number of rows. Find thenumber of rows. 7 rows

64. A room contains 54 chairs. The number of rows is 3 lessthan the number of chairs per row. Find the number of rows. 6 rows of chairs


For Problems 65– 68, set up an equation and solve for eachproblem.

65. Suppose that one leg of a right triangle is 7 feet shorterthan the other leg. The hypotenuse is 2 feet longer thanthe longer leg. Find the lengths of all three sides of theright triangle. 8 feet, 15 feet, and 17 feet

66. Suppose that one leg of a right triangle is 7 meters longerthan the other leg. The hypotenuse is 1 meter longer thanthe longer leg. Find the lengths of all three sides of theright triangle. 5 meters, 12 meters, and 13 meters

67. Suppose that the length of one leg of a right triangle is2 inches less than the length of the other leg. If the lengthof the hypotenuse is 10 inches, find the length of each leg.6 inches and 8 inches

68. The length of one leg of a right triangle is 3 centimetersmore than the length of the other leg. The length of thehypotenuse is 15 centimeters. Find the lengths of the two legs. 9 centimeters and 12 centimeters


THOUGHTS INTO WORDS

69. What does the expression “not factorable using integers”mean to you?

70. Discuss the role that factoring plays in solving equations.

71. Explain how you would solve the equation (x � 3) �(x � 4) � 0, and also how you would solve (x � 3) �(x � 4) � 8.

5�9, 865�8, 765�4, 1265�4, 1065�6, 365�9, 465�3, 465�2, 56


Therefore, we can solve the given equation as follows:

or n � �14

The solution set is {�14, �12}.

Solve each of the following equations.

a. {�18, �12}

b. {�21, �14}

c. {16, 24}

d. {15, 25}

e. {�24, 18}

f. {�16, 32}n2 � 16n � 512 � 0

n2 � 6n � 432 � 0

n2 � 40n � 375 � 0

n2 � 40n � 384 � 0

n2 � 35n � 294 � 0

n2 � 30n � 216 � 0

n � �12

n � 12 � 0 or n � 14 � 0

1n � 12 2 1n � 14 2 � 0

n2 � 26n � 168 � 0

For Problems 72–75, factor each trinomial and assume that all variables appearing as exponents represent positiveintegers.

72. 73.

74. 75.

76. Suppose that we want to factor so that we can solve the equation � 0. We needto find two positive integers whose product is 168 andwhose sum is 26. Because the constant term, 168, is rather large, let’s look at it in prime factored form:

Now we can mentally form two numbers by using all ofthese factors in different combinations. Using two 2sand the 3 in one number and the other 2 and the 7 in another number produces � 12 and � 14.2 # 72 # 2 # 3

168 � 2 # 2 # 2 # 3 # 7

n2 � 26n � 168n2 � 26n � 168

x2a � 6xa � 27x2a � 2xa � 8

x2a � 13xa � 40x2a � 10xa � 24

5.4 Factoring Trinomials of the Form ax2 � bx � c 265


Answers to the Example Practice Skills1. (y � 5)(y � 4) 2. (m � 3)(m � 5) 3. (y � 8)(y � 3) 4. (y � 2)(y � 6) 5. (y � 1)(y � 6)6. (a � 4)(a � 3) 7. Not factorable 8. {2, 12} 9. {�7, 4} 10. {�2, 20} 11. �11, �10 or 10, 1112. 8 yards, 10 yards 13. 5 inches and 12 inches

Answers to the Concept Quiz1. False 2. True 3. True 4. False 5. True 6. True 7. True 8. False 9. True 10. False

1x a � 9 2 1x

a � 3 21x a � 2 2 1x

a � 4 21x

a � 8 2 1x a � 5 21x

a � 4 2 1x a � 6 2

5.4 Factoring Trinomials of the Form ax2 � bx � cO B J E C T I V E S

1 Factor Trinomials Where the Leading Coefficient Is Not 1

2 Solve Equations That Involve Factoring

1 Factor Trinomials Where the Leading Coefficient Is Not 1

Now let’s consider factoring trinomials where the coefficient of the squared term isnot 1. We first illustrate an informal trial-and-error technique that works well for cer-tain types of trinomials. This technique relies on our knowledge of multiplication ofbinomials.

Factor .

Solution

By looking at the first term, 2x2, and the positive signs of the other two terms, we knowthat the binomials are of the form

(2x � _____)(x � _____)

2x2 � 7x � 3E X A M P L E 1


Because the factors of the constant term 3 are 1 and 3, we have only two possibilitiesto try:

(2x � 3)(x � 1) or (2x � 1)(x � 3)

By checking the middle term of both of these products, we find that the second oneyields the correct middle term of 7x. Therefore,


Factor 3y2 � 16y � 5. (3y � 1)(y � 5) ■

Factor .

Solution

First, we note that can be written as or . Second, because the middleterm of the trinomial is negative, and the last term is positive, we know that the bino-mials are of the form

(2x � _____)(3x � _____) or (6x � _____)(x � _____)

The factors of the constant term 5 are 1 and 5, so we have the following possibilities:

By checking the middle term for each of these products, we find that the product (2x � 5)(3x � 1) produces the desired term of �17x. Therefore,


Factor 4a2 � 8a � 3. (2a � 1)(2a � 3) ■

Factor .

Solution

First, we note that the polynomial 8x2 � 8x � 30 has a common factor of 2. Factor-ing out the common factor gives us 2(4x2 � 4x � 15). Now we need to factor 4x2 � 4x � 15.

Now, we note that can be written as or . Also, the lastterm, �15, can be written as (1)(�15), (�1)(15), (3)(�5), or (�3)(5). Thus we cangenerate the possibilities for the binomial factors as follows:

Using 1 and �15 Using �1 and 15(4x � 15)(x � 1) (4x � 1)(x � 15)

(4x � 1)(x � 15) (4x � 15)(x � 1)

(2x � 1)(2x � 15) (2x � 1)(2x � 15)

2x # 2x4x # x4x2

8x2 � 8x � 30

6x2 � 17x � 5 � 12x � 5 2 13x � 1 2

16x � 5 2 1x � 1 2 16x � 1 2 1x � 5 212x � 5 2 13x � 1 2 12x � 1 2 13x � 5 2

6x # x2x # 3x6x2

6x2 � 17x � 5

2x2 � 7x � 3 � 12x � 1 2 1x � 3 2


E X A M P L E 2

E X A M P L E 3

05-W4801-AM1.qxd 8/19/08 8:46 PM 6REVISED PAGES

Using 3 and �5 Using �3 and 5(4x � 3)(x � 5) (4x � 3)(x � 5)

(4x � 5)(x � 3) (4x � 5)(x � 3)

✓ (2x � 5)(2x � 3) (2x � 5)(2x � 3)

By checking the middle term of each of these products, we find that the product indi-cated with a check mark produces the desired middle term of �4x. Therefore,


Factor 6y2 � 27y � 30. 3(y � 2)(2y � 5) ■

Let’s pause for a moment and look back over Examples 1, 2, and 3. Example 3clearly created the most difficulty because we had to consider so many possibilities.We have suggested one possible format for considering the possibilities, but as youpractice such problems, you may develop a format of your own that works better foryou. Whatever format you use, the key idea is to organize your work so that you con-sider all possibilities. Let’s look at another example.

Factor .

Solution

First, we note that can be written as or . Second, because the middleterm is positive and the last term is positive, we know that the binomials are of theform

(4x � _____)(x � _____) or (2x � _____)(2x � _____)

Because 9 can be written as or , we have only the following five possibilitiesto try:

When we try all of these possibilities, we find that none of them yields a middle termof 6x. Therefore, 4x2 � 6x � 9 is not factorable using integers.


Factor 6a2 � 3a � 7. Not factorable ■

Remark: Example 4 illustrates the importance of organizing your work so thatyou try all possibilities before you conclude that a particular trinomial is not factorable.

2 Solve Equations That Involve FactoringThe ability to factor certain trinomials of the form ax2 � bx � c provides us withgreater equation-solving capabilities. Consider the next examples.

12x � 3 2 12x � 3 2 12x � 1 2 12x � 9 214x � 3 2 1x � 3 2 14x � 1 2 1x � 9 214x � 9 2 1x � 1 2

3 # 39 # 1

2x # 2x4x # x4x2

4x2 � 6x � 9

8x2 � 8x � 30 � 212x � 5 2 12x � 3 2


E X A M P L E 4


Solve .

Solution

Factoring 3x2 � 17x � 10 as (x � 5)(3x � 2) mayrequire some extra work on scratch paper


The solution set is . Check it!


Solve 2y2 � 7y � 6 � 0. ■

Solve .

Solution



Solve 2a2 � 5a � 12 � 0. ■


1. Any trinomial of the form ax2 � bx � c can be factored (using integers) into theproduct of two binomials.

2. To factor 2x2 � x � 3, we look for two numbers whose product is �3 and whosesum is �1.

3. A trinomial of the form ax2 � bx � c will never have a common factor other than 1.

4. The factored form (x � 3)(2x � 4) is factored completely.5. The difference-of-squares polynomial 9x2 � 25 could be written as the trinomial

9x2 � 0x � 25.6. The polynomial 12x2 � 11x � 12 is not factorable.7. The solution set of the equation 6x2 � 13x � 5 � 0 is .

8. The solution set of the equation 18x2 � 39x � 20 � 0 is .e 56

, 43f

e 13

, 25f

e�32

, 4 f

e�56

, 34f

x �34

or x � �56

4x � 3 or 6x � �5

4x � 3 � 0 or 6x � 5 � 0

14x � 3 2 16x � 5 2 � 0

24x2 � 2x � 15 � 0

24x2 � 2x � 15 � 0

e�2, �32f

e�5, �23f

x � �5 or x � �23

x � �5 or 3x � �2

x � 5 � 0 or 3x � 2 � 0

1x � 5 2 13x � 2 2 � 0

3x2 � 17x � 10 � 0

3x2 � 17x � 10 � 0



E X A M P L E 6

E X A M P L E 5


33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

49.

50. e�45

, 25f5x15x � 2 2 � 8

e�54

, 14f16x1x � 1 2 � 5

e�34

, �23f12x2 � 17x � �6

e�52

, �23f6x2 � 19x � �10

e32

, 52f4x2 � 16x � 15 � 0

e 25

, 52f10x2 � 29x � 10 � 0

e�52

, 16f12n2 � 28n � 5 � 0

e�23

, 43f9x2 � 6x � 8 � 0

e�23

, 73f9y2 � 15y � 14 � 0

e�38

, 32f16y2 � 18y � 9 � 0

e�9, 32f2t2 � 15t � 27 � 0

e�7, 56f6t2 � 37t � 35 � 0

e 52

, 83f6n2 � 31n � 40 � 0

e 25

, 73f15n2 � 41n � 14 � 0

e 34

, 7 f4x2 � 31x � 21 � 0

e 13

, 8 f3x2 � 25x � 8 � 0

e�103

, �25f15x2 � 56x � 20 � 0

e�23

, �14f12x2 � 11x � 2 � 01 Factor Trinomials Where the Leading

Coefficient Is Not 1

For Problems 1–30, factor each of the trinomials completely.Indicate any that are not factorable using integers.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

27. 28.

29. 30.

2 Solve Equations That Involve Factoring


31.

32. e�5, �13f3x2 � 16x � 5 � 0

e�6, �12f2x2 � 13x � 6 � 0

8t2 � 3t � 420x2 � 31x � 12

15x2 � 34x � 1514x2 � 55x � 21

20n2 � 27n � 96n2 � 2n � 5

9y2 � 52y � 1212y2 � 79y � 35

12t3 � 20t2 � 25t9t2 � 15t � 14

9x2 � 15x � 148x2 � 2x � 21

6x2 � 17x � 1221x2 � 90x � 24

10x2 � x � 518x2 � 45x � 7

7x2 � 19x � 102x2 � x � 7

4n2 � 17n � 154n2 � 26n � 48

6y2 � 4y � 165y2 � 33y � 14

8x2 � 30x � 712x2 � 31x � 20

5x2 � 22x � 84x2 � 25x � 6

12x2 � 19x � 46x2 � 19x � 10

2x2 � 9x � 43x2 � 7x � 2


Problem Set 5.4

THOUGHTS INTO WORDS

51. Explain your thought process when factoring

� 17x � 20

52. Your friend factors � 32x � 32 as follows:

Is she correct? Do you have any suggestions for her?

� 81x � 2 2 1x � 2 2 � 41x � 2 2 12 2 1x � 2 2

8x2 � 32x � 32 � 14x � 8 2 12x � 4 28x2

24x2

53. Your friend solves the equation � 32x � 32 � 0 asfollows:

The solution set is {2}. Is your friend correct? Do you haveany changes to recommend?

x � 2 or x � 2

4x � 8 or 2x � 4

4x � 8 � 0 or 2x � 4 � 0

14x � 8 2 12x � 4 2 � 0

8x2 � 32x � 32 � 0

8x2

Not factorable

Not factorable

Not factorable

Not factorable14x � 3 2 15x � 4 215x � 3 2 13x � 5 217x � 3 2 12x � 7 215n � 3 2 14n � 3 219y � 2 2 1y � 6 2112y � 5 2 1y � 7 2t 16t � 5 2 12t � 5 213t � 2 2 13t � 7 213x � 2 2 13x � 7 214x � 7 2 12x � 3 213x � 4 2 12x � 3 2317x � 2 2 1x � 4 2

13x � 7 2 16x � 1 217x � 5 2 1x � 2 214n � 3 2 1n � 5 2212n � 3 2 1n � 8 2213y � 4 2 1y � 2 215y � 2 2 1y � 7 214x � 1 2 12x � 7 214x � 5 2 13x � 4 215x � 2 2 1x � 4 214x � 1 2 1x � 6 214x � 1 2 13x � 4 212x � 5 2 13x � 2 212x � 1 2 1x � 4 213x � 1 2 1x � 2 2



1 Factor Perfect-Square TrinomialsBefore we summarize our work with factoring techniques, let’s look at two morespecial factoring patterns. These patterns emerge when multiplying binomials.Consider the following examples.

In general, . Also,

In general, . Thus we have the followingpatterns.

1a � b 2 2 � 1a � b 2 1a � b 2 � a2 � 2ab � b2

15x � 2 2 2 � 15x � 2 2 15x � 2 2 � 25x2 � 20x � 4

13x � 4 2 2 � 13x � 4 2 13x � 4 2 � 9x2 � 24x � 16

1x � 6 2 2 � 1x � 6 2 1x � 6 2 � x2 � 12x � 36

1a � b 2 2 � 1a � b 2 1a � b 2 � a2 � 2ab � b2

14x � 7 2 2 � 14x � 7 2 14x � 7 2 � 16x2 � 56x � 49

12x � 3 2 2 � 12x � 3 2 12x � 3 2 � 4x2 � 12x � 9

1x � 5 2 2 � 1x � 5 2 1x � 5 2 � x2 � 10x � 25


5.5 Factoring, Solving Equations, and Problem SolvingO B J E C T I V E S

1 Factor Perfect-Square Trinomials

2 Recognize the Different Types of Factoring Patterns

3 Use Factoring to Solve Equations

4 Solve Word Problems That Involve Factoring

Perfect-Square Trinomials

a2 � 2ab � b2 � 1a � b 2 2a2 � 2ab � b2 � 1a � b 2 2

Trinomials of the form or are called perfect-squaretrinomials. They are easy to recognize because of the nature of their terms. For ex-ample, � 30x � 25 is a perfect-square trinomial for these reasons:9x2

a2 � 2ab � b2a2 � 2ab � b2

Answers to the Example Practice Skills1. (3y � 1)(y � 5) 2. (2a � 1)(2a � 3) 3. 3(y � 2)(2y � 5) 4. Not factorable 5.

6.

Answers to the Concept Quiz1. False 2. False 3. False 4. False 5. True 6. True 7. False 8. True

e�32

, 4 fe�2, �

32f


1. The first term is a square:

2. The last term is a square:

3. The middle term is twice the product of the quantities being squared in thefirst and last terms:

Likewise, is a perfect-square trinomial for these reasons:

1. The first term is a square:

2. The last term is a square:

3. The middle term is twice the product of the quantities being squared in thefirst and last terms:

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

Here are some additional examples of perfect-square trinomials and theirfactored forms.

Factor the following.

(a) x2 � 16x � 64 (b) 16x2 � 56x � 49 (c) 25x2 � 20xy � 4y2

(d) 1 � 6y � 9y2 (e) 4m2 � 4mn � n2

Solution

(a)

(b)

(c)

(d)

(e)

You may want to do the middle part mentally, after you feel comfortable with theprocess.


Factor the following.

(a) a2 � 10a � 25 (b) 36x2 � 12x � 1 (c) 49m2 � 56mn � 16n2 ■

(a) (a � 5)2 (b) (6x � 1)2 (c) (7m � 4n)2


In this chapter, we have considered some basic factoring techniques one at a time, butyou must be able to apply them as needed in a variety of situations. Let’s first sum-marize the techniques and then consider some examples. These are the techniques wehave discussed in this chapter:

1. Factoring by using the distributive property to factor out the greatest com-mon monomial or binomial factor

2. Factoring by grouping

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

4m2 � 4mn � n2 � 12m 22 � 212m 2 1n 2 � 1n 22 � 12m � n 221 � 6y � 9y2 � 11 22 � 211 2 13y 2 � 13y 22 � 11 � 3y 2225x2 � 20xy � 4y2 � 15x 22 � 215x 2 12y 2 � 12y 22 � 15x � 2y 2216x2 � 56x � 49 � 14x 22 � 214x 2 17 2 � 17 22 � 14x � 7 22x2 � 16x � 64 � 1x 22 � 21x 2 18 2 � 18 22 � 1x � 8 22

25x2 � 40xy � 16y2 � 15x � 4y 229x2 � 30x � 25 � 13x � 5 2 2

215x 2 14y 214y 2215x 2 2

25x2 � 40xy � 16y2

213x 2 15 215 2213x 22

5.5 Factoring, Solving Equations, and Problem Solving 271

E X A M P L E 1


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

5. Factoring trinomials of the form into the product of two binomials

6. Factoring trinomials of the form into the product of two binomials

As a general guideline, always look for a greatest common monomial factor first, andthen proceed with the other factoring techniques.

In each of the following examples, we have factored completely wheneverpossible. Study these examples carefully and note the factoring techniques wehave used.

Factor completely 2x2 � 12x � 10.

Solution

First, factor out the common factor of 2


Factor completely 3x2 � 6x � 72. 3(x � 4)(x � 6) ■

Factor completely 4x2 � 36.

Solution

Remember that the sum of two squares is notfactorable using integers unless there is a common factor


Factor completely 4x2 � 100. 4(x2 � 25) ■

Factor completely 4t2 � 20t � 25.

Solution

If you fail to recognize a perfect-square trinomial, no harm is done; simply proceed tofactor into the product of two binomials, and then you will recognize that the two binomialsare the same


Factor completely 9x2 � 30x � 25. (3x � 5)2 ■

Factor completely x2 � 3x � 8.

Solution

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

4t2 � 20t � 25 � 12t � 5 2 2

4x2 � 36 � 4 1x2 � 9 2

� 21x � 1 2 1x � 5 2 2x2 � 12x � 10 � 21x2 � 6x � 5 2

ax2 � bx � c

x2 � bx � c


E X A M P L E 2

E X A M P L E 3

E X A M P L E 4

E X A M P L E 5


No two factors of �8 produce a sum of �3.


Factor completely x2 � 8x � 24. Not factorable ■

Factor completely 6y2 � 13y � 28.

Solution

.

We found the binomial factors as follows:

(y � _____)(6y � _____)

or or

(y � _____)(6y � _____) or

or or

(2y � _____)(3y � _____)

or

(2y � _____)(3y � _____)


Factor completely 3x2 � 10x � 8. (3x � 2)(x � 4) ■

Factor completely 32x2 � 50y2.

Solution

First, factor out the common factor of 2

Factor the difference of squares


Factor completely 4x2 � 100. 4(x � 5)(x � 5) ■

3 Use Factoring to Solve EquationsEach time we considered a new factoring technique in this chapter, we used that tech-nique to help solve some equations. It is important that you be able to recognize whichtechnique works for a particular type of equation.

� 214x � 5y 2 14x � 5y 2 32x2 � 50y2 � 2116x2 � 25y2 2

7 # 44 # 7

14 # 22 # 14

28 # 11 # 28

6y2 � 13y � 28 � 12y � 7 2 13y � 4 2


Product Sum

�2 � 4 � 2 �214 2 � �8

2 � 1�4 2 � �2 21�4 2 � �8

�1 � 8 � 7 �118 2 � �8

1 � 1�8 2 � �7 11�8 2 � �8

E X A M P L E 6

E X A M P L E 7


Solve .

Solution

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


Solve y2 � 4y. {0, 4} ■

Solve .

Solution

If abc � 0, then a � 0 or b � 0 or c � 0

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


Solve z3 � 25z � 0. {�5, 0, 5} ■

Solve .

Solution

The solution set is . Does it check?


Solve 3y2 � 2y � 8 � 0. ■e�2, 43f

e�15

, 32f

x � �15

or x �32

5x � �1 or 2x � 3

5x � 1 � 0 or 2x � 3 � 0

15x � 1 2 12x � 3 2 � 0

10x2 � 13x � 3 � 0

10x2 � 13x � 3 � 0

x � 0 or x � �6 or x � 6

x � 0 or x � 6 � 0 or x � 6 � 0

x1x � 6 2 1x � 6 2 � 0

x1x2 � 36 2 � 0

x3 � 36x � 0

x3 � 36x � 0

x � 0 or x � 25

x � 0 or x � 25 � 0

x1x � 25 2 � 0

x2 � 25x � 0

x2 � 25x

x2 � 25x


E X A M P L E 8

E X A M P L E 9

E X A M P L E 1 0


Solve .

Solution



Solve 25y2 � 20y � 4 � 0. ■

Pay special attention to the next example. We need to change the form of theoriginal equation before we can apply the property ab � 0 if and only if a � 0 or b � 0. The unique feature of this property is that an indicated product is set equalto zero.

Solve .

Solution

Multiply the binomials

Added �40 to each side

The solution set is {�9, 4}. Check it!


Solve (x � 1)(x � 2) � 20. {�6, 3} ■

Solve .

Solution

Multiplied both sides by

The solution set is {�10, 2}. Does it check?

n � �10 or n � 2

n � 10 � 0 or n � 2 � 0

1n � 10 2 1n � 2 2 � 0

12

n2 � 8n � 20 � 0

21n2 � 8n � 20 2 � 0

2n2 � 16n � 40 � 0

2n2 � 16n � 40 � 0

x � �9 or x � 4

x � 9 � 0 or x � 4 � 0

1x � 9 2 1x � 4 2 � 0

x2 � 5x � 36 � 0

x2 � 5x � 4 � 40

1x � 1 2 1x � 4 2 � 40

1x � 1 2 1x � 4 2 � 40

e�25f

e 72f

x �72

or x �72

2x � 7 or 2x � 7

2x � 7 � 0 or 2x � 7 � 0

12x � 7 2 12x � 7 2 � 0

12x � 7 2 2 � 0

4x2 � 28x � 49 � 0

4x2 � 28x � 49 � 0


E X A M P L E 1 1

E X A M P L E 1 2

E X A M P L E 1 3



Solve 3y2 � 12y � 36 � 0. {�2, 6} ■

4 Solve Word Problems That Involve FactoringThe preface to this book states that a common thread throughout the book is to learn a skill, to use that skill to help solve equations, and then to use equations to helpsolve problems. This approach should be very apparent in this chapter. Our new fac-toring skills have provided more ways of solving equations, which in turn gives usmore power to solve word problems. We conclude the chapter by solving a few moreexamples.

Apply Your SkillFind two numbers whose product is 65 if one of the numbers is 3 more than twice theother number.

Solution

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

If , then 2n � 3 � � 3 � �10. If n � 5, then 2n � 3 �

2(5) � 3 � 13. Thus the numbers are and �10, or 5 and 13.


Find two numbers whose product is 36 if one of the numbers is 1 more than twice theother number. 4, 9 or ■

Apply Your SkillThe area of a triangular sheet of paper is 14 square inches. One side of the triangleis 3 inches longer than the altitude to that side. Find the length of the one side and thelength of the altitude to that side.

Solution

Let h represent the altitude to the side.Then h � 3 represents the length of theside of the triangle (see Figure 5.5). Be-cause the formula for finding the area of

a triangle is A � , we have12

bh

�8,�92

�132

2 a�132bn � �

132

n � �132

or n � 5

2n � �13 or n � 5

2n � 13 � 0 or n � 5 � 0

12n � 13 2 1n � 5 2 � 0

2n2 � 3n � 65 � 0

n12n � 3 2 � 65


E X A M P L E 1 4

E X A M P L E 1 5

Figure 5.5

h + 3

h


17. 18.


21. 22.


25. 26.

27. 28.

29.

30.

31.

32. 33.

34. 35.

36. 37.

38. 39.

40. 12x � 7y 224x2 � 28xy � 49y21x � 6y 223x1x � 1 2 1x � 1 2 1x2 � 1 2 x2 � 12xy � 36y23x5 � 3x51x � 2 2 1x � 2 2 1x2 � 4 2813x2 � x � 4 2 5x4 � 8024x2 � 8x � 32612x2 � x � 5 2513x � 2 2 12x � 5 2 12x2 � 6x � 3030x2 � 55x � 50312x � 3 2 14x � 9 21x � 4y 2 12x � 9y 2 24x2 � 18x � 812x2 � xy � 36y2

15x � y 2 14x � 7y 220x2 � 31xy � 7y2

1x � 3 2 1y � 9 2xy � 3y � 9x � 27

1x � 5 2 1y � 8 2xy � 5y � 8x � 409xy1x � 3 214x � 3 2 16x � 5 29x2y � 27xy24x2 � 2x � 1512x � 5y 2251x � 2 2 13x � 7 24x2 � 20xy � 25y215x2 � 65x � 70

13x � 5 22 5x2 � 5x � 69x2 � 30x � 253y1y � 4 2 1y � 8 281x2 � 9 23y3 � 36y2 � 96y8x2 � 721a � 3 2 1a � 10 2a2 � 7a � 303a2 � 7a � 44n1n � 5 2 1n � 5 21n � 12 2 1n � 5 2 4n3 � 100nn2 � 7n � 601 Factor Perfect-Square Trinomials

For Problems 1–12, factor each of the perfect-square trinomials.

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11.

12.


For Problems 13– 40, factor each polynomial completely. Indi-cate any that are not factorable using integers.

13. 14.

15. 16. 16x � 1 2 15x � 1 22x1x � 6 2 1x � 6 2 30x2 � x � 12x3 � 72x12x � 1 2 1x � 8 2 x1x � 19 2x2 � 19x2x2 � 17x � 8

18x � y 2264x2 � 16xy � y2

14x � 3y 2216x2 � 24xy � 9y2

11 � 2x 221 � 4x � 4x212 � 9x 224 � 36x � 81x2

36a2 � 84a � 4914a � 1 2216a2 � 8a � 1

25n2 � 30n � 913n � 2 229n2 � 12n � 4

x2 � 24x � 1441x � 5 22x2 � 10x � 25

1x � 9 22x2 � 18x � 811x � 2 22x2 � 4x � 4


Multiplied both sides by 2

The solution �7 is not reasonable. Thus the altitude is 4 inches, and the length of theside to which that altitude is drawn is 7 inches.


The area of a triangular piece of glass is 30 square inches. One side of the triangle is4 inches longer than the altitude to that side. Find the length of that side and the lengthof the altitude to that side. Altitude is 6 inches, and the side is 10 inches ■

For Problems 1–7, match each factoring problem with the type of pattern that wouldbe used to factor the problem.

1. x2 � 2xy � y2 A. Trinomial with an x-squared coefficient of one2. x2 � y2 B. Common binomial factor3. ax � ay � bx � by C. Difference of two squares4. x2 � bx � c D. Common factor5. ax2 � bx � c E. Factor by grouping6. ax2 � ax � a F. Perfect-square trinomial7. (a � b)x � (a � b)y G. Trinomial with an x-squared coefficient of not one

h � �7 or h � 4

h � 7 � 0 or h � 4 � 0

1h � 7 2 1h � 4 2 � 0

h2 � 3h � 28 � 0

h2 � 3h � 28

h1h � 3 2 � 28

12

h1h � 3 2 � 14

Problem Set 5.5


16a � 7 2215n � 3 221x � 12 22



3 Use Factoring to Solve Equations


41. 42.

43. 44.

45. 46.

47.

48.

49.

50.

51.

52.

53. 54.

55. 56.

57.

58.

59.

60.

61.

62.

63. 64.

65.

66.

67.

68.

69.

70.

4 Solve Word Problems That Involve Factoring

For Problems 71–86, set up an equation and solve eachproblem.

71. Find two numbers whose product is 15 such that one ofthe numbers is seven more than four times the othernumber. �5 and �3 or and 12

54

e�72

, 5

12f24x2 � 74x � 35 � 0

e�43

, 58f24x2 � 17x � 20 � 0

50, 96x3 � 18x2 � 81x � 0

5�5, 06x3 � 10x2 � 25x � 0

e�65f25x2 � 60x � 36 � 0

e 43f9x2 � 24x � 16 � 0

5�4, 06x3 � �4x250, 66x3 � 6x2

e�72

, �3 f12n � 5 2 1n � 4 2 � �1

e�103

, 1 f13n � 1 2 1n � 2 2 � 12

5�18, �156n2 � 33n � 270 � 0

512, 166n2 � 28n � 192 � 0

e�43

, 34f12 � 7x � 12x2 � 0

e 25

, 65f12 � 40x � 25x2 � 0

5�1, 0, 965�4, 0, 66 2t3 � 16t2 � 18t � 0t3 � 2t2 � 24t � 0

5�5, 06�3x2 � 15x50, 662x2 � 12x

5261n � 3 2 1n � 7 2 � �25

5�3, �161n � 2 2 1n � 6 2 � �15

e�17

, 32f12n � 3 2 17n � 1 2 � 0

e 13

, 34f13n � 1 2 14n � 3 2 � 0

e�16

, 15f30n2 � n � 1 � 0

e�23

, 112f6n2 � 29n � 22 � 0

5�3, 0, 365�2, 0, 26 4x3 � 36x � 0�2x3 � 8x � 05�10, 265�3, 126x2 � 8x � 20 � 0x2 � 9x � 36 � 05�8, 0650, 56�3x2 � 24x � 04x2 � 20x � 0

72. Find two numbers whose product is 12 such that one of the numbers is four less than eight times the othernumber. and 8, or �1 and �12

73. Find two numbers whose product is �1. One of the num-bers is three more than twice the other number.

and 2 or �1 and 1

74. Suppose that the sum of the squares of three consecutiveintegers is 110. Find the integers.

�7, �6, and �5, or 5, 6, and 7

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

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

numbers. and or 4 and 9

76. One number is one less than three times another num-ber. If the product of the two numbers is 102, find thenumbers. and �18, or 6 and 17

77. In an office building, a room contains 54 chairs. The numberof chairs per row is three less than twice the number of rows.Find the number of rows and the number of chairs per row.6 rows and 9 chairs per row

78. An apple orchard contains 85 trees. The number of treesin each row is three less than four times the number ofrows. Find the number of rows and the number of treesper row. 5 rows with 17 trees per row

79. Suppose that the combined area of two squares is 360square feet. Each side of the larger square is three times aslong as a side of the smaller square. How big is each square?6 feet by 6 feet and 18 feet by 18 feet

80. The area of a rectangular slab of sidewalk is 45 square feet.Its length is 3 feet more than four times its width. Find thelength and width of the slab.Width of 3 feet and length of 15 feet

81. The length of a rectangular sheet of paper is 1 centimetermore than twice its width, and the area of the rectangle is 55 square centimeters. Find the width and length of therectangle. 5 centimeters by 11 centimeters

82. Suppose that the length of a certain rectangle is threetimes its width. If the length is increased by 2 inches andthe width increased by 1 inch, the newly formed rectanglehas an area of 70 square inches. Find the width and lengthof the original rectangle. 4 inches by 12 inches

83. The area of a triangle is 51 square inches. One side of thetriangle is 1 inch less than three times the length of the alti-tude to that side. Find the length of that side and the lengthof the altitude to that side.Side is 17 inches, altitude is 6 inches

84. Suppose that a square and a rectangle have equal areas.Furthermore, suppose that the length of the rectangle istwice the length of a side of the square, and the width of therectangle is 4 centimeters less than the length of a side ofthe square. Find the dimensions of both figures.

16 centimeters by 4 centimeters and 8 centimeters by 8 centimeters

�173

�435

�245

�12

32



86. The sum of the areas of two circles is 100p square centi-meters. The length of a radius of the larger circle is 2 cen-timeters more than the length of a radius of the smallercircle. Find the length of a radius of each circle.6 centimeters and 8 centimeters

85. The sum of the areas of two circles is 180p square inches.The length of a radius of the smaller circle is 6 inches lessthan the length of a radius of the larger circle. Find thelength of a radius of each circle. 12 inches and 6 inches


THOUGHTS INTO WORDS

87. When factoring polynomials, why do you think that it isbest to look for a greatest common monomial factor first?

88. Explain how you would solve (4x � 3)(8x � 5) � 0 andalso how you would solve (4x � 3)(8x � 5) � �9.

89. Explain how you would solve

(x � 2)(x � 3) � (x � 2)(3x � 1)

Do you see more than one approach to this problem?

Answers to the Example Practice Skills1. (a) (a � 5)2 (b) (6x � 1)2 (c) (7m � 4n)2 2. 3(x � 4)(x � 6) 3. 4(x2 � 25) 4. (3x � 5)2

5. Not factorable 6. (3x � 2)(x � 4) 7. 4(x � 5)(x � 5) 8. {0, 4} 9. {�5, 0, 5} 10.

11. 12. {�6, 3} 13. {�2, 6} 14. 4, 9 or �8, 15. Altitude is 6 inches and side is 10 inches

Answers to the Concept Quiz1. F or A 2. C 3. E 4. A 5. G 6. D 7. B

�92

e�25f

e�2, 43f


280

Chapter 5 SummaryCHAPTERREVIEW

OBJECTIVES SUMMARY EXAMPLE PROBLEMS

(continued)

Find the greatestcommon factor.

To find the greatest common factorof two or more monomials, deter-mine the greatest common factor ofthe numerical coefficients of thegiven monomials. Then for eachvariable that is common to all thegiven monomials, raise that variableto the lowest of its exponents in thegiven monomials.

Find the greatest common factorof 12a3b3, 18a2b2, and 54ab4.

SolutionThe greatest common factor ofthe numerical coefficient of allthree terms is 6. For the variable,a, a1 is the greatest common fac-tor of the given monomials. Forthe variable, b, b2 is the greatestcommon factor of the givenmonomials. So the greatest com-mon factor is 6ab2.

Problems 1– 4

Factor the differenceof two squares.

The factoring pattern a2 � b2 � (a � b)(a � b) is called“the difference of two squares.” Becareful not to apply the pattern tothe sum of two squares such as a2 � b2. There is no pattern for the sum of two squares.

Factor 4x2 � 81y2.

Solution4x2 � 81y2 � (2x � 9y)(2x � 9y)

Problems 13–16

Factor by grouping. Rewriting an expression such as ab � 3a � bc � 3c as a(b � 3) � c(b � 3) and then factoring out the common binomial factor of b � 3 so that a(b � 3) � c(b � 3) becomes (b � 3)( a � c) is called “factoringby grouping.”

Factor 15xy � 6x � 10y2 � 4y.

Solution15xy � 6x � 10y2 � 4y

� 3x(5y � 2) � 2y(5y � 2)� (5y � 2)(3x � 2y)

Problems 9–12

Factor out the greatestcommon factor.

The distributive property in theform ab � ac � a(b � c) providesthe basis for factoring out acommon factor.

Factor 8x5y2 � 20x3y4.

Solution8x5y2 � 20x3y4 � 4x3y2(2x2 � 5y2)

Problems 5– 8

Factor trinomials ofthe form x2 � bx � c.

For trinomials of the form x2 � bx � c, we want two factorsof c whose sum will be equal to b.

Factor a2 � 2a � 24.

SolutionThe factors of �24 that sum to �2 are �6 and 4.

a2 � 2a � 24 � (a � 6)(a � 4)

Problems 17–20

Factor trinomialswhere the leading coefficient is not 1.

To factor trinomials of the form ax2 � bx � c, the process is typically trial and error. Make anorganized listing of possible factorcombinations. Then multiply to seeif they are the correct factors. Besure to factor out any commonfactors first.

Factor 3x2 � x � 14.

Solution3x2 � x � 14 � (3x � 7)(x � 2)

Problems 21–24


281

(continued)

CHAPTERREVIEW

OBJECTIVE SUMMARY EXAMPLE PROBLEMS

Factor perfect-square trinomials.

Perfect-square trinomials are easy to rec-ognize because of the nature of theirterms. The first term and the last term willbe the squares of a quantity. The middleterm is twice the product of the quantitiesbeing squared in the first and last terms.

Factor 16x2 � 56x � 49.

Solution16x2 � 56x � 49

� (4x)2 � 2(4x)(7) � (7)2

� (4x � 7)2

Problems 25–28

Recognize the different types of factoring.

As a general guideline for factoring com-pletely, always look for a greatest com-mon factor first, and then proceed withone or more of the following techniques.1. Apply the difference-of-squares

pattern.2. Apply the perfect-square pattern.3. Factor a trinomial of the form

x2 � bx � c into the product of two binomials.

4. Factor a trinomial of the formax2 � bx � c into the product of twobinomials.

Factor 3x2 � 12xy � 12y2.

Solution3x2 � 12xy � 12y2

� 3(x2 � 4xy � 4y2) � 3(x � 2y)2

Problems 29–52

Use factoring to solve equations.

Property 5.1 states that for all real num-bers a and b, ab � 0 if and only if a � 0or b � 0. To solve equations by applyingthis property, first set the equation equalto 0. Proceed by factoring the other sideof the equation. Then set each factorequal to 0 and solve the equations.

Solve x2 � 7x � 8.

Solutionx2 � 7x � 8

x2 � 7x � 8 � 0(x � 8)(x � 1) � 0x � 8 � 0 or x � 1 � 0

x � �8 or x � 1The solution set is {�8, 1}.

Problems 53–72

Solve word problems that involve factoring.

Knowledge of factoring will expand thetechniques available for solving wordproblems. This chapter introduced thePythagorean theorem. This theorempertains to right triangles and states thatin any right triangle, the square of thelongest side is equal to the sum of thesquares of the other two sides. The formula for the theorem is written as a2 � b2 � c2, where a and b are the legsof the triangle, and c is the hypotenuse.

The length of one leg of a righttriangle is 1 inch longer than thelength of the other leg. The lengthof the hypotenuse is 5 inches. Findthe length of the two legs.

SolutionLet x represent one leg of the tri-angle. Then x � 1 will representthe other leg. We know that thehypotenuse is equal to 5. Applythe Pythagorean theorem.

x2 � (x � 1)2 � 5x2 � x2 � 2x � 1 � 25

2x2 � 2x � 24 � 02(x2 � x � 12) � 0

x2 � x � 12 � 0(x � 4)( x � 3) � 0

x � 4 � 0 or x � 3 � 0x � � 4 or x � 3

Because the length of a side ofthe triangle cannot be negative,the only viable answer is 3.Therefore, one leg of the triangleis 3 and the other leg is 4.

Problems 73–84



For Problems 29–52, factor completely. Indicate any poly-nomials that are not factorable using integers.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.


41. 42.

43. 44.

45.

46. 47.

48. 49.

50.

51. 52.

For Problems 53– 72, solve each equation.

53. 54.{�6, 2} {0, 11}

55. 56.

57. 58.{�2, 2}

59. 60.{�1, 0, 1}

61. 62.{�7, 4} {�5, 5}

63. 64.

65. 66.{�2, 0, 2} {8, 12}

67.

68. {�2, 3}

69. e�73

, 52f12x � 5 2 13x � 7 2 � 0

31x � 2 2 � x1x � 2 2 � 0

e�5, 34f4t2 � 17t � 15 � 0

x2 � 20x � 96 � 02x3 � 8x � 0

e�72

, 1 fe�6, 35f

4n2 � 10n � 145n2 � 27n � 18

1x � 2 2 1x � 2 2 � 21x2 � 3x � 28 � 0

e�94

, �27f

28x2 � 71x � 18 � 0t3 � t � 0

e�54f

16y2 � 40y � 25 � 06n2 � 24

e�83

, 33fe�4,

52f

9n2 � 21n � 8 � 02x2 � 3x � 20 � 0

x2 � 11xx2 � 4x � 12 � 0

n212n � 1 2 13n � 1 215x � y 2 13x � 2y 26n4 � 5n3 � n215x2 � 7xy � 2y2

1y � 3 2 1x � 2 2xy � 3x � 2y � 614x � 5 223x1x � 6 2 1x � 6 216x2 � 40x � 253x3 � 108x

2x1x � 1 2 1x � 1 217t � 4 2 13t � 1 22x3 � 2x21t2 � 5t � 4

1x � y 2 15 � a 25x � 5y � ax � ay

21n � 4 2 12n � 5 212x � y 2 1x � 2y 24n2 � 6n � 402x2 � 3xy � 2y2

x1x � y 2 1x � y 231n � 6 2 1n � 5 2x3 � xy23n2 � 3n � 9014x � 7 2 1x � 1 24x2 � 3x � 7x2 � 7x � 2416n � 5 2 13n � 1 21x � 1 2 1x � 1 2 1x2 � 1 218n2 � 9n � 5x4 � 13xy1y � 2x 21 y � 12 2 1y � 1 23xy2 � 6x2yy2 � 11y � 12n1n � 5 2 1n � 8 215x � 6 22n3 � 13n2 � 40n25x2 � 60x � 3612x � 1 2 12x � 5 213x � 2 2 13x � 2 24x2 � 8x � 59x2 � 43x1x � 7 21x � 2 2 1x � 7 23x2 � 21xx2 � 9x � 14

For Problems 1– 4, find the greatest common factor of theexpressions.

1. 6x2y2, 4x3y, and 14x4y3 2x2y

2. 9xy2, 21x3y2, and 15x4y3 3xy2

3. 24a3b2, 8a2b3, and 12ab3 4ab2

4. 36m5n3, 24m3n4, and 60m4n2 12m3n2

For Problems 5–8, factor out the greatest common factor.

5. 15x4 � 21x2 3x2(5x2 � 7)

6. 24a3 � 20a6 4a3(6 � 5a3)

7. 10x4y � 50x3y2 � 5x2y3 5x2y(2x2 � 10xy � y2)

8. 12m2n � 20m3n2 � 24m4n3 4m2n(3 � 5mn � 6m2n2)

For Problems 9–12, factor completely by grouping.

9. ab � 4a � 3b � 12 (a � 3)(b � 4)

10. 2xy � 12x � 5y � 30 (2x � 5)(y � 6)

11. x3 � x2y � xy2 � y3 (x2 � y2)(x � y)

12. 8m2 � 2m � 12mn � 3n (2m � 3n)(4m � 1)

For Problems 13–16, factor the difference of squares. Be sureto factor completely.

13. 25x2 � 16y2 (5x � 4)(5x � 4)

14. 18x2 � 50 2(3x � 5)(3x � 5)

15. x4 � 16 (x2 � 4)(x � 2)(x � 2)

16. x3 � 49x x(x � 7)(x � 7)

For Problems 17–28, factor completely.

17. x2 � 11x � 24 (x � 3)(x � 8)

18. x2 � 13x � 12 (x � 12)(x � 1)

19. x2 � 5x � 24 (x � 8)(x � 3)

20. x2 � 8x � 20 (x � 10)(x � 2)

21. 2x2 � 5x � 3 (2x � 3)(x � 1)

22. 5x2 � 17x � 6 (5x � 2)(x � 3)

23. 4x2 � 5x � 6 (4x � 3)(x � 2)

24. 3x2 � 10x � 8 (3x � 2)(x � 4)

25. 9x2 � 12x � 4 (3x � 2)2

26. 4x2 � 20x � 25 (2x � 5)2

27. 16x2 � 8x � 1 (4x � 1)2

28. x2 � 12xy � 36y2 (x � 6y)2

Chapter 5 Review Problem Set


78. Find three consecutive positive odd whole numbers suchthat the sum of the squares of the two smaller numbers isnine more than the square of the largest number.7, 9, and 11

79. The number of books per shelf in a bookcase is one lessthan nine times the number of shelves. If the bookcasecontains 140 books, find the number of shelves. 4 shelves

80. The combined area of a square and a rectangle is 225square yards. The length of the rectangle is eight timesthe width of the rectangle, and the length of a side of thesquare is the same as the width of the rectangle. Find thedimensions of the square and the rectangle.A 5-by-5-yard square and a 5-by-40-yard rectangle

81. Suppose that we want to find two consecutive integerssuch that the sum of their squares is 613. What are they?�18 and �17 or 17 and 18

82. If numerically the volume of a cube equals the totalsurface area of the cube, find the length of an edge ofthe cube. 6 units

83. The combined area of two circles is 53p square meters. The length of a radius of the larger circle is 1 meter more than three times the length of a radius of thesmaller circle. Find the length of a radius of each circle.2 meters and 7 meters

84. The product of two consecutive odd whole numbers is oneless than five times their sum. Find the numbers.9 and 11

70. {�9, 6}

71. 72.

Set up an equation and solve each of the following problems.

73. The larger of two numbers is one less than twice thesmaller number. The difference of their squares is 33.Find the numbers. and or 4 and 7

74. The length of a rectangle is 2 centimeters less than fivetimes the width of the rectangle. The area of the rectangleis 16 square centimeters. Find the length and width of therectangle.

75. Suppose that the combined area of two squares is 104square inches. Each side of the larger square is five timesas long as a side of the smaller square. Find the size of eachsquare. A 2-by-2-inch square and a 10-by-10-inch square

76. The longer leg of a right triangle is one unit shorter thantwice the length of the shorter leg. The hypotenuse is oneunit longer than twice the length of the shorter leg. Findthe lengths of the three sides of the triangle. 8 by 15 by 17

77. The product of two numbers is 26, and one of the numbersis one larger than six times the other number. Find thenumbers. and �12 or 2 and 13�

136

�193

�83

e 43

, 52fe�5,

32f

�23x � 6x2 � �20�7n � 2n2 � �15

1x � 4 2 1x � 1 2 � 50

Chapter 5 Review Problem Set 283

The length is 8 centimeters and the width is 2 centimeters


284

Chapter 5 Test

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

For Problems 1–10, factor each expression completely.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.


11. {�3, 3}

12. {�6, 1}

13. {0, 8}

14.

15. {�6, 2}

16. {�12, �4, 0}

17. {5, 9}

18.

19.

20. {�5, 0, 5}

21.

For Problems 22–25, set up an equation and solve each problem.

22. The length of a rectangle is 2 inches less than twice its width. If the area of therectangle is 112 square inches, find the length of the rectangle. 14 inches

23. The length of one leg of a right triangle is 4 centimeters more than the length ofthe other leg. The length of the hypotenuse is 8 centimeters more than the lengthof the shorter leg. Find the length of the shorter leg. 12 centimeters

24. A room contains 112 chairs. The number of chairs per row is five less than threetimes the number of rows. Find the number of chairs per row.16 chairs per row

25. Suppose the sum of squares of three consecutive integers is 77. Find the integers.�6, �5, �4 or 4, 5, 6

e 75f25n2 � 70n � 49 � 0

3x3 � 75x

e�4, 23f8 � 10x � 3x2 � 0

e�12, 13f3t2 � 35t � 12

91x � 5 2 � x1x � 5 2 � 0

x3 � 16x2 � 48x � 0

1x � 3 2 1x � 7 2 � �9

e�52

, 23f13x � 2 2 12x � 5 2 � 0

4n2 � 32n

x2 � 5x � 6 � 0

7x2 � 63

17 � 2x 2 14 � 3x 228 � 13x � 6x2

2x15x � 6 2 13x � 4 230x3 � 76x2 � 48x

61x2 � 4 26x2 � 24

14x � 3 2 1x � 5 24x2 � 17x � 15

1x � y 2 1a � 2b 2ax � ay � 2bx � 2by

312n � 1 2 13n � 2 218n2 � 21n � 6

1x � 9 2 1x � 12 2x2 � 21x � 108

2x1x � 1 2 1x � 1 22x3 � 2x

1x � 3 2 1x � 8 2x2 � 5x � 24

1x � 5 2 1x � 2 2x2 � 3x � 10


factoring, solvingequations, and problem solving 5 · pdf file5.2 factoring the difference of...

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