Review of Algebra
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REVIEW OF ALGEBRA
Review of Algebra
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Here we review the basic rules and procedures of algebra that you need to know in
order to be successful in calculus.
Arithmetic Operations
The real numbers have the following properties:
abba
ab ba
a b c a b c
ab c ab ac
abc abc
(Commutative Law)
(Associative Law)
(Distributive law)
In particular, putting a 1 in the Distributive Law, we get
b c 1b c 1b 1c
and so
b c b c
EXAMPLE 1
(a) 3xy4x 34x 2y 12x 2y
(b) 2t7x 2tx 11 14tx 4t 2x 22t
(c) 4 3x 2 4 3x 6 10 3x
If we use the Distributive Law three times, we get
a bc d a bc a bd ac bc ad bd
This says that we multiply two factors by multiplying each term in one factor by each
term in the other factor and adding the products. Schematically, we have
a bc d
In the case where c a and d b, we have
a b2 a 2 ba ab b 2
or
1
a b2 a 2 2ab b 2
Similarly, we obtain
2
a b2 a 2 2ab b 2
REVIEW OF ALGEBRA
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3
EXAMPLE 2
(a) 2x 13x 5 6x 2 3x 10x 5 6x 2 7x 5
(b) x 62 x 2 12x 36
(c) 3x 14x 3 2x 6 34x 2 x 3 2x 12
12x 2 3x 9 2x 12
12x 2 5x 21
Fractions
To add two fractions with the same denominator, we use the Distributive Law:
a
c
1
1
1
ac
a c a c b
b
b
b
b
b
Thus, it is true that
ac
a
c
b
b
b
But remember to avoid the following common error:
a
a
a
bc
b
c
|
(For instance, take a b c 1 to see the error.)
To add two fractions with different denominators, we use a common denominator:
a
c
ad bc
b
d
bd
We multiply such fractions as follows:
a c
ac
b d
bd
In particular, it is true that
a
a
a
b
b
b
To divide two fractions, we invert and multiply:
a
b
a
d
ad
c
b
c
bc
d
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REVIEW OF ALGEBRA
EXAMPLE 3
x3
x
3
3
1
x
x
x
x
3
x
3x 2 xx 1
3x 6 x 2 x
(b)
x1
x2
x 1x 2
x2 x 2
2
x 2x 6
2
x x2
s2t
ut
s 2 t 2u
s2t 2
(c)
u
2
2u
2
x
xy
1
y
y
x
xx y
x 2 xy
xy
(d)
xy
y
xy
yx y
xy y 2
y
1
x
x
(a)
Factoring
We have used the Distributive Law to expand certain algebraic expressions. We sometimes need to reverse this process (again using the Distributive Law) by factoring an
expression as a product of simpler ones. The easiest situation occurs when the expression has a common factor as follows:
Expanding
3x(x-2)=3x@-6x
Factoring
To factor a quadratic of the form x 2 bx c we note that
x rx s x 2 r sx rs
so we need to choose numbers r and s so that r s b and rs c.
EXAMPLE 4 Factor x 2 5x 24.
SOLUTION The two integers that add to give 5 and multiply to give 24 are 3 and 8.
Therefore
x 2 5x 24 x 3x 8
EXAMPLE 5 Factor 2x 2 7x 4.
SOLUTION Even though the coefficient of x 2 is not 1, we can still look for factors of the
form 2x r and x s, where rs 4. Experimentation reveals that
2x 2 7x 4 2x 1x 4
Some special quadratics can be factored by using Equations 1 or 2 (from right to
left) or by using the formula for a difference of squares:
3
a 2 b 2 a ba b
REVIEW OF ALGEBRA
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5
The analogous formula for a difference of cubes is
a 3 b 3 a ba 2 ab b 2 4
which you can verify by expanding the right side. For a sum of cubes we have
a 3 b 3 a ba 2 ab b 2 5
EXAMPLE 6
(a) x 2 6x 9 x 32
(b) 4x 2 25 2x 52x 5
(c) x 3 8 x 2x 2 2x 4
EXAMPLE 7 Simplify
(Equation 2; a x, b 3)
(Equation 3; a 2x, b 5)
(Equation 5; a x, b 2)
x 2 16
.
x 2 2x 8
SOLUTION Factoring numerator and denominator, we have
x 2 16
x 4x 4
x4
2
x 2x 8
x 4x 2
x2
To factor polynomials of degree 3 or more, we sometimes use the following fact.
6 The Factor Theorem If P is a polynomial and Pb 0, then x b is a factor
of Px.
EXAMPLE 8 Factor x 3 3x 2 10x 24.
SOLUTION Let Px x 3 3x 2 10x 24. If Pb 0, where b is an integer, then
b is a factor of 24. Thus, the possibilities for b are 1, 2, 3, 4, 6, 8, 12,
and 24. We find that P1 12, P1 30, P2 0. By the Factor Theorem,
x 2 is a factor. Instead of substituting further, we use long division as follows:
x 2 x 12
x 2 x 3 3x 2 10 x 24
x 3 2x 2
x 2 10x
x 2 2x
12x 24
12x 24
Therefore
x 3 3x 2 10x 24 x 2x 2 x 12
x 2x 3x 4
Completing the Square
Completing the square is a useful technique for graphing parabolas or integrating
rational functions. Completing the square means rewriting a quadratic ax 2 bx c
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REVIEW OF ALGEBRA
in the form ax p2 q and can be accomplished by:
1. Factoring the number a from the terms involving x.
2. Adding and subtracting the square of half the coefficient of x.
In general, we have
ax 2 bx c a x 2 a x2 a x
b
x c
a
b
x
a
b
2a
b
2a
2
2
c
2
b
2a
c
b2
4a
EXAMPLE 9 Rewrite x 2 x 1 by completing the square.
1
SOLUTION The square of half the coefficient of x is 4. Thus
2
x 2 x 1 x 2 x 14 14 1 (x 12 ) 34
EXAMPLE 10
2x 2 12x 11 2x 2 6x 11 2x 2 6x 9 9 11
2x 32 9 11 2x 32 7
Quadratic Formula
By completing the square as above we can obtain the following formula for the roots
of a quadratic equation.
2
7 The Quadratic Formula The roots of the quadratic equation ax bx c 0
are
x
b sb 2 4ac
2a
EXAMPLE 11 Solve the equation 5x 2 3x 3 0.
SOLUTION With a 5, b 3, c 3, the quadratic formula gives the solutions
x
3 s32 453
3 s69
25
10
The quantity b 2 4ac that appears in the quadratic formula is called the
discriminant. There are three possibilities:
1. If b 2 4ac 0, the equation has two real roots.
2. If b 2 4ac 0, the roots are equal.
3. If b 2 4ac 0, the equation has no real root. (The roots are complex.)
REVIEW OF ALGEBRA
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These three cases correspond to the fact that the number of times the parabola
y ax 2 bx c crosses the x-axis is 2, 1, or 0 (see Figure 1). In case (3) the quadratic ax 2 bx c can’t be factored and is called irreducible.
y
y
0
y
0
x
x
0
FIGURE 1
(a) b@-4ac>0
Possible graphs of y=ax@+bx+c
(b) b@-4ac=0
(c) b@-4ac<0
EXAMPLE 12 The quadratic x 2 x 2 is irreducible because its discriminant is
negative:
b 2 4ac 12 412 7 0
Therefore, it is impossible to factor x 2 x 2.
The Binomial Theorem
Recall the binomial expression from Equation 1:
a b2 a 2 2ab b 2
If we multiply both sides by a b and simplify, we get the binomial expansion
8
a b3 a 3 3a 2b 3ab 2 b 3
Repeating this procedure, we get
a b4 a 4 4a 3b 6a 2b 2 4ab 3 b 4
In general, we have the following formula.
9 The Binomial Theorem If k is a positive integer, then
a bk a k ka k1b kk 1 k2 2
a b
12
kk 1k 2 k3 3
a b
123
kk 1k n 1 kn n
a b
1 2 3 n
kab k1 b k
x
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REVIEW OF ALGEBRA
EXAMPLE 13 Expand x 25.
SOLUTION Using the Binomial Theorem with a x, b 2, k 5, we have
x 25 x 5 5x 42 54 3
543 2
x 22 x 23 5x24 25
12
123
x 5 10x 4 40x 3 80x 2 80x 32
Radicals
The most commonly occurring radicals are square roots. The symbol s1 means “the
positive square root of.” Thus
x sa
x2 a
means
and
x0
Since a x 2 0, the symbol sa makes sense only when a 0. Here are two rules
for working with square roots:
sab sa sb
10
a
sa
b
sb
However, there is no similar rule for the square root of a sum. In fact, you should
remember to avoid the following common error:
sa b sa sb
|
(For instance, take a 9 and b 16 to see the error.)
EXAMPLE 14
(a)
s18
s2
18
s9 3
2
x because s1 indicates the positive square root.
(b) sx 2 y sx 2 sy x sy
2
Notice that sx
(See Appendix A.)
In general, if n is a positive integer,
n
xs
a
xn a
means
If n is even, then a 0 and x 0.
3
4
6
Thus s
8 2 because 23 8, but s
8 and s
8 are not defined. The following rules are valid:
n
n
n
ab s
as
b
s
3
3
3
3
3
EXAMPLE 15 s
x4 s
x 3x s
x3 s
x xs
x
n
n
a
a
s
n
b
sb
REVIEW OF ALGEBRA
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9
To rationalize a numerator or denominator that contains an expression such as
sa sb, we multiply both the numerator and the denominator by the conjugate radical sa sb. Then we can take advantage of the formula for a difference of squares:
(sa sb )(sa sb ) (sa )2 (sb )2 a b
EXAMPLE 16 Rationalize the numerator in the expression
sx 4 2
.
x
SOLUTION We multiply the numerator and the denominator by the conjugate radical
sx 4 2:
sx 4 2
x
sx 4 2
x
sx 4 2
sx 4 2
x 4 4
x (sx 4 2)
x
1
x (sx 4 2)
sx 4 2
Exponents
Let a be any positive number and let n be a positive integer. Then, by definition,
1. a n a a a
n factors
2. a 1
0
1
an
n
4. a1n s
a
m
mn
n
n
a sa m (s
a)
3. an m is any integer
11 Laws of Exponents Let a and b be positive numbers and let r and s be any
rational numbers (that is, ratios of integers). Then
1. a r a s a rs
2.
4. abr a rb r
5.
ar
a rs
as
a
b
r
ar
br
3. a r a rs
s
b0
In words, these five laws can be stated as follows:
1. To multiply two powers of the same number, we add the exponents.
2. To divide two powers of the same number, we subtract the exponents.
3. To raise a power to a new power, we multiply the exponents.
4. To raise a product to a power, we raise each factor to the power.
5. To raise a quotient to a power, we raise both numerator and denominator to
the power.
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REVIEW OF ALGEBRA
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EXAMPLE 17
(a) 28 82 28 232 28 26 214
2
1
1
y2 x2
2 2
x
y
x 2y 2
y2 x2
xy
2 2
1
1
yx
x y
yx
x
y
xy
y xy x
yx
xyy x
xy
2
x y
x1 y1
(b)
3
Alternative solution: 432 (s4 ) 23 8
(c) 432 s43 s64 8
1
1
(d) 3 4 43 x43
x
sx
(e)
Exercises
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x
y
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y 2x
z
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A Click here for answers.
1–16
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3. 2xx 5
4. 4 3xx
5. 24 3a
6. 8 4 x
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8. 53t 4 t 2 2 2tt 3
9. 4x 13x 7
10. xx 1x 2
11. 2x 12
12. 2 3x2
13. y 46 y5 y
14. t 52 2t 38t 1
15. 1 2xx 2 3x 1
17–28
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16. 1 x x 2 2
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29–48
7. 4x 2 x 2 5x 2 2x 1
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1
c1
27.
1
1
c1
Expand and simplify.
2. 2x 2 yxy 4 ■
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1
1. 6ab0.5ac
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x 3 y 8x 4
4 x 7y 5z4
y3
z
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Perform the indicated operations and simplify.
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1
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30. 5ab 8abc
31. x 2 7x 6
32. x 2 x 6
33. x 2 2x 8
34. 2x 2 7x 4
35. 9x 2 36
36. 8x 2 10x 3
37. 6x 2 5x 6
38. x 2 10x 25
39. t 3 1
40. 4t 2 9s 2
41. 4t 2 12t 9
42. x 3 27
43. x 3 2x 2 x
44. x 3 4x 2 5x 2
45. x 3 3x 2 x 3
46. x 3 2x 2 23x 60
47. x 3 5x 2 2x 24
49.
50.
2
3
4
2
a2
ab
b
x2 x 2
x 2 3x 2
2x 2 3x 2
x2 4
51.
52.
x 3 5x 2 6x
x 2 x 12
24.
x
yz
x2 1
x 9x 8
53.
26.
a
b
bc
ac
1
1
2
x3
x 9
25.
xy
z
2r
s
s2
6t
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49–54
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48. x 3 3x 2 4x 12
1
1
20.
x1
x1
23.
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29. 2x 12x 3
1
2
19.
x5
x3
22.
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Factor the expression.
9b 6
18.
3b
u
u1
●
1
1x
2 8x
17.
2
21. u 1 ●
1
28. 1 ■
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Simplify the expression.
2
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REVIEW OF ALGEBRA
x
2
2
x2 x 2
x 5x 4
54.
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55–60
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x 92x4
x3
86.
a n a 2n1
a n2
87.
a3b 4
a5b 5
88.
x1 y1
x y1
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Complete the square.
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85.
55. x 2 2x 5
56. x 2 16x 80
89. 312
90. 9615
57. x 2 5x 10
58. x 2 3x 1
91. 125 23
92. 6443
59. 4x 2 4x 2
60. 3x 2 24x 50
93. 2x 2 y 4 32
94. x5 y 3z 10 35
5
y6
95. s
4
a)
96. (s
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61–68
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Solve the equation.
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97.
61. x 9x 10 0
62. x 2x 8 0
63. x 9x 1 0
64. x 2 2x 7 0
65. 3x 2 5x 1 0
66. 2x 2 7x 2 0
67. x 3 2x 1 0
68. x 3 3x 2 x 1 0
2
2
2
99.
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69–72
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70. 2x 2 9x 4
71. 3x 2 x 6
72. x 2 3x 6
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73–76
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81. s16a 4b 3
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3
2
s
3
54
s
80. sxy sx 3 y
83–100
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Rationalize the expression.
(1sx ) 1
sx 3
x9
102.
103.
x sx 8
x4
104.
s2 h s2 h
h
105.
2
3 s5
106.
1
sx sy
107. sx 2 3x 4 x
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79.
4
32x 4
s
4
2
s
82.
5
96a6
s
5
s3a
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84. 216 410 16 6
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x1
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108. sx 2 x sx 2 x
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109–116
■ State whether or not the equation is true for all
values of the variable.
109. sx 2 x
110. sx 2 4 x 2
a
16 a
1
111.
16
16
1
xy
112. 1
x y1
113.
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Use the Laws of Exponents to rewrite and simplify
the expression.
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83. 310 9 8
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4
4
r 2n1 s
r 1
100. s
101.
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Simplify the radicals.
78.
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t 12sst
s 23
2 5
77. s32 s2
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101–108
76. 3 x 4
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77–82
■
4
8
5
sx
4
sx 3
98.
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74. a b7
75. x 1
2
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Use the Binomial Theorem to expand the expression.
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73. a b6
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Which of the quadratics are irreducible?
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69. 2x 2 3x 4
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3
1
(st ) 5
11
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x
1
xy
1y
2
1
2
4x
2
x
114.
115. x 34 x 7
116. 6 4x a 6 4x 4a
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12
ANSWERS
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Answers
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1. 3a 2bc
2. 2x 3 y 5
3. 2x 2 10x
4. 4x 3x 2
2
5. 8 6a
6. 4 x
7. x 6x 3
8. 3t 2 21t 22
9. 12x 2 25x 7
11. 4x 2 4x 1
10. x 3 x 2 2x
12. 9x 12x 4
13. 30y y y
2
4
14. 15t 56t 31
5
16. x 4 2x 3 x 2 2x 1
19.
3x 7
x 2 2x 15
2b 2 3ab 4a 2
22.
a 2b 2
26.
a2
b2
27.
c
c2
30. ab5 8c
17. 1 4x
21.
x
23.
yz
zx
24.
y
28.
3 2x
2x
18. 3 2b
39. t 1t t 1
32. x 3x 2
40. 2t 3s2t 3s
41. 2t 32
42. x 3x 2 3x 9
43. xx 12
44. x 12x 2
46. x 3x 5x 4
47. x 2x 3x 4
48. x 2x 3x 2
53.
2x 1
50.
x2
x2
x2 9
54.
55. x 12 4
58. ( x 3 2
2
)
54
60. 3x 4 2
2
x1
51.
x8
xx 2
52.
x4
x 2 6x 4
x 1x 2x 4
56. x 82 16
57. ( x 59. 2x 12 3
61. 1, 10
62. 2, 4
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9 s85
2
64. 1 2s2
66.
7 s33
4
67. 1,
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65.
1 s5
2
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5 2
2
)
●
5 s13
6
68. 1, 1 s2
70. Not irreducible
71. Not irreducible (two real roots)
72. Irreducible
73. a 6 6a 5b 15a 4b 2 20a 3b 3 15a 2b 4 6ab 5 b 6
76. 243 405x 2 270x 4 90x 6 15x 8 x 10
154
78. 3
1
81. 4a 2bsb
a2
b
90. 2 5s3
91. 25
t 14
s 124
83. 3 26
82. 2a
87.
99.
80. x 2 y
79. 2 x
86. a 2n3
x3
94. 95 6
y z
45. x 1x 1x 3
x2
49.
x2
●
63.
77. 8
38. x 52
2
●
75. x 8 4x 6 6x 4 4x 2 1
36. 4x 32x 1
37. 3x 22x 3
●
21a 2b 5 7ab 6 b 7
rs
25.
3t
34. 2x 1x 4
35. 9x 2x 2
●
74. a 7 7a 6b 21a 5b 2 35a 4b 3 35a 3b 4
29. 2x1 6x 2
31. x 6x 1
33. x 4x 2
●
69. Irreducible
u 2 3u 1
u1
2x
x2 1
20.
●
6
15. 2x 3 5x 2 x 1
2
●
88.
92.
95. y 65
x y2
xy
101.
103.
x 2 4x 16
xsx 8
105.
3 s5
2
107.
3x 4
sx 2 3x 4 x
104.
89.
85. 16x 10
1
s3
1
256
93. 2s2 x 3 y 6
96. a 34
100. r n2
106.
84. 2 60
1
sx 3
97. t 52
102.
98.
1
sx x
2
s2 h s2 h
sx sy
xy
108.
1
x 18
2x
sx 2 x sx 2 x
109. False
110. False
111. True
112. False
113. False
114. False
115. False
116. True