E. Polynomial Equations and the Zero Product Property
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Section R.4 Factoring Polynomials and Solving Polynomial Equations by Factoring
Quadratic Equations
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
Latin word quadratum, meaning
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
the leading term is squared.
ᮣ
Standard Form
Quadratic
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
xϭ
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
xϭ
or
xϭ
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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CHAPTER R A Review of Basic Concepts and Skills
DEVELOPING YOUR SKILLS
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
answers in the original equation.
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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CHAPTER R A Review of Basic Concepts and Skills
WORKING WITH FORMULAS
101. Surface area of a cylinder: 2r2 ؉ 2rh
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
radius and r is the inner radius of the cone. Write
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 3R Ϫ 3r ,
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
Add and subtract rational
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