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8/13/2019 Appendix B PreCalculus
1/30
REVIEW OF INTEGER EXPONENTS
I write a1, a2, a3, etc., for etc. Isaac Newton (June 13, 1676)
In basic algebra you learned the exponential notation an, where a is a real num
n is a natural number. Because that definition is basic to all that follows in tthe next two sections of this appendix, we repeat it here.
Definition 1 Base and Exponent
Given a real number a and a natural number n, we define an by
n factors
In the expression an the number a is the base, and n is the exponent or po
which the base is raised.
an a # a # a p a
1
aaa,
1
aa,
1
a,
APPENDIX
BB.1 Review of Integer
Exponents
B.2 Review ofnth Roots
B.3 Review of Rational
Exponents
B.4 Review of FactoringB.5 Review of Fractional
Expressions
B.6 Properties of the Real
Numbers
B.1
EXAMPLE
SOLUTION
1 Using Algebraic Notation
Rewrite each expression using algebraic notation:
(a) xto the fourth power, plus five;
(b) the fourth power of the quantityxplus five;
(c) xplusy, to the fourth power;
(d) xplusy to the fourth power.
(a) x4 5 (b) (x 5)4 (c) (xy)4 (d) xy4
Caution: Note that parts (c) and (d) of Example 1 differ only in the use of the c
So for spoken purposes the idea in (c) would be more clearly conveyed by
the fourth power of the quantityxplusy. In the case of (d), if you read it a
another student or your instructor, chances are youll be asked, Do you mean
or do you mean (xy)4?
Moral:Algebra is a precise language; use it carefully. Learn to ask yourself w
what youve written will be interpreted in the manner you intended.
8/13/2019 Appendix B PreCalculus
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B-2 Appendix B
In basic algebra the four properties in the following box are developed for work-
ing with exponents that are natural numbers. Each of these properties is a direct con-
sequence of the definition of an. For instance, according to the first property, we have
a2a3 a5. To verify that this is indeed correct, we note that
a2a3 (aa)(aaa) a5
PROPERTY SUMMARY Properties of Exponents
Property Examples
1. aman amn a5a6 a11; (x 1)(x 1)2 (x 1)3
2. (am)n amn (23)4 212; [(x 1)2]3 (x 1)6
3. a4; 1
4. (ab)m
am
bm
; (2x2)3 23 (x2)3 8x6;
x8
y12ax2
y3b4
#
am
bmaa
b bm
1
a4;
a5
a5a2
a6
a6
a2am
an damn if m n1
anmif m n
1 if m n
Of course, we assume that all of these expressions make sense; in particular, we as-
sume that no denominator of a fraction is zero.
Now we want to extend our definition of an to allow for exponents that are inte-
gers but not necessarily natural numbers. We begin by defining a0.
Definition 2 Zero Exponent
For any nonzero real number a, EXAMPLES
(a) 20 1(b) (p)0 1
(00 is not defined.)(c)
Its easy to see the motivation for defining a0 to be 1. Assuming that the exponent zero
is to have the same properties as exponents that are natural numbers, we can write
That is,
Now we divide both sides of this last equation by an to obtain a0 1, which agrees
with our definition.
Our next definition (see the box that follows) assigns a meaning to the expression
an when n is a natural number. Again, its easy to see the motivation for this definition.
We would like to have
anan an(n) a0 1
a0an an
a0an a0n
a 31 a2 b2
b 0 1a0 1
8/13/2019 Appendix B PreCalculus
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That is,
Now we divide both sides of this last equation by an to obtain an 1an, in
ment with Definition 3.
Definition 3 Negative Exponent
For any nonzero real number a
and natural number n,
an 1
an
anan 1
B.1 Review of Integer Exponents
EXAMPLE
SOLUTION
EXAMPLE
EXAMPLES
(a) 21
(b) 10
(c) x2
(d) (a2b)3
(e) 23 81
23
1
123
1
(a2b)3
1
a6b3
1
x2
a 110
b1 11 110 21
1
21
1
2
It can be shown that the four properties of exponents that we listed earli
tinue to hold now for all integer exponents. We make use of this fact in the ne
examples.
2 Using Properties of Exponents
Simplify the following expression. Write the answer in such a way that only p
exponents appear.
First Method Alternative Method
3 Using Properties of Exponents
Simplify the following expression, writing the answer so that negative expon
not used.
a a5b2c0a3b1
b 3
as obtained
previously
b5
a
(a2b3)2(a5b)1 a45b61 a1b5
(a2b3)2(a5b)1 a4b6
a5b
b61
a54
b5
a
(a2b3)2(a5b)1 (a4b6)(a5b1)(a2b3)2(a5b)1 (a2b3)2 #1
a5b
(a2b3)2(a5b)1
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B-4 Appendix B
We show two solutions. The first makes immediate use of the property (am)n amn.
In the second solution we begin by working within the parentheses.
First Solution Alternative Solution
( a5b2c0 )3 b9a24
( a5b2c0 )3 b6(3)a9 (15)
b9
a24
a a5b2c0a3b1
b 3 a b3a8b 3a a5b2c0
a3b1b 3 a15b6
a9b3
SOLUTION
EXAMPLE
SOLUTION
EXAMPLE
SOLUTION
4 Simplifying with Variable Exponents
Simplify the following expressions, writing the answers so that negative exponents
are not used. (Assume thatp and q are natural numbers.)
(a) (b) (apbq)2(a3pbq)1
(a)
(b)
b3q
ap
apb3q(apbq)2(a3pbq)1 (a2pb2q)(a3pbq)
a4pqpq a3p2q
a4pq
apq
a4pq(pq)
a4pq
apq
5 Simplifying Numerical Quantities Involving Exponents
Use the properties of exponents to compute the quantity
As an application of some of the ideas in this section, we briefly discuss scientific
notation, which is a convenient form for writing very large or very small numbers. Such
numbers occur often in the sciences. For instance, the speed of light in a vacuum is
As written, this number would be awkward to work with in calculating. In fact, thenumber as written cannot even be displayed on some hand-held calculators, because
there are too many digits. To write the number 29,979,000,000 (or any positive num-
ber) in scientific notation, we express it as a number between 1 and 10, multiplied by
an appropriate power of 10. That is, we write it in the form
where 1 a 10 and nis an integer.
a 10n
29,979,000,000 cm/sec
1
(31513)(21210)
1
32 # 22
1
36
210 # 313
33 # 312 # 212
210 # 313
315 # 212
210 #
313
27# 612 2
10 #3
13
(33)(3# 2)12
210 # 313
27# 612.
8/13/2019 Appendix B PreCalculus
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According to this convention, the number 4.03 106 is in scientific notation,
same quantity written as 40.3 105 is not in scientific notation. To convert
number into scientific notation, well rely on the following two-step procedu
1. First move the decimal point until it is to the immediate right of the first n
digit.
2. Then multiply by 10n or 10n, depending on whether the decimal po
moved n places to the left or to the right, respectively.
For example, to express the number 29,979,000,000 in scientific notatio
we move the decimal point 10 places to the left so that its located betwee
and the 9, then we multiply by 1010. The result is
or, more simply,
As additional examples we list the following numbers expressed in both o
and scientific notation. For practice, you should verify each conversion for y
using our two-step procedure.
All scientific calculators have keys for entering numbers in scientific no
(Indeed, many calculators will convert numbers into scientific notation for y
you have questions about how your own calculator operates with respect to sc
notation, you should consult the users manual. Well indicate only one ehere, using a Texas Instruments graphing calculator. The keystrokes on other
are quite similar. Consider the following number expressed in scientific no
3.159 1020. (This gargantuan number is the diameter, in miles, of a galaxy
center of a distant star cluster known as Abell 2029.) The sequence of keystro
entering this number on the Texas Instruments calculator is
3.159 20EE
0.0000002 2 1070.000099 9.9 105
55,708 5.5708 104
29,979,000,000 2.9979 1010
29,979,000,000 2.9979000000 1010
To Express a Number Using Scientific Notation
B.1 Review of Integer Exponents
EXERCISE SET B.1
AIn Exercises 14, evaluate each expression using the givenvalue of x.
1. 2x3 x 4;x 2 2. 1 x 2x2 3x3;x 1
3. x 4. x 11 (x 1)2
1 (x 1)2;
1
2
1 2x2
1 2x3;
In Exercises 516, use the properties of exponents to simeach expression.
5. (a) a3a12 6. (a) (32)3 (23)2
(b) (a 1)3(a 1)12 (b) (x2)3 (x3)2
(c) (a 1)12(a 1)3 (c) (x2)a (xa)2
7. (a) yy2y8
(b) (y 1)(y 1)2(y 1)8
(c) [(y 1)(y 1)8]2
8/13/2019 Appendix B PreCalculus
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B-6 Appendix B
33. 34. (2x2)3 2(x2)3
35. 36.
37. (xp)(xp) 38. (2pq)(2qp1)
In Exercises 3942, use the properties of exponents in comput-
ing each quantity, as in Example 5. (The point here is to do as
little arithmetic as possible.)
39. 40.
41. 42.
For Exercises 4350, express each number in scientific notation.
43. The average distance (in miles) from Earth to the Sun:
44. The average distance (in miles) from the planet Pluto to the
Sun:
45. The average orbital speed (in miles per hour) of Earth:
46. The average orbital speed (in miles per hour) of the planet
Mercury:
47. The average distance (in miles) from the Sun to the nearest
star:
48. The equatorial diameter (in miles) of
(a) Mercury: 3031 (c) Jupiter: 88,733
(b) Earth: 7927 (d) the Sun: 865,000
49. The time (in seconds) for light to travel
(a) one foot:
(b) across an atom:
(c) across the nucleus of an atom:
50. The mass (in grams) of
(a) a proton:
(b) an electron:
0.000000000000000000000000000911
0.00000000000000000000000167
0.000000000000000000000001
0.000000000000000001
0.000000001
25,000,000,000,000,000,000
107,300
66,800
3,666,000,000
92,900,000
a 144 # 12523 # 32
b124532 # 124
212 # 513
101228 # 315
9 # 310 # 12
bp2q
(b2)qbp1
bp
x2
y3
x2
y38. (a) 9. (a)
(b) (b)
(c) (c)
10. (a) (y2y3)2 11. (a)
(b) y2(y3)2 (b)
(c) 2m(2n)2 (c)
12. (a) 13. (a)
(b) (b)
(c) (c)
14. (a) (x2y3z)4 15. (a) 4(x3)2
(b) 2(x2y3z)4 (b) (4x3)2
(c) (2x2y3z)4
(c)
16. (a) 2(x 1)7 (x 1)7
(b) [2(x 1)]7 (x 1)7
(c) 2(x 1)7 [2(x 1)]7
For Exercises 1738, simplify each expression. Write the
answers in such a way that negative exponents do not appear.
In Exercises 3538, assume that the letters p and q represent
natural numbers.
17. (a) 640 18. (a) 20 30
(b) (643)0 (b) (20 30)3
(c) (640)3 (c) (23 33)0
19. (a) 101 102 20. (a) 42 41
(b) (101 102)1 (b)
(c) [(101)(102)]1 (c)
21. 22.
23. (a) 52 102 24. (a)
(b) (5 10)2 (b)
25. (a
2
bc
0
)
3
26. (a
3
b)
3
(a
2
b
4
)
1
27. (a2b1c3)2 28. (22 21 20)2
29. 30.
31. 32.a a3b9c2a5b2c4
b 0ax4y8z2xy2z6
b2ax4y8z2
xy2z6b 2ax3y2z
xy2z3b3
1 14 34 221 14 22 1 34 22
1 13 14 211 13 21 1 14 21[ 1 14 21 1 14 22]1[ 1 14 21 1 14 22]1
(4x2)3
(4x3)2
ax2
y
20
x6y15b2
(x2
3x 2)6
x2 3x 2
x2y20
x6y15x
x6
x6y15
x2y20x6
x
(t2 3)15
(t2 3)9
t9
t15
t15
t9
1210
129
210
212
(x2 3)9
(x2 3)10x10
x12
(x2 3)10
(x2 3)9x12
x10
8/13/2019 Appendix B PreCalculus
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REVIEW OF nTH ROOTS
In this section we generalize the notion of square root that you studied in elem
algebra. The new idea is that of an nth root; the definition and some basic ex
are given in the box that follows.
B.2 Review of nth Roots
EXAMPLES
B5
132
12
B41
16
1
2;13 8 2
125 5;125 5; 125
Hindu mathematicians first recognized
negative roots, and the two square roots
of a positive number, . . . though they
were suspicious also. David Wells in
The Penguin Dictionary of Curious andInteresting Numbers (Harmondsworth,
Middlesex, England: Penguin Books,
Ltd., 1986)
B.2
Definition 2 Principal nth Root
1. Let n be a natural number. If a and b
are nonnegative real numbers, then
if and only if b an
The number a is the principalnth
root of b.
2. If a and b are negative and n is anodd natural number, then
if and only if b an1n b a
1n b a
Definition 1 nth Roots
Let n be a natural number. If a
and b are real numbers and
an b
then we say that a is an nth root of b.
When n 2 and when n 3, we
refer to the roots as square roots
and cube roots, respectively.
As the examples in the box suggest, square roots, fourth roots, and all eve
of positive numbers occur in pairs, one positive and one negative. In these ca
use the notation to denote the positive, or principal, nth root of b. As ex
of this notation, we can write
The symbol is called a radical sign, and the number within the radical sig
radicand. The natural number n used in the notation is called the index
radical. For square roots, as you know from basic algebra, we suppress the in
simply write rather than So, for example, 5.
On the other hand, as we saw in the examples, cube roots, fifth roots, andall oddroots occur singly, not in pairs. In these cases, we again use the notat
for the nth root. The definition and examples in the following box summar
discussion up to this point.
12512 .11n 1
14 81 3 (The negative of the principal fourth root of 81 is 3.14
81 3 (The principal fourth root of 81 is not 3.)
14 81 3 (The principal fourth root of 81 is 3.)2n b
EXAMPLESBoth 3 and 3 are square roots of
because 32 9 and (3)2 9.
Both 2 and 2 are fourth roots of
because 24 16 and (2)4 16.
2 is a cube root of 8 because 23
3 is a fifth root of 243 becaus
(3)5 243.
There are five properties of nth roots that are frequently used in simplifyi
tain expressions. The first four are similar to the properties of square roots t
developed in elementary algebra. For reference we list these properties side
8/13/2019 Appendix B PreCalculus
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(a)
(b) B162
491162149
181127
912
7
713 213 513 (2 5)131413 12513
112
175
1(4)(3)
1(25)(3)SOLUTION
B-8 Appendix B
in the following box. Property 5 is listed here only for the sake of completeness;
well postpone discussing it until Appendix B.3.
PROPERTY SUMMARY Properties of nth Roots
Suppose thatxandy are real numbers and that m and n are
natural numbers. Then each of the following properties holds, CORRESPONDINGprovided only that the expressions on both sides of the PROPERTIES FORequation are defined (and so represent real numbers). SQUARE ROOTS
1.
2.
3.
4. n even:
n odd:
5. m
21nx mn
1x
2n xn x2x2 0x02n xn 0x0B
x
y1x1yBn
x
y1n x1n y
1xy 1x1y1n xy 1n x1n y11x22 x11n x2n x
Our immediate use for these properties will be in simplifying expressions in-
volving nth roots. In general, we try to factor the expression under the radical so
that one factor is the largest perfect nth power that we can find. Then we apply
Property 2 or 3. For instance, the expression is simplified as follows:
In this procedure we began by factoring 72 as (36)(2). Note that 36 is the largest fac-
tor of 72 that is a perfect square. If we were to begin instead with a different factor-
ization, for example, 72 (9)(8), we could still arriveat thesame answer, but it would
take longer. (Check this for yourself.) As another example, lets simplify First,what (if any) is the largest perfect-cube factor of 40? The first few perfect cubes are
so we see that 8 is a perfect-cube factor of 40, and we write
13 40 13 (8)(5) 13 813 5 213 513 1 23 8 33 27 43 64 1
3
40.
172 1(36)(2) 13612 612172
EXAMPLE 1 Simplifying Square Roots
Simplify: (a) (b) B162
49.112 175;
8/13/2019 Appendix B PreCalculus
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EXAMPLE
SOLUTION
EXAMPLE
SOLUTION
B.2 Review of nth Roots
EXAMPLE
SOLUTION
2 Simplifying nth Roots
Simplify:
3
13
2
213 2 513 2 413 213 16 13 250 13 128 13 813 2 13 12513 2 13 6413 2
13 16 13 250 13 128.
3 Simplifying Radicals Containing Variables
Simplify each of the following expressions by removing the largest possible p
square or perfect-cube factor from within the radical:
(a)
(b) wherex 0;
(c) where a 0;
(d)
(a)
(b)
(c)
(d) 23 16y5 23 8y323 2y2 2y23 2y2 3a312a
218a7 2(9a6)(2a) 29a612a 212x 2x2 xbecausex 028x
214122x2
28x2 2(4)(2)(x2) 14122x2 212 0x023 16y5.2
18a7,
28x2,28x2;
4 Simplifying nth Roots Containing Variables
Simplify each of the following, assuming that a, b, and c are positive:
(a) (b)
(a)
(b)
24 16a4b424 2a2b
c2
2ab24 2a2bc2
B432a6b5
c8
24 32a6b524 c8
2bc2
12ac
28ab2c5 2(4b2c4)(2ac) 24b2c412acB4
32a6b5
c8.28ab2c5;
8/13/2019 Appendix B PreCalculus
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B-10 Appendix B
In Examples 1 through 4 we used the definitions and propertiesofnth roots to sim-
plify certain expressions. The box that follows shows some common errors to avoid
in working with roots. Use the error box to test yourself: Cover up the columns labeled
Correction and Comment, and try to decide for yourself where the errors lie.
Error Correction Comment
2 Although 2 is one of the fourth
roots of 16, it is not theprincipal
fourth root. The notation isreserved for the principal fourth root.
5 Although 5 is one of the square
roots of 25, it is not theprincipal
square root.
The expression The properties of roots differ with
cannot respect to addition versus multi-
be simplified. plication. For we do have thesimplification
(assuming that a and b are
nonnegative).
The expression For we do have
cannot
be simplified.
There are times when it is convenient to rewrite fractions involving radicals in al-
ternative forms. Suppose, for example, that we want to rewrite the fraction in
an equivalent form that does not involve a radical in the denominator. This is calledrationalizing the denominator. The procedure here is to multiply by 1 in this way:
That is, as required.
To rationalize a denominator of the form a we need to multiply the fraction
not by but rather by 1a 21a 2( 1). To see why this is necessary,notice that
which still contains a radical, whereas
which is free of radicals. Note that we multiply a by a to obtain an
expression that is free of square roots by taking advantage of the form of a difference
1b1b a2 b
1a 1b 2 1a 1b 2 a2 a1b a1b b1a 1b
21b a1b b
1b1b1b,1b 1b,5
13 513
3,
5
13 5
13 # 1 5
13 #1313
5133
513
13 a b13 ab 13 a13 b.13 ab13 a b 13 a 13 b
1ab 1a1b1ab
1a b1a b 1a 1b
125125 514 16
14 1614 16 2Errors to Avoid
8/13/2019 Appendix B PreCalculus
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of two squares. Similarly, to rationalize a denominator of the form a , w
tiply the fraction by 1a 21a 2. (The quantities a and a to be conjugates of each other.) The next two examples make use of these ide
1b1b1b1b 1b
B.2 Review of nth Roots
EXAMPLE
5 Rationalizing a Square Root Denominator
Simplify: 3150.112SOLUTION First, we rationalize the denominator in the fraction 1
Next, we simplify the expression
Now, putting things together, we have
2912
2
12 3012
2
(1 30)122
1
12 3150 12
2 1512 1
2
2
30
12
2
3150 31(25)(2) 312512 (3)(5)12 15123150:
1
12 1
12 # 1 1
12 #1212
122
12:
EXAMPLE 6 Using the Conjugate to Rationalize a Denominator
Rationalize the denominator in the expression 42 13
.
SOLUTION We multiply by 1, writing it as
Note: Check for yourself that multiplying the original fraction by d
eliminate radicals in the denominator.
2
13
2 13
8 413
4 12 13 24 113 22
4 12 13 24 3
8 413
1
4
2 13 # 1 4
2 13 #2 132 13
2 132 13.
In the next example we are asked to rationalize the numerator rather th
denominator. This is useful at times in calculus.
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B-12 Appendix B
EXAMPLE 7 Rationalizing a Numerator Using the Conjugate
Rationalize the numeratorof: wherex 0,x 3.1x 13
x 3,
SOLUTION
1
1x13
11x22 113 22
(x 3)11x13 2x 3
(x 3)11x13 2
1x13
x 3
# 1 1x13
x 3
#1x13
1x
13
The strategy for rationalizing numerators or denominators involving nth roots is
similar to that used for square roots. To rationalize a numerator or a denominator in-
volving an nth root, we multiply the numerator or denominator by a factor that yields
a product that itself is a perfect nth power. The next example displays two instances
of this.
EXAMPLE 8 Rationalizing Denominators Containing nth Roots
(a) Rationalize the denominator of:
(b) Rationalize the denominator of: where a 0, b 0.ab
24 a2b3,
6
13 7.
Finally, just as we used the difference of squares formula to motivate the method
of rationalizing certain two-term denominators containing a square root term, we can
use the difference or sum of cubes formula (see Appendix B.4) to rationalize a
denominator involving a cube root term. In particular, in the sum of cubes formula
x3 y3 (xy)(x2 xy y2), the expression (x2 xy y2) is the conjugate of
the expressionxy.
SOLUTION (a)
(b)
ab24 a2b24 a4b4
ab24 a2bab
24 a2b
ab
24 a2b3 # 1 ab
24 a2b3 #24 a2b24 a2b
613 49
7 as required
62
372
23 73
6
13 7 # 1 6
13 7 #23 7223 72
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B.2 Review of nth Roots
EXAMPLE 9 Using the Conjugate to Rationalize a Denominator Involving a Cube Ro
Rationalize the denominator in the expression1
23 a 23 b .SOLUTION If we letx andy the conjugate of the denominatorxy
is
where, for example, Then
23 a2 23 ab 23 b2
123 a 23 123 b 23 23 a2 23 ab 23
a b
1
23 a 23 b 1
23 a 23 b #23 a2 23 ab 23 b223 a2 23 ab 23 b2
123 a 22 23 a23 a 23 a2.x2 xy y2 123 a 22 23 a23 b 123 b 22 23 a2 23 ab 23 b2
23 a23 b,
23 a
EXERCISE SET B.2
AIn Exercises 18, determine whether each statement is TRUE
or FALSE.
1. 2.
3. 4.
5. 6.
7. 8.
For Exercises 918, evaluate each expression. If the expression
is undefined (i.e., does not represent a real number), say so.
9. (a) 10. (a)
(b) (b)
11. (a) 12. (a)
(b) (b)
13. (a) 14. (a)
(b) (b)
15. (a) 16. (a)
(b) (b)
17. (a) 18. (a)
(b) (b)
In Exercises 1944, simplify each expression. Unless otherwise
specified, assume that all letters in Exercises 3544 represent posi-
tive numbers.
19. (a) 20. (a)
(b) (b)13 37513 54 1150118
23 (10)3
15 32
24 (10)415 3216 64322712516 64422568114 1614 1614 1611626 1100023 812523 1100023 812515 3214 64 15
3213 64
2(5)2 5113 10 23 10110 6 110 16243 213 149 71100 101256 16181 9
21. (a) 22. (a)
(b) (b)
23. (a) 24. (a)
(b) (b)
25. (a) 26. (a)
(b) (b)
27. (a)
(b)
28. (a)
(b)
29. (a)
(b)
30. (a)
(b)
31. 32.
33. 34.
35. (a) 36. (a)
(b) wherey 0 (b) where
37. (a) 38. (a)
(b) (b) where
39. 40.
41. 42.
43. 44.24 ab324 a3b23 16a12b2c923 8a4b624 16a4b52(a b)5(16a2b272a3b4c5 2
464y4,2ab32a3b 2
3 125x6
2ab2
2a2b
24 16a4,236y2,2225x4y3236x2
23 14096216413 192 13 81 4124 8154 216
281121 23 8133113 2 13 213 0.00810.0915 2 15 64 15 4861
3 112 14814 32
14 162
4150 31128213 81 313 213 2 13 16413 212712 1825 2562434216625222549225413 10815 64127198
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By replacing 5 and 3 with b and n, respectively, we can see that we want to
b1n to mean Also, by thinking of bmn as (b1n)m, we see that the defini
bmn ought to be These definitions are formalized in the box that follo11n b 2m.1n
b.
B.3 Review of Rational Exponents
Definition Rational Exponents
1. Let b denote a real number and n a
natural number. We define b1n by
(If n is even, we require that b 0.)2. Let mn be a rational number reduced
to lowest terms. Assume that n ispositive and that exists. Then
or, equivalently,
bmn 2n bmbmn 11n b 2m
1n b
b1n 1n b
It can be shown that the four properties of exponents that we listed in Appen
(on page B-2) continue to hold for rational exponents in general. In fact, we
this for granted rather than follow the lengthy argument needed for its verifi
We will also assume that these properties apply to irrational exponents. For in
we have
(The definition of irrational exponents is discussed in Section 5.1.) In the nex
examples we display the basic techniques for working with rational exponen
1215 215 25 32
EXAMPLES
412 2
(8)13
823 22 4
or, equivalently,
823 413 6423 82113 8 2213 8 21
4
EXAMPLE 1 Evaluating Rational Exponents
Simplify each of the following quantities. Express the answers using positiv
nents. If an expression does not represent a real number, say so.
(a) 4912 (b) 4912 (c) (49)12 (d) 4912
SOLUTION (a) 4912 7
(b) 4912 (4912) 7
(c) The quantity (49)12 does not represent a real number because there i
real numberxsuch thatx2
49.
(d) 4912
Alternatively, we have
4912 (4912)1 71 1
7
1
711
14921492491
149149
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B-16 Appendix B
EXAMPLE 2 Simplifying Expressions Containing Rational Exponents
Simplify each of the following. Write the answers using positive exponents. (Assume
that a 0.)
(a) (5a23)(4a34) (b) (c) (x2 1)15(x2 1)45B516a13
a14
SOLUTION (a)
because
(b)
because
(c) (x2 1)15(x2 1)45 (x2 1)1 x2 1
1615a160
13
14
112 (16a
112)15
B516a13
a14 a 16a13
a14b 15
23
34
1712 20a
1712
(5a23)(4a34) 20a(23)(34)
EXAMPLE3 Evaluating Rational Exponents
Simplify: (a) 3225; (b) (8)43.
SOLUTION (a) 3225 22
Alternatively, we have
3225 (25)25 22
(b) (8)43 (2)4 16
Alternatively, we can write
(8)43 [(2)3]43 (2)4 16
113 8 241
4
1
22
1
4
1
22115 32 22
Rational exponents can be used to simplify certain expressions containing radi-
cals. For example, one of the properties of nth roots that we listed but did not discuss
in Appendix B.2 is Using exponents, it is easy to verify this property.
We have
x1mm mn1x as we wished to show
m21nx (x1n)1mm21nx mn1x.
EXAMPLE 4 Using Rational Exponents to Combine Radicals
Consider the expression wherexandy are positive.
(a) Rewrite the expression using rational exponents.
(b) Rewrite the expression using only one radical sign.
1x23 y2,
SOLUTION (a)
(b) rewriting the fractions usinga common denominator
26x326y4 26x3y4x36y46
2x23y2 x12y232x23y2 x12y23
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B-18 Appendix B
REVIEW OF FACTORING
There are many cases in algebra in which the process offactoring simplifies the work
at hand. To factor a polynomial means to write it as a product of two or more non-
constant polynomials. For instance, a factorization ofx2 9 is given by
In this case,x 3 andx 3 are the factors ofx2 9.
There is one convention that we need to agree on at the outset. If the polynomial
or expression that we wish to factor contains only integer coefficients, then the fac-
tors (if any) should involve only integer coefficients. For example, according to this
convention, we will not consider the following type of factorization in this section:
because it involves coefficients that are irrational numbers. (We should point out,
however, that factorizations such as this are useful at times, particularly in calculus.)
As it happens,x2 2 is an example of a polynomial that cannot be factored using in-
teger coefficients. In such a case we say that the polynomial is irreducible over theintegers.
If the coefficients of a polynomial are rational numbers, then we do allow factors
with rational coefficients. For instance, the factorization of y2 over the rational
numbers is given by
Well consider five techniques for factoring in this section. These techniques will
be applied in the next two sections and throughout the text. In Table 1 we summarize
these techniques. Notice that three of the formulas in the table are just restatements
y2 1
4ay 1
2b ay 1
2b
14
x2 2 1x 12 2 1x 12 2
x2
9 (x 3)(x 3)
B.4fac # tor (fakt r), n. . . . 2. Math.one of two or more numbers, algebraic
expressions, or the like, that when
multiplied together produce a given
product; . . . v.t. 10. Math. to express
(a mathematical quantity) as a product
of two or more quantities of like kind,
as 30 2 # 3 # 5, or x2 y2 (xy) (xy). The Random House
Dictionary of the English Language,
2nd ed. (New York: Random House,
1987)
0
TABLE 1 Basic Factoring Techniques
Technique Example or Formula Remark
Common factor 3x4 6x3 12x2 3x2(x2 2x 4) In any factoring problem, the first step always is
4(x2 1) x(x2 1) (x2 1)(4 x) to look for the common factor of highest degree.
Difference of x2 a2 (x a)(x a) There is no corresponding formula for a sum of
squares squares;x2 a2 is irreducible over the integers.
Trial and error x2 2x 3 (x 3)(x 1) In this example the only possibilities, or trials, are
(a) (x 3)(x 1) (c) (x 3)(x 1)
(b) (x 3)(x 1) (d) (x 3)(x 1)
By inspection or by carrying out the indicated
multiplications, we find that only case (c) checks.Difference of cubes x3 a3 (x a)(x2 ax a2) Verify these formulas for yourself by carrying out
Sum of cubes x3 a3 (x a) (x2 ax a2) the multiplications. Then memorize the formulas.
Grouping This is actually an application of the common
factor technique.
(x 1)(x2 1)
x2(x 1) (x 1)# 1x3 x2 x 1 (x3 x2) (x 1)
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of Special Products formulas. Remember that it is the form or pattern in the f
that is important, not the specific choice of letters.
The idea in factoring is to use one or more of these techniques until each of
tors obtained is irreducible. Theexamples that followshow howthis works in p
B.4 Review of Factoring
EXAMPLE 1 Factoring Using the Difference of Squares Technique
Factor: (a) x2 49; (b) (2a 3b)2 49.
SOLUTION (a)
difference of squares
(b) Notice that the pattern or form is the same as in part (a): Its a differenc
squares. We have
difference of squ
(2a 3b 7)(2a 3b 7)
[(2a 3b) 7][(2a 3b) 7]
(2a 3b)2 49 (2a 3b)2 72
(x 7)(x 7)
x2 49 x2 72
EXAMPLE 2 Factoring Using Common Factor and Difference of Squares
Factor: (a) 2x3 50x; (b) 3x5 3x.
SOLUTION (a) common factor
difference of squares
(b)
3x(x 1)(x 1)(x2 1) difference of squares, again 3x(x2 1)(x2 1) difference of squares 3x[(x2)2 12]
3x5 3x 3x(x4 1) common factor
2x(x 5)(x 5)
2x3 50x 2x(x2 25)
EXAMPLE 3 Factoring by Trial and Error
Factor: (a) x2 4x 5; (b) (a b)2 4(a b) 5.
SOLUTION (a) x2 4x 5 (x 5)(x 1) trial and error
(b) Note that the form of this expression is
This is the sameform as the expression in part (a), so we need only replac
the quantity a b in the solution for part (a). This yields
(a b 5)(a b 1)
(a b)2 4(a b) 5 [(a b) 5][(a b) 1]
( )2 4( ) 5
EXAMPLE 4 Factoring by Trial and Error
Factor: 2z4 9z2 4.
SOLUTION We use trial and error to look for a factorization of the form (2z2 ?)(z2 ?)
are three possibilities:
(i) (2z2 2)(z2 2) (ii) (2z2 1)(z2 4) (iii) (2z2 4)(z2 1)
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B-20 Appendix B
Each of these yields the appropriate first term and last term, but (after checking) only
possibility (ii) yields 9z2 for the middle term. The required factorization is then
2z4 9z2 4 (2z2 1)(z2 4).
Question: Why didnt we consider any possibilities with subtraction signs in placeof addition signs?
EXAMPLE 5 Two Irreducible Expressions
Factor: (a) x2 9; (b) x2 2x 3.
SOLUTION (a) The expressionx2 9 is irreducible over the integers. (This can be discovered
by trial and error.) If the given expression had instead beenx2 9, then it could
have been factored as a difference of squares. Sums of squares, however, cannot
in general be factored over the integers.
(b) The expression x2 2x 3 is irreducible over the integers. (Check this for
yourself by trial and error.)
EXAMPLE 6 Factoring Using Grouping and a Special Product
Factor:x2 y2 10x 25.
SOLUTION Familiarity with the special product (a b)2 a2 2ab b2 suggests that we try
grouping the terms this way:
Then we have
difference of squares
(xy 5)(xy 5)
[(x 5) y][(x 5) y]
(x 5)2 y2x2 y2 10x 25 (x2 10x 25) y2
(x2 10x 25) y2
EXAMPLE 7 Factoring Using Grouping
Factor: ax ay2 bx by2.
SOLUTION We factor a from the first two terms and b from the second two to obtain
We now recognize the quantityxy2 as a common expression that can be factored
out. We then have
The required factorization is therefore
Alternatively,
(a b)(xy2)
x(a b) y2(a b)
(ax bx) (ay2 by2)
ax ay2 bx by2
ax ay2 bx by2 (x y2)(a b)
a(xy2) b(xy2) (xy2)(a b)
ax ay2 bx by2 a(xy2) b(xy2)
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B.4 Review of Factoring
EXAMPLE 8 Factoring Sum and Difference of Cubes
Factor: (a) t3 125; (b) 8 (a 2)3.
SOLUTION (a)
difference of cubes
(b)
sum of c
As you can check now, the expression a2 6a 12 is irreducible over the in
Therefore the required factorization is a(a2 6a 12).
For some calculations (particularly in calculus) its helpful to be able to fa
expression involving fractional exponents. For instance, suppose that we wan
tor the expression
The common expression to factor out here is (2x 1)12; the technique is to
the expression with the smaller exponent.
x(2x 1)12 (2x 1)32
a(a2 6a 12)
a(4 2a 4 a2 4a 4)
[2 (a 2)][22 2(a 2) (a 2)2]
8 (a 2)3 23 (a 2)3
(t 5)(t2 5t 25)
t3 125 t3 53
EXAMPLE 9 Factoring with Rational Exponents
Factor:x(2x 1)12 (2x 1)32.
SOLUTION
We conclude this section with examples of four common errors to avoid in
ing. The first two errors in the box that follows are easy to detect; simply multiplythe supposed factorizations shows that they do not check. The third error ma
from a lack of familiarity with the basic factoring techniques listed on page B-
fourth error indicates a misunderstanding of what is required in factoring; the fin
tity in a factorization must be expressed as aproduct, not a sum, of terms or expr
(2x 1)12(4x2 3x 1)
(2x 1)12[x (2x 1)2]
x(2x 1)12 (2x 1)32 (2x 1)12[x (2x 1)3212]
Error Correction Comment
x2 6x 9 (x 3)2 x2 6x 9 (x 3)2 Check the middle term.
x2 64 (x 8)(x 8) x2 64 is irreducible A sum of squares is, in general, irreducible ove
over the integers. the integers. (A difference of squares, howeveralways be factored.)
x3 64 is irreducible x3 64 (x 4)(x2 4x 16) Although a sum of squares is irreducible, a sum
over the integers of cubes can be factored.
x2 2x 3 factors as x2 2x 3 is irreducible over The polynomialx2 2x 3 is thesum of the e
x(x 2) 3 the integers. sionx(x 2) and the constant 3. By definition,
ever, the factored form of a polynomial must b
product(of two or more nonconstant polynom
Errors to Avoid
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B-22 Appendix B
EXERCISE SET B.4
AIn Exercises 166, factor each polynomial or expression. If a
polynomial is irreducible, state this. (In Exercises 16 the fac-
toring techniques are specified.)
1. (Common factor and difference of squares)
(a) x2 64 (c) 121z z3
(b) 7x4 14x2 (d) a2b2 c2
2. (Common factor and difference of squares)
(a) 1 t4 (c) u2v2 225
(b) x6 x5 x4 (d) 81x4 x2
3. (Trial and error)
(a) x2 2x 3 (c) x2 2x 3
(b) x2 2x 3 (d) x2 2x 3
4. (Trial and error)
(a) 2x2 7x 4 (c) 2x2 7x 4
(b) 2x2 7x 4 (d) 2x2 7x 4
5. (Sum and difference of cubes)
(a) x3 1 (c) 1000 8x6
(b) x3 216 (d) 64a3x3 125
6. (Grouping)
(a) x4 2x3 3x 6
(b) a2x bx a2z bz
7. (a) 144 x2 8. (a) 4a2b2 9c2
(b) 144 x2 (b) 4a2b2 9c2
(c) 144 (y 3)2 (c) 4a2b2 9(ab c)2
9. (a) h3 h5
(b) 100h3 h5
(c) 100(h 1)3 (h 1)5
10. (a) x4 x2
(b) 3x4 48x2
(c) 3(x h)4 48(x h)2
11. (a) x2 13x 40 12. (a) x2 10x 16
(b) x2 13x 40 (b) x2 10x 16
13. (a) x2 5x 36 14. (a) x2 x 6
(b) x2 13x 36 (b) x2 x 6
15. (a) 3x2 22x 16 16. (a) x2 4x 1
(b) 3x2 x 16 (b) x2 4x 3
17. (a) 6x2 13x 5 18. (a) 16x2 18x 9
(b) 6x2 x 5 (b) 16x2 143x 9
19. (a) t4 2t2 1 20. (a) t4 9t2 20
(b) t4 2t2 1 (b) t4 19t2 20
(c) t4 2t2 1 (c) t4 19t2 20
21. (a) 4x3 20x2 25x 22. (a) x3 x2 x
(b) 4x3 20x2 25x (b) x3 2x2 x
23. (a) ab bc a2 ac
(b) (u v)xxy (u v)2 (u v)y
24. (a) 3(x 5)3 2(x 5)2
(b) a(x 5)3 b(x 5)2
25. x2z2 xztxyz yt 26. a2t2 b2t2 cb2 ca2
27. a4 4a2b2c2 4b4c4 28. A2 B2 16A 64
29. A2 B2 30. x2 64
31. x3 64 32. 27 (a b)3
33. (xy)3 y3 34. (a b)3 8c3
35. x3 y3 xy 36. 8a3 27b3 2a 3b
37. (a) p4 1 38. (a) p4 4
(b) p8 1 (b) p4 439. x3 3x2 3x 1 40. 1 6x 12x2 8x3
41. x2 16y2 42. 4u2 25v2
43. 44.
45. 46.
47. 48.
49. xy y2 50. x2 x 1
51. 64(x a)3 x a 52. 64(x a)4 x a
53. x2 a2 y2 2xy 54. a4 (b c)4
55. 21x3 82x2 39x
56. x3a2 8y3a2 4x3b2 32y3b2
57. 12xy 25 4x2 9y2
58. ax2 (1 ab)xy by2
59. ax2 (a b)x b
60. (5a2 11a 10)2 (4a2 15a 6)2
61. (x 1)12 (x 1)32
62. (x2 1)32 (x2 1)72
63. (x 1)12 (x 1)32
64. (x2 1)23 (x2 1)53
65. (2x 3)12 (2x 3)32
66. (ax b)12
In Exercises 67 and 68, evaluate the given expressions usingfactoring techniques. (The point here is to do as little actual
arithmetic as possible.)
67. (a) 1002 992 68. (a) 103 93
(b) 83 63 (b) 502 492
(c) 10002 9992(c)
BIn Exercises 6973, factor each expression.
69. A3 B3 3AB(A B)
70. p3 q3 p(p2 q2) q(p q)2
71. 2x(a2
x2
)12
x3
(a2
x2
)32
72. (x a)12(x a)12 (x a)12(x a)32
73. y4 (p q)y3 (p2q pq2)y p2q2
74. (a) Factorx4 2x2y2 y4.
(b) Factorx4 x2y2 y4. Hint: Add and subtract a
term. [Keep part (a) in mind.]
(c) Factorx6 y6 as a difference of squares.
(d) Factorx6 y6 as a difference of cubes. [Use the result
in part (b) to obtain the same answer as in part (c).]
1
2
1
2
153 103
152 102
1ax bb13
14x
2
x3
8
512
x3125
m3n3 1
(a b)2
4
a2b2
9z4
81
16
81
4y2
25
16 c2
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But if you do use a rule involving
mechanical calculation, be patient,
accurate, and systematically neat in
the working. It is well known to
mathematical teachers that quite half the
failures in algebraic exercises arise from
arithmetical inaccuracy and slovenly
arrangement. George Chrystal (1893)
B.5 Review of Fractional Expressions
REVIEW OF FRACTIONAL EXPRESSIONS
The rules of basic arithmetic that you learned for working with simple fracti
also used for fractions involving algebraic expressions. In algebra, as in arit
we say that a fraction is reduced to lowest terms or simplifiedwhen the num
and denominator contain no common factors (other than 1 and 1). The fa
techniques that we reviewed in Appendix B.4 are used to reduce fractions.
ample, to reduce the fraction we write
In the box that follows, we review two properties of negatives and fractio
will be useful in this section. Notice that Property 2 follows from Property 1 b
a b
b a
a b
(a b) 1
x2 9
x2 3x
(x 3)(x 3)
x(x 3)
x 3
x
x2 9
x2 3x,
B.5
PROPERTY SUMMARY Negatives and Fractions
Property Example
1. b a (a b) 2 x (x 2)
2.2x 5
5 2x 1
a b
b a 1
EXAMPLE 1 Using Property 2 to Simplify a Fractional Expression
Simplify:4 3x
15x 20.
SOLUTION
using Property 2
In Example 2 we display two more instances in which factoring is used to
a fraction. After that, Example 3 indicates how these skills are used to multip
divide fractional expressions.
1
5a 4 3x
3x 4b 1
5
4 3x
15x 20
4 3x
5(3x 4)
EXAMPLE 2 Using Factoring to Simplify a Fractional Expression
Simplify: (a) (b)x2 6x 8
a(x 2) b(x 2).
x3 8
x2 2x;
SOLUTION (a)
(b)x2 6x 8
a(x 2) b(x 2)
(x 2)(x 4)
(x 2)(a b)
x 4
a b
x3 8
x2 2x
(x 2)(x2 2x 4)
x(x 2)
x2 2x 4
x
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B-24 Appendix B
EXAMPLE 3 Using Factoring to Simplify Products and Quotients
Carry out the indicated operations and simplify:
(a) (b)
(c) x2
144x2 4
x 12x 2
.
2x2 x 6
x2 x 1#x3 1
4 x2;
x3
2x2 3x#12x 18
4x2 6x;
SOLUTION (a)
(b)
using the fact
(c)
As in arithmetic, to combine two fractions by addition or subtraction, the fractions
must have a common denominator, that is, the denominators must be the same. The
rules in this case are
For example, we have
Fractions with unlike denominators are added or subtracted by first converting to
a common denominator. For instance, to add 9a and 10a2, we write
Notice that the common denominator used was a2. This is the least commondenominator. In fact, other common denominators (such as a3 or a4) could be used
here, but that would be less efficient. In general, the least common denominator for a
given group of fractions is chosen as follows. Write down a product involving the
irreducible factors from each denominator. The power of each factor should be equal to
(but not greater than) the highest power of that factor appearing in any of the individ-
ual denominators. For example, the least common denominator for the two fractions
9a
a2
10
a2
9a 10
a2
9
a
10
a2
9
a#a
a
10
a2
4x 1
x 1
2x 1
x 1
4x 1 (2x 1)
x 1
4x 1 2x 1
x 1
2x
x 1
a
b
c
b
a c
b
and a
b
c
b
a c
b
(x 12)(x 12)
(x 2)(x 2)#x 2
x 12
x 12
x 2
x2 144
x2 4
x 12
x 2
x2 144
x2 4#x 2
x 12
thatx 22 x
1
2x2 x 3
2 x
(2x 3)(x 1)
2 x
2x2 x 6
x2 x 1#x3 1
4 x2
(2x 3)(x 2)
(x2 x 1)#(x 1)(x2 x 1)
(2 x)(2 x)
6x3
2x2(2x 3)
3x
2x 3
x3
2x2 3x#12x 18
4x2 6x
x3
x(2x 3)#
6(2x 3)
2x(2x 3)
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and is (x 1)2(x 2). In the example that follows
that the denominators must be in factored form before the least common denom
can be determined.
1
(x 1)(x 2)
1
(x 1)2
B.5 Review of Fractional Expressions
EXAMPLE 4 Using the Least Common Denominator
Combine into a single fraction and simplify:
(a) (b) (c)15
x2 x 6
x
2
x
x2 9
1
x 3;
3
4x
7x
10y2 2;
SOLUTION (a) Denominators: 22 #x; 2# 5#y2; 1
Least common denominator; 22 # 5xy2 20xy2
Note: The final numerator is irreducible over the integers.
(b) Denominators:x2 9 (x 3)(x 3);x 3Least common denominator: (x 3)(x 3)
(c)
Notice in the last line that we were able to simplify the answer by factor
numerator and reducing the fraction. (This is why we prefer to leave thcommon denominator in factored form, rather than multiplying it out, in th
of problem.)
The next four examples illustrate using the least common denomin
simplify compound fractions, or complex fractions, which are fractions
fractions.
(x 2)(x 6)
(x 2)(x 3)
x 6
x 3
x 6
x 3
15 (x2 4x 3)
(x 2)(x 3)
x2 4x 12
(x 2)(x 3)
15
(x 2)(x 3)
x 1
x 2#x 3
x 3
using the fact tha
2 x (x 2
15
x2 x 6
x 1
2 x
15
(x 2)(x 3)
x 1
x 2
x (x 3)
(x 3)(x 3)
3
(x 3)(x 3)
x
(x 3)(x 3)
1
x 3#x 3
x 3
x
x2 9
1
x 3
x
(x 3)(x 3)
1
x 3
15y2
14x2
40xy2
20xy2
3
4x
7x
10y2
2
1
3
4x#5y2
5y2
7x
10y2#2x
2x
2
1#20xy2
20xy2
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B-26 Appendix B
EXAMPLE 5 Simplifying a Compound Fraction
Simplify:
1
3a
1
4b
5
6a2
1
b
.
SOLUTION The least common denominator for the four individual denominators 3a, 4b, 6a2, and
b is 12a2b. Multiplying the given expression by 12a2b12a2b, which equals 1, yields
12a2b
12a2b#
1
3a
1
4b
5
6a2
1
b
4ab 3a2
10b 12a2
EXAMPLE 6 Simplifying a Fraction Containing Negative Exponents
Simplify: (x1 y1)1. (The answer is not xy.)
SOLUTION After applying the definition of negative exponent to rewrite the given expression,
well use the method shown in Example 5.
multiplying by 1 xy
xy
xy
xy#
1
1
x
1
y
xy
y x
(x1 y1)1 a 1x
1
yb1 1
1
x
1
y
EXAMPLE 7 Simplifying a Compound Fraction
Simplify: (This type of expression occurs in calculus.)
1
x h
1
x
h.
SOLUTION
multiplying by 1
x (x h)
(x h)xh
h
(x h)xh
1
x(x h)
(x h)x
(x h)x
1
x h
1
x
h
(x h)x
(x h)x#
1
x h
1
x
h
EXAMPLE 8 Simplifying a Compound Fraction
Simplify:
x1
x2
1
x2 1
.
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B.5 Review of Fractional Expressions
SOLUTION
multiplying by 1
(x 1)(x2 x 1)
(x 1)(x 1)
x2 x 1
x 1
x3 1
1 x2
x3 1
x2 1
x2
x2
x1
x2
1
x2 1
x2
x2#x
1
x2
1
x2 1
EXERCISE SET B.5
AIn Exercises 112, reduce the fractions to lowest terms.
1. 2.
3. 4.
5. 6.
7. 8.
9. 10.
11.
12.
In Exercises 1355, carry out the indicated operations and
simplify where possible.
13. 14.
15.
16.
17.
18.
19. 20.
21. 22.1
a
1
b
1
c
6
a
a
6
1
3x
1
5x2
1
30x34
x
2
x2
a2 a 42
a4 216a
a2 49
a3 6a2 36a
x3 y3
x2
4xy 3y2
(xy)3
x2
2xy 3y2
(3t2 4txx2) 3t2 2txx2
t2 x2
x2 x 2
x2 x 12#
x2 3x
x2 4x 4
ax 3
2a 1
a2x2 3ax
4a2 1
2
x 2#x2 4
x 2
x4 y4
(x4y x2y3 x3y2 xy4)(xy)2
x3 y3
(xy)3
(xy)2(a b)
(x2 y2)(a2 2ab b2)
a3 a2 a 1
a2 1
a3b2 27b5
(ab 3b2)29ab 12b2
6a2 8ab
a b
ax2 bx2x2 2x 4
x3 8
x2
x 202x2 7x 4
x 2x4 16
25 x2
x 5
x2 9
x 3
23. 24.
25. 26.
27.
28.
29. 30.
31.
32.
33.
34.
35.
36.
37. 38. 39.
40. 41. 42.
43. 44. 45.
a
x2
a2 a
3
x2 h
3
x2
h
1
2 h
1
2
h
1x2
1y
1
x
1
y
a 1a
1 1
a
1a
1b
1
a
1
b
1
x
1
a
x a
4
a a
2
a 1
1
x 1
1
x 1
1
x 1
1
x2 1
1
x3 1
2q p
2p2 9pq 5q2
p q
p2 5pq
4
6x2 5x 4
1
3x2 4x
1
2x 1
1
x2 x 20
1
x2 8x 16
3
2x 2
5
x2 1
1
x 1
a2 b2
a2 b2
a
a b
b
b a
x
x a
a
a x
4
x 5
4
5 x
a2 5a 4
a2 16
2a
2a2 8a
ax 1
2ax(x 1)2
3ax
2
(x 1)3
1 1
x
1
x23x
x 2
6
x2 4
4
x 4
4
x 1
1
x 3
3
x 2
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B-28 Appendix B
C
55.
56. Consider the three fractions
(a) If a 1, b 2, and c 3, find the sum of the three
fractions. Also compute their product. What do you
observe?
(b) Show that the sum and the product of the three given
fractions are, in fact, always equal.
b c
1 bc , c a
1 ca , and a b
1 ab
a a 1bba a a 1
bbb
a b 1aba a b 1
abb
46. 47. (x1 2)1 48.
49. 50. (a1 a2)1
51. x(xy)1 y(xy)1
52. x(2x 2y)1 y(2y 2x)1
B
53. 54.1x a1x a
1x a1x a
a b
a b
a b
a b
a b
a b
a b
a b
#ab3 a3b
a2 b2
a 1a1
1
a2b1
(x2 2x)1
x2
xxy
y x
y2
x2 y2 1
Todays familiar plus and minus signswere first used in 15th-century Germany
as warehouse marks. They indicated
when a container held something that
weighed over or under a certain
standard weight. Martin Gardner in
Mathematical Games (Scientific
American, June 1977)
PROPERTIES OF THE REAL NUMBERS
I do not like as a symbol for multiplication, as it is easily confounded with x; . . .
often I simply relate two quantities by an interposed dot. . . . G. W. Leibniz in a letter
dated July 29, 1698
In this section we will first list the basic properties for the real number system. Then
we will use those properties to prove the familiar rules of algebra for working with
signed numbers.
The set of real numbers is closed with respect to the operations of addition
and multiplication. This means that when we add or multiply two real numbers, the
result (that is, the sum or the product) is again a real number. Some of the other most
basic properties and definitions for the real-number system are listed in the following
box. In the box, the lowercase letters a, b, and c denote arbitrary real numbers.
B.6
PROPERTY SUMMARY Some Fundamental Properties of the Real NumbersCommutative properties a b b a ab ba
Associative properties a (b c) (a b) c a(bc) (ab)c
Identity properties 1. There is a unique real number 0 (called zero or the additive identity) such that
2. There is a unique real number 1 (called one or themultiplicative identity) such that
Inverse properties 1. For each real number a there is a real number a (called the additive inverse of a
or the opposite of a) such that
2. For each real number a 0 there is a real number denoted by (or 1a or a1)
and called the multiplicative inverse or reciprocal of a, such that
Distributive properties a(b c) ab ac (b c)a ba ca
a#1
a 1 and 1
a# a 1
1
a
a (a) 0 and (a) a 0a# 1 a and 1# a a
a 0 a and 0 a a
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On reading this list of properties for the first time, many students ask the
question, Why do we even bother to list such obvious properties? One re
that all the other laws of arithmetic and algebra (including the not-so-obviou
can be derived from our rather short list. For example, the rule 0 # a 0
proved by using the distributive property, as can the rule that the product of tw
ative numbers is a positive number. We now state and prove those properties a
eral others.
B.6 Properties of the Real Numbers
Theorem
Let a and b be real numbers. Then
(a) a 0 0 (c) (a) a (e) (a)(b) ab
(b) a (1)a (d) a(b) ab
#
Proof of Part (a)
Now since a 0 is a real number, it has an additive inverse, (a 0). Adding
both sides of the last equation, we obtain
Thus , as we wished to show.
Proof of Part (b)
Now, by adding
a to both sides of this last equation, we obtain
This last equation asserts that a (1)a, as we wished to show.
a (1)a additive identity propertya 0 (1)a additive inverse propertyassociative property of addition
a (a a) (1)a additive identity property anda 0 a [a (1)a]multiplicative identity property a (1)a
distributive property 1# a (1)a
additive inverse property [1 (1)]a
of multiplication
using part (a) and the commutative0 0# a
a# 0 00 a
#0 additive identity prop
0 a# 0 0 additive inverse prope
addition
a# 0 [(a# 0)] a# 0 5a# 0 [(a# 0)]6 associative property oa# 0 [(a# 0)] (a# 0 a# 0) [(a# 0)]
##
a# 0 a# 0 a# 0 distributive property
a# 0 a# (0 0) additive identity property
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B-30 Appendix B
Proof of Part (c)
By adding a to both sides of this last equation, we obtain
This last equation states that (a) a, as we wished to show.
Proof of Part (d)
Thus a(b) ab, as we wished to show.
Proof of Part (e)
Weve now shown that (a)(b) ab, as required.
ab commutative property of multiplication
ba using part (c)
[(ba)] using part (d)
[b(a)] commutative property of multiplication
(a)(b) [(a)b] using part (d)
(ab) using part (b)
(1)(ab) associative property of multiplication
[(1)a]b commutative property of multiplication
[a(1)]b associative property of multiplication
a(b) a[(1)b] using part (b)
(a) a additive identity property(a) 0 a additive inverse property and additive identity property
(a) (a a) a associative property of addition
[(a) (a)] a 0 a(a) (a) 0 additive inverse property