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E. Polynomial Equations and the Zero Product Property

# E. Polynomial Equations and the Zero Product Property

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Section R.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

A quadratic equation can be written in the form

ax2 ϩ bx ϩ c ϭ 0,

with a, b, c ʦ ‫ޒ‬, and a

0.

Notice that a is the leading coefficient, b is the coefficient of the linear (first

degree) term, and c is a constant. All quadratic equations have degree two, but can have

one, two, or three terms. The equation n2 Ϫ 81 ϭ 0 is a quadratic equation with two

terms, where a ϭ 1, b ϭ 0, and c ϭ Ϫ81.

EXAMPLE 10

Determining Whether an Equation Is Quadratic

State whether the given equation is quadratic. If yes, identify coefficients a, b, and c.

Ϫ3

a. 2x2 Ϫ 18 ϭ 0

b. z Ϫ 12 Ϫ 3z2 ϭ 0

c.

xϩ5ϭ0

4

d. z3 Ϫ 2z2 ϩ 7z ϭ 8

e. 0.8x2 ϭ 0

Solution

WORTHY OF NOTE

The word quadratic comes from the

square. The word historically refers

to the “four sidedness” of a square,

but mathematically to the area of a

square. Hence its application to

polynomials of the form

ax2 ϩ bx ϩ c, where the variable of

Standard Form

a.

2x Ϫ 18 ϭ 0

yes, deg 2

aϭ2

b.

Ϫ3z ϩ z Ϫ 12 ϭ 0

yes, deg 2

a ϭ Ϫ3

c.

Ϫ3

xϩ5ϭ0

4

no, deg 1

(linear equation)

d.

z3 Ϫ 2z2 ϩ 7z Ϫ 8 ϭ 0

no, deg 3

(cubic equation)

e.

0.8x ϭ 0

yes, deg 2

2

2

2

Coefficients

bϭ0

a ϭ 0.8

c ϭ Ϫ18

bϭ1

bϭ0

c ϭ Ϫ12

cϭ0

Now try Exercises 53 through 64

With quadratic and other polynomial equations, we generally cannot isolate the

variable on one side using only properties of equality, because the variable is raised to

different powers. Instead we attempt to solve the equation by factoring and applying

the zero product property.

Zero Product Property

If A and B represent real numbers or real-valued expressions

and A # B ϭ 0,

then A ϭ 0 or B ϭ 0.

In words, the property says, If the product of any two (or more) factors is equal to

zero, then at least one of the factors must be equal to zero. We can use this property to

solve higher degree equations after rewriting them in terms of equations with lesser

degree. As with linear equations, values that make the original equation true are called

solutions or roots of the equation.

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CHAPTER R A Review of Basic Concepts and Skills

EXAMPLE 11

Solving Equations Using the Zero Product Property

Solve by writing the equations in factored form and applying the zero product property.

a. 3x2 ϭ 5x

b. Ϫ5x ϩ 2x2 ϭ 3

c. 4x2 ϭ 12x Ϫ 9

Solution

3x2 ϭ 5x

3x Ϫ 5x ϭ 0

x13x Ϫ 52 ϭ 0

x ϭ 0 or 3x Ϫ 5 ϭ 0

5

x ϭ 0 or

3

b.

Ϫ5x ϩ 2x2 ϭ 3

2x2 Ϫ 5x Ϫ 3 ϭ 0

12x ϩ 121x Ϫ 32 ϭ 0

or x Ϫ 3 ϭ 0

2x ϩ 1 ϭ 0

1

xϭϪ

or

xϭ3

2

c.

4x2 ϭ 12x Ϫ 9

2

4x Ϫ 12x ϩ 9 ϭ 0

12x Ϫ 3212x Ϫ 32 ϭ 0

2x Ϫ 3 ϭ 0 or 2x Ϫ 3 ϭ 0

3

3

or

2

2

a.

2

given equation

standard form

factor

set factors equal to zero (zero product property)

result

given equation

standard form

factor

set factors equal to zero (zero product property)

result

given equation

standard form

factor

set factors equal to zero (zero product property)

result

3

This equation has only the solution x ϭ , which we call a repeated root.

2

Now try Exercises 65 through 88

CAUTION

EXAMPLE 12

Consider the equation x2 Ϫ 2x Ϫ 3 ϭ 12. While the left-hand side is factorable, the result

is 1x Ϫ 321x ϩ 12 ϭ 12 and finding a solution becomes a “guessing game” because the

equation is not set equal to zero. If you misapply the zero factor property and say that

x Ϫ 3 ϭ 12 or x ϩ 1 ϭ 12, the “solutions” are x ϭ 15 or x ϭ 11, which are both incorrect!

After subtracting 12 from both sides, x2 Ϫ 2x Ϫ 3 ϭ 12 becomes x2 Ϫ 2x Ϫ 15 ϭ 0

giving 1x Ϫ 521x ϩ 32 ϭ 0 with solutions x ϭ 5 or x ϭ Ϫ3.

Solving Polynomials by Factoring

Solve by factoring: 4x3 Ϫ 40x ϭ 6x2.

Solution

4x3 Ϫ 40x ϭ 6x2

4x Ϫ 6x2 Ϫ 40x ϭ 0

2x 12x2 Ϫ 3x Ϫ 202 ϭ 0

2x12x ϩ 521x Ϫ 42 ϭ 0

2x ϭ 0 or 2x ϩ 5 ϭ 0 or x Ϫ 4 ϭ 0

x ϭ 0 or

x ϭ Ϫ5

or

xϭ4

2

3

given equation

standard form

common factor is 2x

factored form

zero product property

result — solve for x

Substituting these values into the original equation verifies they are solutions.

Now try Exercises 89 through 92

Example 12 reminds us that in the process of factoring polynomials, there may be

a common monomial factor. This factor is also set equal to zero in the solution process

(if the monomial is a constant, no solution is generated).

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Section R.4 Factoring Polynomials and Solving Polynomial Equations by Factoring

EXAMPLE 13

Solving Higher Degree Equations

Solve each equation by factoring.

a. x3 Ϫ 4x ϩ 20 ϭ 5x2

Solution

49

a.

b. x4 Ϫ 10x2 ϩ 9 ϭ 0

x3 Ϫ 4x ϩ 20 ϭ 5x2

x Ϫ 5x2 Ϫ 4x ϩ 20 ϭ 0

original equation

3

x 1x Ϫ 52 Ϫ 41x Ϫ 52 ϭ 0

1x Ϫ 521x2 Ϫ 42 ϭ 0

1x Ϫ 52 1x ϩ 221x Ϫ 22 ϭ 0

or x Ϫ 2 ϭ 0

x Ϫ 5 ϭ 0 or x ϩ 2 ϭ 0

x ϭ Ϫ2 or

xϭ2

x ϭ 5 or

2

standard form; factor by grouping

remove common factors from each group

factor common binomial

factored form

zero product property

solve

The solutions are x ϭ 5, x ϭ Ϫ2, and x ϭ 2.

b. The equation appears to be in quadratic form and we begin by substituting u

for x2 and u2 for x4.

original equation

x4 Ϫ 10x2 ϩ 9 ϭ 0

substitute u for x 2 and u 2 for x 4

u2 Ϫ 10u ϩ 9 ϭ 0

factored form

1u Ϫ 921u Ϫ 12 ϭ 0

zero product property

u Ϫ 9 ϭ 0 or u Ϫ 1 ϭ 0

2

2

substitute x 2 for u

x Ϫ 9 ϭ 0 or x Ϫ 1 ϭ 0

1x ϩ 321x Ϫ 32 ϭ 0 or 1x ϩ 12 1x Ϫ 12 ϭ 0 factor

x ϭ Ϫ1 or x ϭ 1 zero product property

x ϭ Ϫ3 or x ϭ 3 or

The solutions are x ϭ Ϫ3, x ϭ 3, x ϭ Ϫ1, and x ϭ 1.

Now try Exercises 93 through 100

E. You’ve just seen how

we can solve polynomial

equations by factoring

In Examples 12 and 13, we were able to solve higher degree polynomial equations

by “breaking them down” into linear and quadratic forms. This basic idea can be

applied to other kinds of equations as well.

R.4 EXERCISES

CONCEPTS AND VOCABULARY

Fill in each blank with the appropriate word or phrase. Carefully reread the section, if necessary.

1. To factor an expression means to rewrite the

expression as an equivalent

.

2. If a polynomial will not factor, it is said to be a(n)

polynomial.

3. The difference of two perfect squares always

factors into the product of a(n)

and its

.

4. The expression x2 ϩ 6x ϩ 9 is said to be a(n)

trinomial, since its factored

form is a perfect (binomial) square.

5. Discuss/Explain why

4x2 Ϫ 36 ϭ 12x Ϫ 62 12x ϩ 62 is not written in

completely factored form, then rewrite it so it is

factored completely.

6. Discuss/Explain why a3 ϩ b3 is factorable, but

a2 ϩ b2 is not. Demonstrate by writing x3 ϩ 64 in

factored form, and by exhausting all possibilities for

x2 ϩ 64 to show it is prime.

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Factor each expression using the method indicated.

Greatest Common Factor

7. a. Ϫ17x2 ϩ 51

c. Ϫ3a4 ϩ 9a2 Ϫ 6a3

b. 21b3 Ϫ 14b2 ϩ 56b

8. a. Ϫ13n2 Ϫ 52

b. 9p2 ϩ 27p3 Ϫ 18p4

c. Ϫ6g5 ϩ 12g4 Ϫ 9g3

Common Binomial Factor

9. a. 2a1a ϩ 22 ϩ 31a ϩ 22

b. 1b2 ϩ 323b ϩ 1b2 ϩ 322

c. 4m1n ϩ 72 Ϫ 111n ϩ 72

10. a. 5x1x Ϫ 32 Ϫ 21x Ϫ 32

b. 1v Ϫ 522v ϩ 1v Ϫ 523

c. 3p1q2 ϩ 52 ϩ 71q2 ϩ 52

Trinomial Factoring where ͦaͦ ‫ ؍‬1

13. a. Ϫp2 ϩ 5p ϩ 14

c. n2 ϩ 20 Ϫ 9n

b. q2 Ϫ 4q ϩ 12

14. a. Ϫm2 ϩ 13m Ϫ 42

c. v2 ϩ 10v ϩ 15

b. x2 ϩ 12 ϩ 13x

1

15. a. 3p Ϫ 13p Ϫ 10

c. 10u2 Ϫ 19u Ϫ 15

b. 4q2 ϩ 7q Ϫ 15

16. a. 6v ϩ v Ϫ 35

c. 15z2 Ϫ 22z Ϫ 48

b. 20x ϩ 53x ϩ 18

2

Difference of Perfect Squares

17. a. 4s2 Ϫ 25

c. 50x2 Ϫ 72

e. b2 Ϫ 5

b. 9x2 Ϫ 49

d. 121h2 Ϫ 144

18. a. 9v Ϫ

c. v4 Ϫ 1

e. x2 Ϫ 17

b. 25w Ϫ

d. 16z4 Ϫ 81

2

1

25

b. z2 Ϫ 18z ϩ 81

d. 16q2 ϩ 40q ϩ 25

Sum/Difference of Perfect Cubes

21. a. 8p3 Ϫ 27

c. g3 Ϫ 0.027

b. m3 ϩ 18

d. Ϫ2t 4 ϩ 54t

22. a. 27q3 Ϫ 125

c. b3 Ϫ 0.125

8

b. n3 ϩ 27

d. 3r4 Ϫ 24r

b. x4 ϩ 13x2 ϩ 36

25. Completely factor each of the following (recall that

“1” is its own perfect square and perfect cube).

a. n2 Ϫ 1

b. n3 Ϫ 1

3

c. n ϩ 1

d. 28x3 Ϫ 7x

12. a. 6h Ϫ 9h Ϫ 2h ϩ 3

b. 4k3 ϩ 6k2 Ϫ 2k Ϫ 3

c. 3x2 Ϫ xy Ϫ 6x ϩ 2y

2

20. a. x2 ϩ 12x ϩ 36

c. 25p2 Ϫ 60p ϩ 36

24. a. x6 Ϫ 26x3 Ϫ 27

b. 31n ϩ 52 2 ϩ 21n ϩ 52 Ϫ 21

c. 21z ϩ 32 2 ϩ 31z ϩ 32 Ϫ 54

2

2

b. b2 ϩ 10b ϩ 25

d. 9n2 Ϫ 42n ϩ 49

23. a. x4 Ϫ 10x2 ϩ 9

c. x6 Ϫ 7x3 Ϫ 8

11. a. 9q3 ϩ 6q2 ϩ 15q ϩ 10

b. h5 Ϫ 12h4 Ϫ 3h ϩ 36

c. k5 Ϫ 7k3 Ϫ 5k2 ϩ 35

Trinomial Factoring where a

19. a. a2 Ϫ 6a ϩ 9

c. 4m2 Ϫ 20m ϩ 25

u-Substitution

Grouping

3

Perfect Square Trinomials

2

1

49

26. Carefully factor each of the following trinomials, if

possible. Note differences and similarities.

a. x2 Ϫ x ϩ 6

b. x2 ϩ x Ϫ 6

2

c. x ϩ x ϩ 6

d. x2 Ϫ x Ϫ 6

e. x2 Ϫ 5x ϩ 6

f. x2 ϩ 5x Ϫ 6

Factor each expression completely, if possible. Rewrite

the expression in standard form (factor out “Ϫ1” if

needed) and factor out the GCF if one exists. If you

believe the expression will not factor, write “prime.”

27. a2 ϩ 7a ϩ 10

28. b2 ϩ 9b ϩ 20

29. 2x2 Ϫ 24x ϩ 40

30. 10z2 Ϫ 140z ϩ 450

31. 64 Ϫ 9m2

32. 25 Ϫ 16n2

33. Ϫ9r ϩ r2 ϩ 18

34. 28 ϩ s2 Ϫ 11s

35. 2h2 ϩ 7h ϩ 6

36. 3k2 ϩ 10k ϩ 8

37. 9k2 Ϫ 24k ϩ 16

38. 4p2 Ϫ 20p ϩ 25

39. Ϫ6x3 ϩ 39x2 Ϫ 63x

40. Ϫ28z3 ϩ 16z2 ϩ 80z

41. 12m2 Ϫ 40m ϩ 4m3

42. Ϫ30n Ϫ 4n2 ϩ 2n3

43. a2 Ϫ 7a Ϫ 60

44. b2 Ϫ 9b Ϫ 36

45. 8x3 Ϫ 125

46. 27r3 ϩ 64

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47. m2 ϩ 9m Ϫ 24

48. n2 Ϫ 14n Ϫ 36

49. x3 Ϫ 5x2 Ϫ 9x ϩ 45

50. x3 ϩ 3x2 Ϫ 4x Ϫ 12

51. Match each expression with the description that fits best.

a. prime polynomial

b. standard trinomial a ϭ 1

c. perfect square trinomial

d. difference of cubes

e. binomial square

f. sum of cubes

g. binomial conjugates

h. difference of squares

i. standard trinomial a 1

3

A. x ϩ 27

B. 1x ϩ 32 2

C. x2 Ϫ 10x ϩ 25

D. x2 Ϫ 144

2

E. x Ϫ 3x Ϫ 10

F. 8s3 Ϫ 125t3

G. 2x2 Ϫ x Ϫ 3

H. x2 ϩ 9

I. 1x Ϫ 72 and 1x ϩ 72

52. Match each polynomial to its factored form. Two

of them are prime.

a. 4x2 Ϫ 9

b. 4x2 Ϫ 28x ϩ 49

c. x3 Ϫ 125

d. 8x3 ϩ 27

e. x2 Ϫ 3x Ϫ 10

f. x2 ϩ 3x ϩ 10

g. 2x2 Ϫ x Ϫ 3

h. 2x2 ϩ x Ϫ 3

i. x2 ϩ 25

A. 1x Ϫ 521x2 ϩ 5x ϩ 252

B. 12x Ϫ 321x ϩ 12

C. 12x ϩ 3212x Ϫ 32

D. 12x Ϫ 72 2

E. prime trinomial

F. prime binomial

G. 12x ϩ 321x Ϫ 12

2

H. 12x ϩ 3214x Ϫ 6x ϩ 92

I. 1x Ϫ 521x ϩ 22

Determine whether each equation is quadratic. If so,

identify the coefficients a, b, and c. If not, discuss why.

53. 2x Ϫ 15 Ϫ x2 ϭ 0

55.

2

xϪ7ϭ0

3

54. 21 ϩ x2 Ϫ 4x ϭ 0

56. 12 Ϫ 4x ϭ 9

63. 1x Ϫ 12 2 ϩ 1x Ϫ 12 ϩ 4 ϭ 9

64. 1x ϩ 52 2 Ϫ 1x ϩ 52 ϩ 4 ϭ 17

Solve using the zero factor property. Be sure each

equation is in standard form and factor out any

common factors before attempting to solve. Check all

65. x2 Ϫ 15 ϭ 2x

66. z2 Ϫ 10z ϭ Ϫ21

67. m2 ϭ 8m Ϫ 16

68. Ϫ10n ϭ n2 ϩ 25

69. 5p2 Ϫ 10p ϭ 0

70. 6q2 Ϫ 18q ϭ 0

71. Ϫ14h2 ϭ 7h

72. 9w ϭ Ϫ6w2

73. a2 Ϫ 17 ϭ Ϫ8

74. b2 ϩ 8 ϭ 12

75. g2 ϩ 18g ϩ 70 ϭ Ϫ11

76. h2 ϩ 14h Ϫ 2 ϭ Ϫ51

77. m3 ϩ 5m2 Ϫ 9m Ϫ 45 ϭ 0

78. n3 Ϫ 3n2 Ϫ 4n ϩ 12 ϭ 0

79. 1c Ϫ 122c Ϫ 15 ϭ 30

80. 1d Ϫ 102d ϩ 10 ϭ Ϫ6

81. 9 ϩ 1r Ϫ 52r ϭ 33

82. 7 ϩ 1s Ϫ 42s ϭ 28

83. 1t ϩ 421t ϩ 72 ϭ 54

84. 1g ϩ 1721g Ϫ 22 ϭ 20

85. 2x2 Ϫ 4x Ϫ 30 ϭ 0

86. Ϫ3z2 ϩ 12z ϩ 36 ϭ 0

87. 2w2 Ϫ 5w ϭ 3

88. Ϫ3v2 ϭ Ϫv Ϫ 2

89. 22x ϭ x3 Ϫ 9x2

90. x3 ϭ 13x2 Ϫ 42x

91. 3x3 ϭ Ϫ7x2 ϩ 6x

92. 7x2 ϩ 15x ϭ 2x3

93. p3 ϩ 7p2 Ϫ 63 ϭ 9p

94. q3 Ϫ 4q ϩ 24 ϭ 6q2

95. x3 Ϫ 25x ϭ 2x2 Ϫ 50

96. 3c2 ϩ c ϭ c3 ϩ 3

1

57. x2 ϭ 6x

4

58. 0.5x ϭ 0.25x

59. 2x2 ϩ 7 ϭ 0

60. 5 ϭ Ϫ4x2

2

61. Ϫ3x2 ϩ 9x Ϫ 5 ϩ 2x3 ϭ 0

62. z2 Ϫ 6z ϩ 9 Ϫ z3 ϭ 0

97. x4 Ϫ 29x2 ϩ 100 ϭ 0

98. z4 Ϫ 20z2 ϩ 64 ϭ 0

99. 1b2 Ϫ 3b2 2 Ϫ 141b2 Ϫ 3b2 ϩ 40 ϭ 0

100. 1d2 Ϫ d 2 2 Ϫ 81d2 Ϫ d2 ϩ 12 ϭ 0

51

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WORKING WITH FORMULAS

101. Surface area of a cylinder: 2␲r2 ؉ 2␲rh

102. Volume of a cylindrical shell: ␲R2h ؊ ␲r2h

The surface area of a cylinder is given by the formula

shown, where h is the height of the cylinder and r is the

radius. Factor out the GCF and use the result to find the

surface area of a cylinder where r ϭ 35 cm and h ϭ 65 cm.

Answer in exact form and in approximate form rounded to

the nearest whole number.

r

The volume of a cylindrical shell (a

larger cylinder with a smaller cylinder

removed) can be found using the

formula shown, where R is the radius of

the larger cylinder and r is the radius of

the smaller. Factor the expression

R

completely and use the result to find the

volume of a shell where

R ϭ 9 cm, r ϭ 3 cm, and h ϭ 10 cm. Answer in exact

form and in approximate form rounded to the nearest

whole number.

APPLICATIONS

In many cases, factoring an expression can make it easier to evaluate as in the following applications.

103. Conical shells: The volume of a conical shell (like

the shell of an ice cream cone) is given by the

1

1

formula V ϭ ␲R2h Ϫ ␲r2h, where R is the outer

3

3

the formula in completely factored form, then find

the volume of a shell when R ϭ 5.1 cm,

r ϭ 4.9 cm, and h ϭ 9 cm. Answer in exact form

and in approximate form rounded to the nearest

tenth.

104. Spherical shells: The volume of

a spherical shell (like the outer

r

shell of a cherry cordial) is given

R

4

4

3

3

by the formula V ϭ 3␲R Ϫ 3␲r ,

where R is the outer radius and r

is the inner radius of the shell.

Write the right-hand side in

completely factored form, then find the volume of

a shell where R ϭ 1.8 cm and r ϭ 1.5 cm. Answer

in exact form and in approximate form rounded to

the nearest tenth.

105. Volume of a box: The volume of a rectangular box

x inches in height is given by the relationship

V ϭ x3 ϩ 8x2 ϩ 15x. Factor the right-hand side to

determine: (a) The number of inches that the width

exceeds the height, (b) the number of inches the

length exceeds the height, and (c) the volume given

the height is 2 ft.

106. Shipping textbooks: A publisher ships paperback

books stacked x copies high in a box. The total

number of books shipped per box is given by the

relationship B ϭ x3 Ϫ 13x2 ϩ 42x. Factor the

right-hand side to determine (a) how many more or

fewer books fit the width of the box (than the

height), (b) how many more or fewer books fit the

length of the box (than the height), and (c) the

number of books shipped per box if they are

stacked 10 high in the box.

107. Space-Time relationships: Due to the work of

Albert Einstein and other physicists who labored

on space-time relationships, it is known that the

faster an object moves the shorter it appears to

become. This phenomenon is modeled by the

v 2

1Ϫa b,

c

B

where L0 is the length of the object at rest, L is the

relative length when the object is moving at velocity

v, and c is the speed of light. Factor the radicand and

use the result to determine the relative length of a

12-in. ruler if it is shot past a stationary observer at

0.75 times the speed of light 1v ϭ 0.75c2 .

Lorentz transformation L ϭ L0

108. Tubular fluid flow: As a fluid flows through a

tube, it is flowing faster at the center of the tube

than at the sides, where the tube exerts a backward

drag. Poiseuille’s law gives the velocity of the flow

G 2

1R Ϫ r2 2 ,

at any point of the cross section: v ϭ

4␩

where R is the inner radius of the tube, r is the

distance from the center of the tube to a point in

the flow, G represents what is called the pressure

gradient, and ␩ is a constant that depends on the

viscosity of the fluid. Factor the right-hand side

and find v given R ϭ 0.5 cm, r ϭ 0.3 cm, G ϭ 15,

and ␩ ϭ 0.25.

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Section R.5 Rational Expressions and Equations

53

Solve by factoring.

109. Envelope sizes: Large mailing envelopes often come

in standard sizes, with 5- by 7-in. and 9- by 12-in.

envelopes being the most common. The next larger

size envelope has an area of 143 in2, with a length

that is 2 in. longer than the width. What are the

dimensions of the larger envelope?

a length that is 6 in. longer than the width. What are

the dimensions of the Ledger size paper?

Letter

Legal

Ledger

110. Paper sizes: Letter size paper is 8.5 in. by 11 in.

Legal size paper is 812 in. by 14 in. The next larger

(common) size of paper has an area of 187 in2, with

EXTENDING THE CONCEPT

111. Factor out a constant that leaves integer

coefficients for each term:

a. 12x4 ϩ 18x3 Ϫ 34x2 ϩ 4

b. 23b5 Ϫ 16b3 ϩ 49b2 Ϫ 1

112. If x ϭ 2 is substituted into 2x3 ϩ hx ϩ 8, the result

is zero. What is the value of h?

113. Factor the expression: 192x3 Ϫ 164x2 Ϫ 270x.

114. As an alternative to evaluating polynomials by direct

substitution, nested factoring can be used. The

method has the advantage of using only products

and sums — no powers. For P ϭ x3 ϩ 3x2ϩ

1x ϩ 5, we begin by grouping all variable terms

and factoring x: P ϭ 3 x3 ϩ 3x2 ϩ 1x 4 ϩ 5 ϭ

R.5

In Section R.5 you will review how to:

A. Write a rational

C.

D.

E.

Factor each expression completely.

115. x4 Ϫ 81

116. 16n4 Ϫ 1

117. p6 Ϫ 1

118. m6 Ϫ 64

119. q4 Ϫ 28q2 ϩ 75

120. a4 Ϫ 18a2 ϩ 32

Rational Expressions and Equations

LEARNING OBJECTIVES

B.

x 3 x2 ϩ 3x ϩ 14 ϩ 5. Then we group the inner

terms with x and factor again:

P ϭ x 3 x2 ϩ 3x ϩ 14 ϩ 5 ϭ x3x1x ϩ 32 ϩ 14 ϩ 5.

The expression can now be evaluated using any input

and the order of operations. If x ϭ 2, we quickly

find that P ϭ 27. Use this method to evaluate

H ϭ x3 ϩ 2x2 ϩ 5x Ϫ 9 for x ϭ Ϫ3.

expression in simplest

form

Multiply and divide

rational expressions

expressions

Simplify compound

fractions

Solve rational equations

A rational number is one that can be written as the quotient of two integers. Similarly,

a rational expression is one that can be written as the quotient of two polynomials. We

can apply the skills developed in a study of fractions (how to reduce, add, subtract,

multiply, and divide) to rational expressions, sometimes called algebraic fractions.

A. Writing a Rational Expression in Simplest Form

A rational expression is in simplest form when the numerator and denominator have

no common factors (other than 1). After factoring the numerator and denominator, we

apply the fundamental property of rational expressions.

Fundamental Property of Rational Expressions

If P, Q, and R are polynomials, with Q, R 0,

P

P

P#R

P#R

ϭ

and 122 ϭ #

112

#

Q R

Q

Q

Q R

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CHAPTER R A Review of Basic Concepts and Skills

In words, the property says (1) a rational expression can be simplified by canceling common factors in the numerator and denominator, and (2) an equivalent expression can be formed by multiplying numerator and denominator by the same nonzero

polynomial.

EXAMPLE 1

Simplifying a Rational Expression

Write the expression in simplest form:

Solution

1x Ϫ 121x ϩ 12

x2 Ϫ 1

ϭ

2

1x Ϫ 121x Ϫ 22

x Ϫ 3x ϩ 2

1x Ϫ 121x ϩ 12

x2 Ϫ 1

.

x Ϫ 3x ϩ 2

2

factor numerator and denominator

1

ϭ

1x Ϫ 121x Ϫ 22

xϩ1

ϭ

xϪ2

common factors reduce to 1

simplest form

Now try Exercises 7 through 10

WORTHY OF NOTE

If we view a and b as two points on

the number line, we note that they

are the same distance apart,

regardless of the order they are

subtracted. This tells us the

numerator and denominator will

have the same absolute value but

be opposite in sign, giving a value of

Ϫ1 (check using a few test values).

CAUTION

When simplifying rational expressions, we sometimes encounter expressions of

aϪb

aϪb

. If we factor Ϫ1 from the numerator, we see that

ϭ

the form

bϪa

bϪa

Ϫ11b Ϫ a2

ϭ Ϫ1.

bϪa

When reducing rational numbers or expressions, only common factors can be reduced.

It is incorrect to reduce (or divide out) individual terms:

xϩ1

xϩ2

1

(except for x ϭ 0)

2

Note that after simplifying an expression, we

are actually saying the resulting (simpler)

expression is equivalent to the original expression for all values where both are defined. The

first expression is not defined when x ϭ 1 or

x ϭ 2, the second when x ϭ 2 (since the

denominators would be zero). The calculator

screens shown in Figure R.5 help to illustrate

this fact, and it appears that we would very

much prefer to be working with the simpler

expression!

Ϫ6 ϩ 423

2

Ϫ3 ϩ 423, and

Figure R.5

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E. Polynomial Equations and the Zero Product Property

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