Review of Algebra
2
s
REVIEW OF ALGEBRA
Review of Algebra
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
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: a b b a ab a b c a b ab c ab ac In particular, putting a b and so b c b c ba c (Commutative Law) (Associative Law) (Distributive law)
ab c
a bc
1 in the Distributive Law, we get c 1 b c 1b 1c
EXAMPLE 1
(a) 3xy 4x 3 4 x 2y 12x 2y (b) 2t 7x 2tx 11 14tx 4t 2x 22t (c) 4 3 x 2 4 3x 6 10 3x If we use the Distributive Law three times, we get a b c 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 In the case where c or
1
b c
d
a and d a b
b, we have
2
a2
ba
ab
b2
a
b
2
a2
2ab
b2
Similarly, we obtain
2
a
b
2
a2
2ab
b2
REVIEW OF ALGEBRA
x
3
EXAMPLE 2
6x 2 3x (a) 2x 1 3x 5 (b) x 6 2 x 2 12x 36 2x 6 (c) 3 x 1 4x 3
10x 3 4x 2 12x 2 12x 2
5 x 3x 5x
6x 2
7x
5 12 12
3 2x 9 2x 21
Fractions
To add two fractions with the same denominator, we use the Distributive Law: a b Thus, it is true that a b c a b c b c b 1 b a 1 b c 1 a b c a b c
But remember to avoid the following common error:
|
b
a c
a b
a 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 b c d ad bd bc
We multiply such fractions as follows: a b In particular, it is true that a b a b a b c d ac bd
To divide two fractions, we invert and multiply: a b c d
a b
d c
ad bc
4
s
REVIEW OF ALGEBRA
EXAMPLE 3
3 x x 3 x 3x 2 xx 1 3x 6 x 2 x (b) x 1 x 2 x 1 x 2 x2 x 2 2 x 2x 6 2 x x 2 s2t ut s 2 t 2u s2t 2 (c) u 2 2u 2 x x y 1 y y x xx y x 2 xy x y (d) x y y x y yx y xy y 2 y 1 x x (a) 1
Factoring
x
3
x x
3 x
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 x r x s
bx x2
c we note that r s sx rs c.
so we need to choose numbers r and s so that r
EXAMPLE 4 Factor x 2
b and rs
5x
24. 24 are 3 and 8.
SOLUTION The two integers that add to give 5 and multiply to give
Therefore x2
EXAMPLE 5 Factor 2x 2
5x 4.
24
x
3 x
8
7x
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 2x 2 7x
4. Experimentation reveals that 4 2x 1 x 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
a2
b2
a
b a
b
REVIEW OF ALGEBRA
x
5
The analogous formula for a difference of cubes is
4
a3
b3
a
b a2
ab
b2
which you can verify by expanding the right side. For a sum of cubes we have
5
a3
b3
a
b a2
ab
b2
EXAMPLE 6
(a) x 2 6x 9 x 32 2 (b) 4x 25 2x 5 2x 5 (c) x 3 8 x 2 x 2 2x 4 x2 x
2
(Equation 2; a (Equation 3; a (Equation 5; a
x, b 3) 2x, b 5) x, b 2)
EXAMPLE 7 Simplify
16 . 2x 8
SOLUTION Factoring numerator and denominator, we have
x2 x
2
16 2x 8
x x
4 x 4 x
4 2
x x
4 2
To factor polynomials of degree 3 or more, we sometimes use the following fact.
6 The Factor Theorem If P is a polynomial and P b
0, then x
b is a factor
of P x .
EXAMPLE 8 Factor x 3
3
3x 2
2
10x
24.
SOLUTION Let P x x 3x 10x 24. If P b 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 P 1 12, P 1 30, P 2 0. By the Factor Theorem, x 2 is a factor. Instead of substituting further, we use long division as follows:
x
x2 2 x3 x3
x 3x 2 2x 2 x2 x2
12 10 x 10x 2x 12x 12x 2 x2 2 x
24
24 24 x 3 x 12 4
Therefore
x3
3x 2
10x
24
x x
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
6
s
REVIEW OF ALGEBRA
in the form a x p 2 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 x2 b x a b x a b 2a
2
c b 2a c
2
a x2
b 2a b2 4a
2
c
a x
EXAMPLE 9 Rewrite x 2
x
1 by completing the square.
1
SOLUTION The square of half the coefficient of x is 4. Thus
x2
EXAMPLE 10
x
1
x2
x
1 4
1 4
1
(x
1 2 2
)
3 4
2x 2
12x
11
2 x2 2 x
6x 3
2
11 9
2 x2 11
6x 2x 3
9
2
9 7
11
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 2a 4ac
EXAMPLE 11 Solve the equation 5x 2
SOLUTION With a
3x
3
0.
5, b x
3, c 3
3, the quadratic formula gives the solutions s32 4 5 25 3 3 10 s69
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
x
7
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 y
0
x
0
x
0
x
FIGURE 1
Possible graphs of y=ax@+bx+c
(a) b@-4ac>0
(b) b@-4ac=0
(c) b@-4ac<0
EXAMPLE 12 The quadratic x 2
x 4ac
2 is irreducible because its discriminant is 12 41 2 x 2. 7 0
negative: b2
Therefore, it is impossible to factor x 2
The Binomial Theorem
Recall the binomial expression from Equation 1: a If we multiply both sides by a
8
b
2
a2
2ab
b2
b and simplify, we get the binomial expansion
3
a
b
a3
3a 2b
3ab 2
b3
Repeating this procedure, we get a b
4
a4
4a 3b
6a 2b 2
4ab 3
b4
In general, we have the following formula.
9 The Binomial Theorem If k is a positive integer, then
a
b
k
ak
ka k 1b kk
kk 1 k 2 2 a b 1 2 1 k 2 k 3 3 a b 1 2 3 kk 1 k 1 2 3
1
n n
1
a k nb n
kab k
bk
8
s
REVIEW OF ALGEBRA
EXAMPLE 13 Expand x
2 5. x, b 2, k 2
3
SOLUTION Using the Binomial Theorem with a
5, we have 5x 2
4
x
2
5
x5 x5
5x 4 10x 4
2
5 4 3 x 1 2 40x 3 80x 2
2
2
5 4 3 2 x 1 2 3 32
2
5
80x
Radicals
The most commonly occurring radicals are square roots. The symbol s1 means “the positive square root of.” Thus x sa means x2 a and x 0 0. Here are two rules
Since a x 2 0, the symbol sa makes sense only when a for working with square roots: a b sa sb
10
sab
sa 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:
|
(For instance, take a
EXAMPLE 14
sa 9 and b
b
sa
sb
16 to see the error.)
(a)
s18 s2
18 2 sx 2 sy
2
s9
3
(b) sx 2 y
x sy x because s1 indicates the positive square root.
Notice that sx (See Appendix A.)
In general, if n is a positive integer, x
n sa
means
xn 0 and x
a 0.
If n is even, then a
3 Thus s 8 2 because lowing rules are valid:
2
3
4 6 8, but s 8 and s 8 are not defined. The fol-
n sab
n n sa sb
n
a b
n sa n sb
3 EXAMPLE 15 sx 4
3 sx 3x
3 3 sx 3 sx
3 xsx
REVIEW OF ALGEBRA
x
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
sx
a 4 x
b 2
EXAMPLE 16 Rationalize the numerator in the expression
.
SOLUTION We multiply the numerator and the denominator by the conjugate radical
sx
4 sx
2: 4 x 2 sx 4 x x x (sx 4 2) sx 2 sx sx 4 4 1 4 2 2 2 x 4 x (sx 4 4 2)
Exponents
Let a be any positive number and let n be a positive integer. Then, by definition,
1. a n
a a
n factors
a
2. a 3. a
0
1
n
4. a1
n n
am
1 an n sa n sa m
(sa )m
n
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
as
ar
s
2.
ar as a b
r
ar
s
3. a r
s
a rs
4. ab
r
a rb r
5.
ar br
b
0
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.
10
s
REVIEW OF ALGEBRA
EXAMPLE 17
(a) 28 x x
2 1
82 y y
2 1
28 1 x2 1 x y
23
2
28
26
214 xy y x
(b)
1 y2 x2 2 y x 2y 2 y2 x2 1 y x x 2y 2 y xy x y x y x xy y x xy 8
3
(c) 43 2 1 (d) 3 4 sx (e) x y
3
s43 s64 1 x 4 x4 3 y 2x z
4
Alternative solution: 43 2
(s4 )3
23
8
x3 y3
y 8x 4 z4
x 7y 5z
4
Exercises
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
A Click here for answers. 1–16 1.
s
1
27. 2. 4. 4 6. 8
1 c 1 c
s s s
1 1
s s s s s
Expand and simplify. 2x 2 y 3x x 4 x xy 4
28. 1
1 1 1 1
s s
6ab 0.5ac 5 3a x 4 1 3x 1
2
1
s s
x
s s s s
3. 2x x 5.
s
s
s
s
24
29–48 29. 2x 31. x 2
s
Factor the expression. 12x 3 7x 2x 36 5x 1 12t 2x 2 3x 2 5x 2 9 x x 2x
s s s
7. 4 x 2 8. 5 3t 9. 4x 11. 2x 13. y 4 6 14. t 15. 1
s s s
2 t2 7
5 x2 2
2x 2t t
1 3
10. x x 12. 2
30. 5ab
8abc x 7x 10x 10x 9s 2 27 4x 2 2x 2 3x 2
s s
6 8
32. x 2 34. 2x 2 36. 8x 2
6 4 3 25
1 x 3x
2
2
33. x 2 35. 9x 2 37. 6x 2
y 5 5
2
y 3 8t 3x
s s s
6
38. x 2 40. 4t 2 42. x 3 44. x 3
2t
1
16. 1
s s s s s
39. t 3
2x x 2
s s
1
x
s s
x2
s
2
41. 4t 2 43. x 3
s s s s
5x 23x 4x
s s
2 60 12
s s s
17–28 17. 19.
45. x 3
s
3 24
s s s
46. x 3 48. x 3
s s s
Perform the indicated operations and simplify. 8x
9b 6 18. 3b 2 5 1 x u u 1 3 20. 22. 24. 1 x 2 a2 x y z 1 3 ab x 1 1 4 b2
47. x 3
s s s
2 2
1 x
s
49–54 49. 51. 53.
s
Simplify the expression. x 3x 2 2 1 9x 8 1 3 x2 9
50. 52.
x2 x2 x2 x x
2
2x 2
3x 2 x 4
2
21. u 23. 25. x y z
x 3 5x 2 6x x 2 x 12
1
2r s
s2 6t
26.
a bc
b ac
REVIEW OF ALGEBRA
x
11
54.
s s
x2
s
x x
s s
2
s
x2
s s
2 5x
s s
4
s s s s s s s s s s
85. 87. 56. x 2 58. x 2 60. 3x 2
x 9 2x x3 a b a 5b 5
1 2 3 3 4
4
86. 88.
an a x x
1
a 2n
n 2
1
55–60 55. x 2 57. x 2
s
Complete the square. 2x 5x 4x
s s
y y
5 4 3
1 1
5 10 2
s s s s s
16x 3x 24x
s s s
80 1 50
s s s s
89. 3
90. 961 92. 64
3 2
91. 125 2
59. 4x 2
s s s
93. 2x 2 y 4
5 95. sy 6
94. x 5 y 3z 10
4 96. (sa ) 8 sx 5 4 sx 3
3 5
s
s
s
3
61–68 61. x 63. x
2 2
s
Solve the equation. 9x 9x 5x 2x 1
s s
10 1 1 0
s
0 0 0
62. x 64. x
2 2
2x 2x 7x 3x 2
s s
8 7 2 x
s s
0 0 0
s
97. 99.
s
1 (st ) 5
4
98.
65. 3x 2 67. x 3
s s s
66. 2x 2 68. x 3
s s s s s s
t 1 2sst s2 3
s s s s s s s
4 100. sr 2n
1
4 sr
1
s
s
s
s
s
s
s
s
s
s
s
1
s
0
s s
101–108 101.
s
Rationalize the expression. 3 9
102. 104. 106.
s
69–72
s
Which of the quadratics are irreducible? 3x x
s s
sx x
(1 sx )
x s2 1 sx sy x
s s
1 1 s2 h h
69. 2x 2 71. 3x
s s
4 6
s s s s s
70. 2x 2 72. x
s s
9x 3x
s s
4 6
s s s s s
2
2
103. 105.
x sx 8 x 4 2 3 s5 3x
s s s
h
s
s
73–76 73. a 75. x
s s
s
Use the Binomial Theorem to expand the expression. b
6 4
74. a 76. 3
s s s s s s s s s
b x
s
7
107. sx 2
2 5
s s s s s s s s s
4
s
x
s s s
108. sx 2
s s s
sx 2
s s s
x
s s
2
1
s
s
109–116
77–82
s
Simplify the radicals.
78.
3 s 2 3 s54
s State whether or not the equation is true for all values of the variable.
77. s32 s2 80. sxy sx y
3
s s s s s s s
79.
4 3
4 s32x 4 4 s2 5
109. sx 2 111. 113.
s s s
x 1 1 y 1 x7 4x a
s s s
110. sx 2
4 1 y x
1
x x 1 2 2 x
2 y
81. s16a b
s s s s
s96a6 82. 5 s3a
s s s s s s
16 a 16 x x x3
4
a 16 y
112. 114.
x 4
1
2
83–100 83. 310
Use the Laws of Exponents to rewrite and simplify the expression.
s
115.
116. 6
s s s
6
s
4x
s s
4a
s s s s s s s s s s
98
84. 216
410
16 6
s
12
s
ANSWERS
Answers
1. 5. 8.
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
q
2. 2x 3 y 5 3. 2x 2 4. 4x 3a 2bc 10x 6. 4 7. x 2 8 6a x 6x 3 3t 2 21t 22 25x 7 9. 12x 2 x2
2
3x 2
63.
9 2 7 4
s85
64. 1
2s2 1 2 s5
65.
5 6
68.
s13
10. x 3 12. 9x 14. 16. x 4 19. 22. 26. x2
2x 4
11. 4x 2 13. 30y 31 2x 20. 4a 2 c
4
4x
y
5
1
y 5x 2 4x 21. u2 u
6
66.
s33
67. 1,
1,
1
s2
12x 56t 2x 3
15t 2
15. 2x 3 1 2x x2 1 x 23. yz 28. 3 2 2x x 1
69. Irreducible x 1 18. 3 3u 1 rs 25. 3t 6x 2 3 x 2 1 2 b 73. a 6 74. a 7
70. Not irreducible 72. Irreducible
x2
17. 1
71. Not irreducible (two real roots)
6a 5b 7a 6b 4x 6 405x
78.
15a 4b 2 21a 5b 2 6x 4
2 1 3
20a 3b 3 35a 4b 3
15a 2b 4 35a 3b 4
6ab 5
b6 b7
3x 7 2x 15 3ab a 2b 2 27. 8c 4 x 2 c
2b 2 a b2
2
zx 24. y
21a 2b 5
75. x 8 76. 243 77. 8
7ab 6
4x 2 270x
4
1 90x 6 15x 8
80. x 2 y 84. 2 60 85. 16x 10
2 31. x
29. 2x 1 32. x 4
x 10
79. 2 x 83. 3 26 88. 92.
5 1 256 4
30. ab 5 33. x 35. 9 x 37. 3x 39. t 41. 2t 43. x x 45. x 47. x
6 x
81. 4a 2bsb 86. a 2n
3
82. 2a 87.
34. 2x
1 x
2 x 2 2x 1 t 3 1
2
2 3 t 1
42. x 44. x
36. 4x 38. x 40. 2t
3 2x 5
2
1
a2 b
x xy
y
2
89.
1 s3
90. 2 5s3
91. 25 95. y 6 100. r n
93. 2s2 x 3 y 6 97. t
5 2
3s 2t 3x 2 3 x 2 x 1 8 4 9
3s
94. 99. 103.
2 2
3 x 1 3 4 1 2
2
2
x3 y z t1 4 s 1 24 x2
9 5 6
96. a 3 101. 104.
98.
1 x1 8 x
x
2
1 sx h 3 2 s2
1 x 2 x 2 2 2 9 1
2
1 x 3 x 2x 50. x
54.
46. x 48. x
5 x 3 x x x 52. x
4 2 2 4
102.
1 sx
x 49. x
53.
x 51. x
4x 16 xsx 8 s5 2
106.
s2 sy y
108.
h
x x2
x
x 2 6x 4 1 x 2 x 8 1
2 2
105. 57. ( x
5 2 2
3
sx x x
55. x 58. ( x 60. 3 x
4
5 4 2
56. x 59. 2x
16 3
62.
)
15 4
107.
sx 2
3x 4 3x 4
2x sx 2 x sx 2
112. False 116. True
x
3 2 2
)
109. False
110. False 114. False
111. True 115. False
4
2
61. 1,
10
2, 4
113. False