©
: Pre-Calculus - Chapter 1A
Chapter 1A - Real Numbers
Properties of Real Numbers
Real numbers are used in almost every human endeavor. Whenever we need to quantify objects, we use
numbers. Cooking recipes, prices, interest rates, blood pressure, height, age, voltage, and wind velocity
are a few of the everyday objects that are quantified by real numbers. As we know, the two operations
of addition  and multiplication  are defined for real numbers. In other words, for any two real
numbers a and b, the sum a  b and the product a  b are uniquely defined real numbers. Two special
real numbers are zero ( 0 ) and one ( 1 ). These operations satisfy:
Properties of real numbers:
Commutative:
ab  ba
ab  ba
Example:
7  3  3  7  10
5  6  6  5  30
Associative:
a  b  c  a  b  c
abc  abc
Example:
3  4  7  3  4  7
2  5  3  2  5  3
3  11  7  7  14
2  15  10  3  30
Identity:
a0  0a  a
a1  1a  a
Example:
80  08  8
11  1  1  11  11
For each real number a, there is a real number, denoted by −a, called the negative of a, for which
Inverse:
a  −a  0  −a  a
Example: 7  −7  0
Subtraction, denoted by a − b, is defined as follows:
a − b  a  −b
−1
For each a ≠ 0, there is a real number, denoted by 1
a or 1/a or a , called the reciprocal of a, for which
1
Inverse:
a 1
a 1 a a
Example: 7  1  1  7  1
7
7
Division, denoted by a  b or a or a/b, where b ≠ 0, is defined as follows:
b
a  b  a  a/b  a 1b   ab −1
b
Finally, there is a property which relates addition and multiplication:
Distributive:
ab  c  ab  ac
Example:
−4  3  2  −4  3  −4  2
−4  5  −12  −8  −20
©
: Pre-Calculus
a  bc  ac  bc
©
: Pre-Calculus - Chapter 1A
Types of Real Numbers
Integers
The real numbers are classified into several categories. The positive integers, also called counting
numbers, are
1, 2, 3, 4, …
The negative integers,
… , −4, −3, −2, −1
are the negatives (or additive inverses) of the positive integers. An integer is either a positive integer, a
negative integer, or zero.
Rationals.
The ratio a of any two integers a and b, where b ≠ 0, is called a rational number. Common fractions
b
are rationals. If b  1 then the number is also an integer, so integers are also rational numbers.
Examples:
4 , 121 , − 16 , −7
5
31
54
Irrationals.
Real numbers that are not rational are called irrational. Examples of irrational numbers include , 2 ,
and so on. One distinction between rational numbers and irrational numbers is that rational numbers
have repeating or terminating decimal expansions, whereas irrational numbers do not.
Summary
The diagram below shows the relationship among these types of numbers, all of which are a part of the
set of real numbers: That is, whole numbers are a subset of the integers, which are a subset of the
rational numbers. The irrational numbers are disjoint from all of these and the real numbers are
comprised of the union of the set of irrational numbers and rational numbers.
Real Numbers
Irrational Numbers
l Numb
ers
tiona
Ra
ge
Inte rs
Whole Numbers
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1A
Number Lines and Absolute Value
A number line is a method of picturing the set of real numbers. Each point on the number line
corresponds to exactly one real number, as in the picture below:
P
-5
-4
-3
-3/2
O
-1
0
-2
3/2 5/2 π
1
2
3
Q
4
5
The absolute value of a number x, denoted by |x|, refers to the distance from that number to the origin,
or zero (point O in the picture above). Note that, since distance is always positive or zero, the absolute
value of a number will always be positive or zero. A more precise definition is as follows: if x is
positive or zero, then |x| will be the same as x, but if x is negative, we must change the sign to positive,
indicated by −x. Therefore,
|x| 
x
if x ≥ 0
−x if x  0
For instance, |−7|  7
|10|  10
|0|  0
|3 − 5|  |−2|  2.
Example: Evaluate |6 − 9|.
Solution: |6 − 9|  |−3|  3.
Notice that in this example, we did not simply change the subtraction to addition. The absolute value
bars act as parentheses, so what is inside must be evaluated first. Therefore, in general, it is incorrect to
say that |x − 3|  |x|  3, or |x − 3|  x  3!!
Also notice that |6 − 9| gave us 3, the distance between the numbers 6 and 9 on the number line. In
general, |a − b| gives us the distance between the numbers a and b on the number line.
Example: What is the distance between the numbers 2 and −24?
Solution: The distance between any two numbers is the absolute value of their difference. Thus,
distance between 2 and −24 equals |2 − −24|  |2  24|  |26|  26.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1A
Exercises for Chapter 1A - Real Numbers
For problems 1-6, determine which property of the real numbers is used.
1. −6x  6x  0
2. 17  x  x  17
3. 28  x − 7  28  x − 7
4. 14  x − 3  x − 3  14
5. 13x 1x   13
6. x  3 − y  −y  x  3
For problems 7-10, simplify and name the properties used.
7. x  7 − x
8. 4x − 3  12
9. 3x  4  4x
10. x  7x − 3
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
©
For problems 11-20, evaluate the given expression.
|−18 − 3|
|7 − 14|
−|15 − 4|
−2 − 4 − |−6|
−334 − 25 − |10|
−48  −37  48  38  999
|−24  −24|  75  −49 − |74|
21 − 5  1
12
4
3
2
2
2
2 −51
5
3
8
1
1  1
1  2
2
2
3
3
Choose , , or : −|2| ? |−2|
Describe the following in absolute-value notation: The distance between x and 3 is at most
2.
Describe in words: |y  2|  a (Hint: y  2  y − −2 )
Order the following list of numbers from smallest to largest:
|3 − 5|, |−3 − 5|, |−3  5|, − |3 − −5|, − |3 − 5|, |− |−3 − 5||
: Pre-Calculus
©
: Pre-Calculus - Chapter 1A
Answers to Exercises for Chapter 1A - Real Numbers
1.
2.
3.
4.
5.
6.
7.
Additive Inverse
Commutative Property of Addition
Associative Property of Addition
Commutative Property of Multiplication
Multiplicative Inverse and Multiplicative Identity (Since 13x 1x   131  13 )
Commutative Property of Addition (remember that x  3 − y  x  3  −y )
x  7 − x Commutative Property
 7  x − x Associative Property
 7  x − x Additive Inverse Property
 7  0 Identity for Addition
7
NOTE: While you can simply say that x  7 − x  7, the properties of real numbers help
explain why this simplifying is allowed.
8.
4x − 3  12 Distributive Property
 4x − 12  12 Associative Property
 4x  −12  12 Additive Inverse*
 4x  0 Identity for Addition
 4x
*-remember that 4x − 12  4x  −12
9.
3x  4  4x Commutative Property-last 2 terms
 3x  4x  4 Distributive Property (in reverse)
 3  4x  4
 7x  4
Again you could simplify immediately: 3x  4  4x  7x  4. The properties allow you to
see why this works.
10. Later we will learn another method (FOIL) for doing this. It is based on the distributive
property as shown below:
x  7x − 3
 x  7x  −3
 x  7x  x  7−3
 x 2  7x  x  7−3
 x 2  7x − 3x − 21 then simplify as #9
 x 2  4x − 21
11. 21
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1A
12. 7
13. −11
14. − 2 − 4 − |−6| (work parentheses from the inside out)
 −2 − 4 − 6
 −2 − −2
 −2  2  −4
15. − 334 − 25 − |10| (do multiplication before subtraction)
 −334 − 10 − 10
 −314  −42
16. − 48  −37  48  38  999 ( Here it may be best to rearrange the terms.)
 −48  48  38  −37  999
 0  1  999  1000
Note that the numbers may be added in any order (associative and commutative properties),
but the above is the most expediant way.
17. |−24  −24|  75  −49 − |74| (Again rearranging terms may help.)
 |−48|  75  −49 − 74
 48  −49  75 − 74
18.
19.
20.
21.
22.
23.
24.
©
 −1  1  0
2 1 − 5  1 Find a common denominator
12
3
4
3
5
2
−
 4
Write the mixed number as an improper fraction
12
12
12
 27 − 9  18  3 or 1 1
12
12
12
2
2
2 2 2 2 − 5 1 Start by writing improper fractions:
5
3
8
8 − 41 Common denominator inside absolute value
 12
5
3
8
64
12

− 123
5
24
24
59
12

cross cancel the 12 and the 24
5
24
 59 or 5 9 or 5. 9
10
10
1
Work inside the parentheses first
1  1
1  2
2
2
3
3
1

3  2
3  4
6
6
6
6
1

5
7
6
6
1

 36
35
35
36
−|2|  |−2| Since −|2|  −2 and |−2|  2
|x − 3| ≤ 2
y is more than a units from −2
− |3 − −5|  −8
− |3 − 5|  −2
|3 − 5|  |−3  5|  2
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Chapter 1B - Exponents and Radicals
Properties of Exponents
Exponents can be used to represent repeated multiplication of a single factor. More formally, if n
represents a positive integer, the expression x n means multiply x times itself n times.
If n  0, then we define x n  x 0  1, and for n  0 we define
xn  x  x  x    x
n times
Examples:
 4 3  4  4  4  64
2
 −15  −15  −15  225
 Note that −15 2  −15 2   −225 since exponents are computed before multipliying (unless
there are parentheses as in the previous example)
 y3  y  y  y
 What does 5 4 equal? Answer: 5 4  5  5  5  5  625
We define x to a negative power as follows (we assume that x ≠ 0)
x −1  1x
x −2  12
x
−n
x  1n
x
Examples:
1  1
 4 −3 
64
43
1
−5
 2
 5  1
32
2
The following list of properties of exponentiation should be memorized, as you need to be able to
manipulate expressions involving exponents.
Property
1
2
3
4
5
6
7
Example
a m  a n  a mn
a m  a m−n
an
a −n  1n
a
2 3  2 4  2 34  2 7
x 4  x 4−2  x 2
x2
3 −4  14  1
81
3
0
2 0
a  1 for a ≠ 0 14 − x   1 iff 14 − x 2  ≠ 0
ab n  a n b n
a n  an
b
bn
2y 3  2 3 y 3  8y 3
5 2  5 2  25
x
x2
x2
a m  n  a mn
x 3  −4  x 3−4  x −12  112
x
Although these properties work for any real exponents m and n, the properties are easy to see when m
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
and n are integers.
"Proof" of Example 1(from previous page):
23  24
 2  2  2  2  2  2  2
 2222222
 27
We can use these properties to simplify expressions with exponents in them.
Example 1: Simplify the expression −z 3 3z 4 .
Solution:
−z 3 3z 4  −1 3 z 3 3 4 z 4
 −181z 34
 −81z 7
Example 2: Simplify the expression
x −3 y 4
5y −1
−3
x −3 y 4
5y −1
−3
.
Solution:
x 9 y −12
5 −3 y 3
9 3
 x3 512
y y
9
125x

15
y

Example 3: Simplify the expression 15 4 5 −6 .
Solution:
15 4 5 −6  3  5 4 5 −6
 3 4 5 4 5 −6
 3 4 5 4 5 −6 
 3 4 5 4−6
One could have stopped at 3 4 5 −2
 3 4 5 −2
4
 32
5
and not bothered about rewriting 5 −2 as 1/5 2 .
Example 4: Simplify the expression 5 3 4 −2 10 2 7 3 .
Solution: Before starting, scan the expression to find common terms. Here we note that 4  2  2, and
10  5  2. This means that we should think to combine the common integer factors of the terms 5 3 ,
10 2 , and 4 −2 .
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
5 3 4 −2 10 2 7 3  5 3 2 2  −2 5  2 2 7 3
 5 3 2 −4 5 2 2 2 7 3
 5 32 2 2−4 7 3
 5 5 2 −2 7 3
5 3
 5 72 .
2
5 3
We could have stopped at the line 5 5 2 −2 7 3 , and not rewritten this as 5 72 .
2
If we have a complicated expression involving integers raised to a power, then the easiest first step
would be to write the integers as a product of their prime factors. Remember that a positive integer
(greater than 1) is said to be prime if its only integer divisors are itself and 1. Thus, 2, 3, 7, 23, and 31
are prime, while 4, 9, 28, and 32 are not.
Example 5: The following example is fairly complicated. If you can follow all of the steps and
understand the reasoning, you are well on your way to understanding and being able to use the
exponential properties.
198 3 700 −4
Simplify the expression
5 4 6 −3
Solution: As we noted at the end of Example 4, when simplifying a complicated expression involving
integers, factor each integer into its prime factors.
2  3 2 11 3 2 2 5 2 7 −4
198 3 700 −4

5 4 6 −3
5 4 2  3 −3
2 3 3 6 11 3 2 −8 5 −8 7 −4 

5 4 2 −3 3 −3
 2 3 3 6 11 3 2 −8 5 −8 7 −4 5 −4 2 3 3 3 
 2 3−83 3 63 5 −8−4 7 −4 11 3
 2 −2 3 9 5 −12 7 −4 11 3
9
3
 32 11
12 4
2 5 7
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Radicals and Properties of Radicals
Radicals (or roots) are, in effect, the opposite of exponents. In other words, the n th root of a number a
is a number b such that
b
n
a  a 1/n  b n  a
The number b is called an n th root of a. The number n is referred to as the index of the radical (if no
index appears, n is understood to be 2). The principal n th root of a number is the n th root of a which
has the same sign as a. For example both 2 and −2 satisfy x 2  4, but 2 is the (principal) square root of
4.
Examples:
 3 27  3 since 3 3  27
4
 4 16  2 since 2 4  16 (Note −2  16 also, but 2 is the principal 4 th root
3
 3 −64  −4 since −4  −64
 4 −81 is not a real number and we will say that it does not exist. (In this course we won’t learn
how to take an eventh power of a negative number.)
Radicals are used to define rational exponents:
1
an 
m
a n 
n
n
a
am
The notation a 1/n is extremely useful, and we encourage you to use it whenever you have to simplify
expressions involving radicals.
Examples:
1
 125 3  3 125  5
2

−64 3 

32 − 5
3
−64 2 
1


3
32 5
3
2
−64
 −4 2  16
1
 13  1
3
8
2
5
32
3
Since radicals are nothing more than rational exponents, many of the properties of exponents also apply
to radicals.
Property
Example
1
n
am   n a 
m
2
n
a  nb 
ab
n
3
n
4
m
a
b
n

a 
n
n
a ,b ≠ 0
b
mn
5
32 3 
5
32
27  3 
3
a
27 
8
4
2x 
3
8
3
 23  8
81  9
27
 3
3
2
8
2x
5a If n is odd n a n  a
3
−127 3  −127
5b If n is even n a n  |a|
4
−127 4  |−127|  127
The following list is a restatement of these properties, but in exponential notation. You need to be
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
familiar with both radical and exponential notation, and be able to convert between the two.
Property
Example
3
32 3  1/5  32 1/5   2 3  8
 a m/n
1
a 
2
a 1/n b 1/n  ab 1/n
3
a 1/n 
b 1/n
4
a 1/m 
m 1/n
1/n
a
b
1/n
,b ≠ 0
 a 1/mn
5a If n is odd a n  1/n  a
5b If n is even a n  1/n  |a|
27 1/2 3 1/2  27  3 1/2  81 1/2  9
27
8
1/3
2x 1/4
1/3
 271/3  3
2
8
1/2
 2x 1/8
−127 3
1/3
 −127
−127 4
1/4
 |−127|  127
Examples:





©
−3 2  3 (refer to Property 5b)
3
16 3/2  16 3/2  16 1/2
 4 3  64 (refer to property 1-given the right hand side)
3
1 3
3
3/2
(refer to property 1) 
−16
−16  −16 2  −16 2
There is no answer as we cannot take the square root of −16.
32 1/5 27 1/3  32 1/5 27 1/3  2  3  6
8
x 8  |x|
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Simplifying Radicals
Properties 2 and 3 in the previous section can be used to simplify radical expressions.
A radical expression is simplified when the following conditions hold:
1. All possible factors (”perfect roots”) have been removed from the radical.
2. The index of the radical is as small as possible. Remember, the index of 3 10 is 3.
3. No radicals appear in the denominator.
For example, to simplify the radical 40 , we factor the number into prime factors. Since the index is
2, any square factors are ”pulled out” of the radical:
40 
23  5

22  2  5

2 2  2  5 (Using Property 2)
 2  10 , or 2 10 .
Notes for the above example:



1
1 1
Recall that Property 2 states that n ab  n a n b or, in exponential notation ab n  a n b n
Since 2 3  2 2  2 , the perfect square is pulled out of the radical, while the remaining 2 stays
inside the radical.
The previous example could also be done directly by finding a factor of 40 which is a perfect
square:
40 

4  10
4  10 (by Property 2 again-see previous note)
 2 10
This method works quite well for square roots.
Example1: Simplify 72x 3
Solution:
72x 3  3 2  2 2  2  x 2  x
 3 2 2 2 x 2 2x
 3  2  |x| 2  x
 6|x| 2x
Example 2: Simplify
Solution:
3
3
54xy 4
54xy 4 
3
2  33  x  y4
 3y 32xy
 3y 3 2xy
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Example 3: Simplify
4
80x 2
y4
Solution:
4
80x 2 
y4
4
24  5  x2
y4
 2 4 5x 2 (since 4 y 4  |y| )
|y|
Finally, we can use the fact that radicals can be written as fractional exponents to be sure the radical is
in ”lowest terms”.
Example 4: Simplify 4 9
4
Solution:
9  4 32
2
 34
1
 32
 3
Example 5: Simplify 6 81x 4
6
Solution:
81x 4  6 3 4 x 4
4
4
 36x6
2
2
 33x3
 3 32x2
 3 9x 2
These last two examples demonstrated the utility of exponential notation.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Rationalizing Denominators
Rationalizing the Denominator is a technique used whenever a radical appears in the denominator of an
expression. To begin with, make sure the radical is simplified; that is, perfect roots are pulled out. The
next step depends on how the radical appears in the denominator:
I. The Radical is a single term
Multiply both the numerator and the denominator of the expression by something which will produce a
perfect root in the denominator.
Example 1: Rationalize the denominator: 8
3
54
Solution: First simplify the radical in the denominator:
8
8 
 8 (Recall that 3 3 3  3)
3
3
3
54
332
23
Note that if the 2 was a cube, we would no longer need a radical in the denominator.
To rationalize this denominator, we multiply both the numerator and denominator by 3 2 2 as follows:
8 3 22
834
434
22



32
3
3
22
3 3 23
3
8 
332
Question:
Answer:
Why do we pick 3 2 2 ?
We want to multiply the denominator of
8 by a term which will enable us to remove
332
any radicals. Thus, we need to figure out what to multiply 3 2 by. Well, if we write this in exponential
notation, we have
3
2  2 1/3 .
Clearly if we multiply by 2 2/3 we get something nice.
2 1/3 2 2/3  2 1  2 .
Moreover 2 2/3 
3
22 .
5
10
Example 2: Rationalize the denominator:
Solution:
5 
10
10
5 10

10
10
10

2
Example 3: Rationalize the denominator:
3
Solution:
1

2
3 5x
5x

3
5x
3

3
3
3
5x
5x 3
1
5x 2
(note that you only need one more 5x )
5x
5x
II. The Denominator is a Sum of Terms
In this example the denominator has the form a  b m (here we only concern ourselves with square
roots). The conjugate of this expression is a − b m . Similarly, the conjugate of the expression
a  b is a − b . In the example on the next page, notice what happens when we multiply the
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
numerator and denominator by the conjugate:
5 5− 7
5− 7
5


5 7
5− 7


5 7
5− 7
25 − 5 7
52 −
2
7
25 − 5 7
.
18
Here the multiplication in the denominator is done using a distributive property technique called FOIL
(shown below). You can also use the special product a  ba − b  a 2 − b 2 which we will discuss
in detail in another section.
5 7
5− 7
 5  5  5 − 7
First
5 7 
Outer
Inner
 52 − 5 7  5 7 −
 52 −
2
7
7
7
− 7
Last
2
Note how the inner terms cancel.
 25 − 7  18
Question: What should you multiply 5 − 3 by to rationalize it?
Answer:
Multiply 5 − 3 by 5  3 .
5− 3
5 3
 25 − 9  16.
Remember
a − ba  b  a 2 − b 2 .
Question:
Answer:
What should you multiply a  b c by to rationalize it?
Multiply a  b c by a − b c .
a  b c a − b c   a 2 − b 2  c 
 a 2 − b 2 |c| .
Example 4:
Solution:
2x
5− 3
2x 5  3
Rationalize the denominator:
5 3
2x


5 3
5− 3
52 −
3
2x 5  3
22
x 5 3

11

©
: Pre-Calculus
2
2
©
: Pre-Calculus - Chapter 1B
Example 5:
4
5  6
Rationalize the denominator:
Solution:
4

5  6
©
5 − 6

5 − 6
: Pre-Calculus
4
5
5 − 6
2
−
6
2

4
5 − 6
5−6
 −4
5 − 6
 4 6 −4 5
©
: Pre-Calculus - Chapter 1B
Combining Radical Expressions
Radical expressions can only be combined by addition or subtraction if there are like terms-terms with
the same index and same radicand (the expression inside the radical).
2 2 3 4 2 −4 3 ,
In the above expression, the terms with 2 are alike and the terms with 3 are alike. Therefore, the
above expressions simplifies as follows:
2 2 3 4 2 −4 3 
2 4 2 2 3 −4 3
 1  4 2  2 − 4 3
 5 2 −2 3
Notice that the two remaining terms are not alike and, hence, cannot be simplified.
Some terms which do not look alike at first glance may be alike after simplifying. Therefore, it is
important that you simplify all radicals before combining like terms.
Examples:

50 − 32  2  5 2  2 − 2 5  2
 5 2 −4 2  2
 5 − 4  1 2
2 2
3
3
 7 80x − 2 270x  4 3 10  7 3 2 4  5x − 2 3 3 3  2  5x  4 3 10
 7  2 3 2  5x − 2  3 3 2  5x  4 3 10
 14 3 10x − 6 3 10x  4 3 10
 8 3 10x  4 3 10
(note that these terms are NOT alike even though they both have a 10)
1 − 1 , we begin by using Property 3 ( n a  n a ), then simplifying and
 In simplifying
2
8
n
b
b
rationalizing the denominator before combining:
1 − 1  1 − 1 (Note that 8  4  2)
8
2
2
8
2
 1 − 1 (Multiply both by
)
2
2
2 2
1 2
1 2
−

2 2
2 2 2
2
2

−
(now get a common denominator)
2
4
2
2 2
−

4
4
2 2 − 2

(now combine radicals)
4
2

4
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Exercises for Chapter 1B - Exponents and Radicals
For problems 1-3, evaluate the expression.
−3 4
5 −3
2 −2
3.
5
1.
2.
4.
5.
6.
For problems 4-6, simplify the expression.
−3ab 4 4ab −3 
5a 3 a 2  −1 16a 10  0
−2
15x −2 y
5x 3 y −4
For problems 7-13, compute the following, if possible:
7. 3 −343
8. 4 81
16
9.
−49
10. Write 3 y 4 using exponents
11. Write
1
12. −64 3
125
13.
27
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
©
x
10
− 23
using radicals
− 23
For problems 14-18, simplify the expression.
48
3
54x 4
3
56
4 162x 3 y 6
10
32x 5
For problems 19-26, rationalize the denominator.
3
2
2
18
6
3
16
9
4
8x
2
1 5
3
5 3
: Pre-Calculus
©
25.
26.
: Pre-Calculus - Chapter 1B
4
7 − 3
5
4 − 11
For problems 27-31, simplify the expression.
27 − 5  48  2 45
4
48  5 4 2 − 3 4 3
75x 3 − 27x  3x 5
3 3
30. 2 
− 2
2
6
3
31. 45 − 6 125
32. 16x 3  3 16x 4 y  3 54x 4 y  x 25x
27.
28.
29.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
Answers to Exercises for Chapter 1B - Exponents and
Radicals
1.
2.
3.
4.
5.
6.
7.
8.
−3  −3  −3  −3  81
1  1
125
53
5 2  5 2  25
4
2
22
4
−3ab 4ab −3   −12a 11 b 4−3  −12a 2 b
5a 3 a 2  −1 16a 10  0  125a 3 12 1  125a
a
−2
−2
−2
4
15x y
3y  y
3y 5 −2



x5
5x 3 y −4
x3  x2
(Other methods are possible)
−7 , since −7 3  −343
4
81
 3
4
2
16
x5
3y 5
2
10
 x 10
9y
9. Not possible; there is no real number whose square is −49
4
10. y 3
11.
x
10
3
−2

10
x
3
2

3
100
x2
12. −4
13.
14.
15.
16.
17.
18.
19.
20.
21.
©
2
3
2
2
27
 3
 9
5
3
25
125
48  2 4  3  2 2 3  4 3
3
54x 4  3 2  3 3  x 4  3x 3 2x
3
56  3 2 3  7  2 3 7
4 162x 3 y 6  4 2  3 4  x 3  y 6  3y 4 2x 3 y 2 (Since y 6  y 4  y 2 )
27
125

3
5
5
1
1
32x 5  10 2 5 x 5  2 10 x 10  2 2 x 2  2x
3  2  3 2
2
2
2
2 
2
(simplify radical first)
18
2  32
2
 2 
3 2
2
2 2

32
2

(Note the cancellation of the 2’s)
3
6  6 (simplify radical first)
3
3
16
24
 6 (OK to cancel first)
232
3
22
 3 
3
3
2
22
10
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
3 3 22
3
23
334

2
9 
9
(factor first to see what you need)
4
3
4
8x
2 x
4
2x 3
9


(to get 4th powers you need one 2 and 3 x’s)
4
4
23  x
2x 3
9 4 2x 3

4
24x4
9 4 2x 3

2x
2 1− 5
1− 5
2


2
1− 5
1 5
12 −
5

22.
23.
21 − 5 
−4
1− 5
−
2
3 5 −3
5 −3

2
5 −3
5
− 32

24.
3

5 3
−3 5 − 9
3 5 −9
or
−4
4
7

3
4
7  3

2
2
7  3
7
−
3

25.
4

7 − 3
7  3
4
 7  3 (since the 4’s cancel)
5 4  11
4 5  55
5
4  11



2
5
2
4 − 11
4  11
4 −
11

26.
4
27 − 5  48  2 45  3 2  3 − 5  2 4  3  2 3 2  5
 3 3 − 5 4 3 6 5
 3  4 3  −1  6 5
 7 3 5 5
4
4
4
4
4
28. 48  5 2 − 3 3  2  3  5 4 2 − 3 4 3
 243 542 −343
 5 4 2  2 − 3 4 3
 542 − 43
29. 75x 3 − 27x  3x 5  5 2  3  x 3 − 3 3  x  3x 5
 5x 3x − 3 3x  x 2 3x
 x 2  5x − 3 3x
30. (rationalize the denominators first)
2  6  3 3  2 − 2  3  2 6  3 6 − 2 3
6
2
3
6
2
2
3
3
6
2 6
9 6
4 3


−
6
6
6
27.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1B
2 6 9 6 −4 3
6
11 6 − 4 3

6
3
31. 45 − 6 125  3 2  5 − 6 5 3  3 5 − 5 6  3 5 − 5  2 5
32. 16x 3  3 16x 4 y  3 54x 4 y  x 25x  4x x  2x 3 2xy  3x 3 2xy  5x x  9x x  5x
3 2xy

©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Chapter 1C - Polynomials
Definition of a Polynomial
A polynomial is the most common algebraic expression. Polynomials are expressions which contain
terms of the form ax k , where a is a real constant and k is a nonnegative integer. More formally, a
polynomial expression is an expression of the form
a n x n  a n−1 x n−1    a 1 x  a 0
In the above expression, a n is a nonzero real number, a 0 , a 1 , ... , a n−1 are real numbers, and n is a
positive integer. n is called the degree of the polynomial, a n is called the leading coefficient, and a 0 is
called the constant term.
If n  2 the polynomial is called a binomial or quadratic.
If n  3 the polynomial is called a trinomial or cubic.
Example The height, s (in feet), of a ball thrown in the air at time t  0 seconds is given by the
polynomial equation s  −16t 2  80t  5 . Here the degree of the polynomial is 2 (the highest
exponent), the leading coefficient is −16, and the constant is 5.
Example The volume, V, of a cube with edge length x is given by V  x 3 . Here the degree of the
polynomial is 3, the leading coefficient is 1 (remember that x 3  1x 3 ), and the constant is 0.
Polynomial expressions do not contain negative exponents nor do they contain radicals. So an
expression such as 9 − x 2 is not a polynomial.
Question: Is x 3  1x a polynomial?
Answer: No, the 1x term is the same as x −1 and the exponent is not a positive integer.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Sums and Differences of Polynomials
Addition and Subtraction of Polynomials is done by combining like terms, that is, terms which have
the same variable and exponent.
Example 1: Simplify the expression: 15x 2 − 6  8x 3 − 14x 2  17
Solution:
15x 2 − 6  8x 3 −14x 2 17
 8x 3  15 − 14x 2  −6  17
 8x 3  x 2  11
To subtract polynomials, recall that subtraction is defined as adding the opposite (inverse); that is,
a − b  a  −b . It is important that you distribute the negative sign over the entire polynomial.
Example 2: Simplify the expression: 15x 2 − 6 − 8x 3 − 14x 2  17
Solution:
15x 2 − 6 − 8x 3 − 14x 2  17
 15x 2 − 6  −8x 3  14x 2 − 17 (NOT −8x 3 − 14x 2  17 !!)
 15x 2 −6 − 8x 3 14x 2 − 17
 −8x 3  29x 2 − 23
Example 3: Simplify the expression: 3x 3 − 2x 2  8  3x 4  2x 2 − 5
Solution:
3x 3 − 2x 2  8  3x 4  2x 2 − 5
 3x 3 − 2x 2  8  3x 4  2x 2 − 5
 3x 4  3x 3  −2  2x 2  8 − 5
 3x 4  3x 3  3
Example 4: Simplify the expression: −5x 2 − 1 − −3x 2  5
Solution:
− 5x 2 − 1 − −3x 2  5 (distribute negative signs)
 −5x 2  1  3x 2 − 5
 −2x 2 − 4
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Products of Polynomials
Mulitplication of Polynomials is based on the Distributive Property of Real Numbers as illustrated in
the example below:
Example 1: Expand the following product x − 68x  7.
Solution:
x − 68x  7  x8x  7 − 68x  7 (treat 8x  7 as a single number first)
 x8x  x7 − 68x − 67
 8x 2  7x − 48x − 42
 8x 2 − 41x − 42
Notice, in the underlined step, that the distributive properties follow the familiar pattern FOIL, which is
explained below.
x8x  x7 − 68x − 67
First
Outer Inner Last
terms terms terms
terms
Although FOIL only works with binomials, the distributive property method works on any
polynomials.
Example 2: Expand 5x − 6x 3 − 4x 2  2.
Solution 1
5x − 6x 3 − 4x 2  2  5xx 3 − 4x 2  2 − 6x 3 − 4x 2  2
 5x 4 − 20x 3  10x − 6x 3  24x 2 − 12
 5x 4 − 26x 3  24x 2  10x − 12
(Don’t forget to distribute the negative sign as well!!!)
Solution 2: The distributive method is easy to organize and picture using a rectangle, or box, whose
lengths are the two factors as illustrated below:
−4x 2
x3
5x 5x
4
−20x
2
3
10x
−6 −6x 3 24x 2 −12
 5x 4 − 20x 3 − 6x 3  24x 2  10x − 12
 5x 4 − 26x 3  24x 2  10x − 12
Example 3: Expand x 2  9x 2 − 6x  9.
Solution:
x 2  9x 2 − 6x  9  x 2 x 2 − 6x  9  9x 2 − 6x  9
 x 4 − 6x 3  9x 2  9x 2 − 54x  81
 x 4 − 6x 3  18x 2 − 54x  81
Example 4: Expand a  ba − b.
Solution:
a  ba − b  aa − b  ba − b
 a 2 − ab  ba − b 2
 a 2 − ab  ab − b 2 (By the commutative property)
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
 a2 − b2
The previous example a  ba − b  a 2 − b 2 is an example of a Special Product. Special products
can not only make multiplication of certain polynomials easier, they are also useful when you want to
factor a polynomial; that is, given a polynomial, find two or more polynomials whose product is the
given polynomial. The most common special products are listed below (here a and b represent any
numbers, variables, or algebraic expressions).
a  ba − b  a 2 − b 2

a  b 2  a 2  2ab  b 2
a − b 2  a 2 − 2ab  b 2
 Sum of Cubes:
a  ba 2 − ab  b 2   a 3  b 3
 Difference of Cubes:
a − ba 2  ab  b 2   a 3 − b 3
 Cube of a Binomial:
a  b 3  a 3  3a 2 b  3ab 2  b 3
a − b 3  a 3 − 3a 2 b  3ab 2 − b 3
NOTE that there is no special product for the sum of squares that involves only real numbers..

Difference of Squares:
Square of a Binomial:
Historical Note: The French mathematician Blaise Pascal (1623-1662) discovered an easy way to
remember the coefficients of a binomial power using what is now called Pascal’s Triangle. After the
first two rows, each succesive row is obtained by adding the two numbers immediately above it as
shown below:
1
a  b 0  1
1 1
a  b 1  1a  1b
1 2 1
a  b 2  1a 2  2ab  1b 2
1 3 3 1 etc. a  b 3  1a 3  3a 2 b  3ab 2  1b 3
Using this triangle, we can see that
a  b 4  1a 4  4a 3 b  6a 2 b 2  4ab 3  1b 4
The numbers in Pascal’s Triangle are also useful in Probability.
Example 5: Expand
Solution:
x − 2 3 .
x − 2 3  x 3 − 3x 2 2  3x2 2  − 2 3
 x 3 − 6x 2  12x − 8
Example 6: Expand x  2yx − 2y 2 .
Solution:
x  2yx − 2y 2 



©
: Pre-Calculus
2
x 2 − 2y 2 (work inside the parenthesis first)
x 2 − 4y 2  2
x 2  2 − 2x 2 4y 2   4y 2  2
x 4 − 8x 2 y 2  16y 4
©
: Pre-Calculus - Chapter 1C
Quotients of Polynomials
Division of Polynomials is done using standard long division techniques.
Example 1: Review basic long division for numbers. Use long division to find 1350  18.
Solution: The algorithm for long division is:
  Divide
 Multiply
 Subtract
 Bring Down.
 Repeat these steps as long as it is possible to divide.
Note that 18  7  126, so 18 divides 135 seven times:
7
18
(divide)
1
3 5 0
1
2 6 (multiply)
9 (subtract)
Now bring down the 0 and repeat, noting that 18  5  90:
18
7
5
0
1
3
5
1
2
6
9
0
9
0
0
So 1350  18  75.
Example 2: Now use long division to divide polynomials: 8x 2  13x − 6  x  2
2
Solution:
First, divide the highest terms. 8xx  8x. Then follow the steps discussed in the above
example.
8x
−3
x  2 8x 2 13x −6
8x 2 16x
−3x −6
−3x −6
0
so 8x 2  13x − 6  x  2  8x − 3
Another way of writing this is to say that 8x 2  13x − 6  x  28x − 3.
Example 3: Perform the following operation: x 4 − 3x 2 − 4  x 2  x
Solution:
We begin by writing out the long division, using 0’s for any missing powers of x:
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
x 2  x x 4 0x 3 −3x 2 0x −4
4
We need to determine how many times x 2 will divide into x 4 . Since x 2  x 2 , the first term of our
x
quotient is x 2 :
Divide x 2
x 2  x x 4 0x 3 −3x 2 0x −4
Now multiply and subtract:
x2
x 2  x x 4 0x 3 −3x 2 0x −4
Multiply x 4
x 3
Subtract
−x 3
Bring down the next term:
x2
x 2  x x 4 0x 3 −3x 2 0x −4
x 3
x4
−x 3 −3x 2
Bring down
The next term of the quotient will be −x2  −x, so we repeat the process:
x
3
−x
x2
Divide
x 2  x x 4 0x 3 −3x 2 0x −4
x 3
x4
−x 3 −3x 2
x2
x x x
2
4
x4
−x
0x 3 −3x 2 0x −4
x 3
−x 3 −3x 2
−x 3
Multiply
x2
x x x
2
4
x4
−x 2
−x
0x 3 −3x 2 0x −4
x 3
−x 3 −3x 2
−x 3
Subtract
−x 2
−2x 2 0x Bring down
2
The next term of our quotient will be −2x2  −2, so bring down the next term and repeat:
x
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
−x
x2
x x x
2
0x
4
−2
−3x
3
2
Divide
0x −4
x 3
x4
−x 3 −3x 2
−x 3
−x 2
−2x 2 0x
x2
x x x
2
4
x4
−x
0x
3
−2
−3x 2 0x −4
x 3
−x 3 −3x 2
−x 3
−x 2
−2x 2 0x
−2x 2 −2x
Multiply
x2
−x
−2
x 2  x x 4 0x 3 −3x 2 0x −4
x4
x 3
−x 3 −3x 2
−x 3
−x 2
−2x 2 0x
−2x 2 −2x
Subtract
2x −4 Bring down
Since x 2 does not evenly divide 2x, the long division is complete.
We can say that x 4 − 3x 2 − 4  x 2  x  x 2 − x − 2 with a remainder of 2x − 4. We could also say
either of the following:
x 4 − 3x 2 − 4  x 2 − x − 2  2x − 4
x2  x
x2  x
or:
x 4 − 3x 2 − 4  x 2  xx 2 − x − 2  2x − 4
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Factoring
The process of factoring is a type of ”reverse-multiplication”, where you are given a polynomial and
have to write it as a product of factors. A polynomial is completely factored when each factor is prime,
or cannot be factored again. The idea is very similar to factoring numbers:
60  5  12 is one factoring of 60
 562
 5232
 5  2 2  3 is the prime factorization of 60
There are several strategies to determine how to factor polynomials.
I. Common Factors
A common factor is a factor of every term of an expression. Common factors can be pulled out of an
expression using the distributive property in reverse:
ab  ac  ab  c
Examples:
 x 3  4x  xx 2  4
 −2x 2  6x  −2xx − 3 (Note the negative in the second term)
 6mn 2  15m 2 n − 30m 3 n 3  3mn2n  5m − 10m 2 n 2 
Finding common factors should always be the first step in factoring an expression.
II. Factoring by Grouping
Factoring by Grouping is especially useful when you have more than three terms in the polynomial.
The technique of factoring by grouping is really using common factors creatively, as shown in the
following example:
Example 1:
Factor 2x 3 − x 2  6x − 3
Solution:
Although there is no factor common to all terms (except 1, which we ignore), we can
”group” the polynomial by twos, each of which have a common factor:
 2x 3 − x 2  6x − 3
 x 2 2x − 1  32x − 1
Now notice that there are two terms, each of which has the factor 2x − 1.
 2x − 1x 2  3
It is very easy to check your answer by mulitplying it out:
2x − 1x 2  3
 2xx 2  3 − 1x 2  3
 2x 3  6x − x 2 − 3
 2x 3 − x 2  6x − 3 ✓
III. Factoring Using Special Products
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
The special products we learned in the section ”Products of Polynomials” can be used to help factor
expressions which have an appropriate form:
Example 2:
Solution:
Factor completely x 3 − 9x
As mentioned in the previous section, the first thing we do is look for a common factor:
x 3 − 9x  xx 2 − 9
Notice how the second factor can now be written as x 2 − 3 2 , a difference of two squares. Recall that
a 2 − b 2  a  ba − b.
x 3 − 9x  xx 2 − 9
 xx  3x − 3
The expression is now completely factored.
Example 3:
Factor completely x 3  8y 3
Solution:
Notice how the expression can be rewritten as x 3  2y 3 , a sum of two cubes.
Recall that a  ba 2 − ab  b 2   a 3  b 3
x 3  8y 3
 x  2y x 2  x2y  2y 2
 x  2yx 2  2xy  4y 2 
Example 4: Factor completely x 2  25
Solution:
Recall that there is no special product for a sum of squares a 2  b 2 . In fact, this
expression cannot be factored; it is prime over the real numbers.
IV. Factoring Binomials (x 2  bx  c )
If possible, a binomial of this form must factor as x  mx  n , where m and n are integers. Note
that if this is true:
x  mx  n  x 2  bx  c
x 2  nx  mx  mn  x 2  bx  c
x 2  m  nx  mn  x 2  bx  c
We can see in the last statement that the numbers m and n that we seek must add up to b and multiply to
c.
Example 5:
Factor completely x 2  7x  12
Solution:
We are looking for two numbers which will add up to 7 and multiply to 12. The
numbers (obtained by trial and error) are 3 and 4. Therefore,
x 2  7x  12  x  3x  4
Example 6:
©
Factor completely x 2 − 4x − 21
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Solution:
We are looking for two numbers which will add up to −3 and multiply to −21. Note
that since the product is negative, the numbers must be different signs (one negative and one positive).
The numbers are −7 and 3 (Note that if we had chosen 7 and −3, the numbers add up to 4. In general,
if the signs are different, the larger number will have the same sign as b, the middle term). Therefore,
x 2 − 3x − 21  x − 7x  3
Example 7:
Factor completely x 2 − 6x  9
Solution:
We are looking for two numbers which will add up to −6 and multiply to 9. Note that
since the product is positive, the numbers must be the same sign (both positive or both negative). Since
the middle term is negative, the numbers must both be negative. The numbers are −3 and −3.
Therefore
x 2 − 6x  9  x − 3x − 3  x − 3 2
Note that we could have factored this using the special product a − b 2  a 2 − 2ab  b 2
V. Factoring Trinomials (ax 2  bx  c )
If possible, a trinomial of this form (where the leading coefficient is not 1) must factor in the form
mx  snx  t , where m, n, s, and t are all integers. It is possible to find these numbers by trial and
error; however, such a method may at times be haphazard or tedious. The following is a more
straightforward approach utilizing the technique of factoring by grouping.
First, we need to split the middle term bx into two terms which will allow us to factor by grouping. To
do this, we use a similar technique as before, only now we look for two numbers whose sum is b and
whose product is ac .
If you are curious as to why, see below:
As before, multiply the desired form out:
mx  snx  t  ax 2  bx  c
mnx 2  mtx  nsx  st  ax 2  bx  c
mnx 2  mt  nsx  st  ax 2  bx  c
Now note that b  mt  ns and ac  mnst  mtns . The numbers we are
looking for are mt and ns. How they split up into m, n, s, and t are determined when we
factor by grouping.
Example 8:
Factor completely 9x 2 − 3x − 2
Solution:
We first must find two numbers whose sum is −3 and whose product is
9−2  −18. The numbers are −6 and 3. Now we rewrite −3x as −6x  3x and factor by grouping:
9x 2 − 3x − 2
 9x 2 − 6x 3x − 2
 3x3x − 2  13x − 2
 3x  13x − 2
(remember that you can check your answer by multiplying)
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Example 9:
Factor completely 6x 2 − x − 15
Solution:
We must first find two numbers whose sum is −1 and whose product is
6−15  −90. The numbers are 9 and −10. Now we rewrite −x as 9x − 10x and factor by grouping:
6x 2 − x − 15
 6x 2  9x −10x − 15
 3x2x  3 − 52x  3
(Notice that both signs change in the last term)
 3x − 52x  3
If you are afraid of getting the middle terms in the wrong order, not to worry...
6x 2 − x − 15
 6x 2 − 10x  9x − 15
 2x3x − 5  33x − 5
 2x  33x − 5
the same as above.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Exercises for Chapter 1C - Polynomials
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
©
Determine which of the following are polynomials:
4x 3 − 8x 2  17x − 12 , −6x  6x −1 , −24 , 3 − x 2 , 16 − x 2
For the polynomials in #1, give the degree, leading coefficient, and constant term
For problems 3-6, simplify the expression.
−5x 7  2x 4  3x 2 − 2  −9x 2  4
2x 3 − 5x 2  1  −3x 3  7x − 6 − 3x 3  6x 2 − 7x
y 3  1 − y 3 − 1  y 2 − 2y  1
y 3  1 − y 3 − 1 − y 2 − 2y  1
Find a polynomial which, when added to x 3  3x 2 − 3x  2, yields 2x 2  4x  7
For problems 8-14, simplify the expression.
3x − 25x  7
−9x 2  4x − 5
3x − 2x 2 − 2x  2
4x  5 2
3x  7 2 3x − 7
2x  1x − 3  5x − 3
2x − 1 3
An open box made from a 9-inch square piece of material by cutting equal squares from all
corners and folding the sides. The volume of the box is given by x9 − 2x 2 . Give the
degree, leading coefficient, and constant term of this polynomial.
For problems 16-23, divide by long division.
x 2 − 6x  10  2x  3
x 3 − 1  x − 1
x4 − 1
x−1
x5 − 1
x−1
n
If n is a positive integer, what do you think x − 1 equals ?
x−1
x2  1
x−1
x 3 − 6x 2  12x − 8
x−2
x 4 − 2x 3  4x 2 − 2x − 7
x 2 − 2x  3
For problems 24-38, factor each expression as completely as possible.
6x 3 − 4x
x − 22x − 3x − 2
y 3 − 27z 3
2x 3  16
x 4 − 2x 3 − 8x  16
a 3 − 27a
: Pre-Calculus
©
30.
31.
32.
33.
34.
35.
36.
37.
38.
©
: Pre-Calculus - Chapter 1C
4x 3  4x 2 − 9x − 9
x 2  11x  10
y 2 − 5y  6
x2  x − 2
x 2  8x − 20
y 3 − 2y 2 − 3y
2x 2  x − 15
3x 2  10x  8
15x 2 − 11x  2
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Answers to Exercises for Chapter 1C - Polynomials
1.
2.
3.
4.
5.
6.
7.
The polynomials are 4x 3 − 8x 2  17x − 12 , −24, and 3 − x 2
− 6x  6x −1 is not a polynomial because of the negative exponent
16 − x 2 is not a polynomial because of the radical.
4x 3 − 8x 2  17x − 12 : degree 3, leading coefficient 4, and constant term −12
−24 : degree 0, leading coefficient −24, and constant term −24
3 − x 2  −x 2  3 : degree 2, leading coefficient −1, and constant term 3
−5x 7  2x 4  3x 2 − 2  −9x 2  4  −5x 7  2x 4 3x 2 −2−9x 2 4
 −5x 7  2x 4 − 6x 2  2
2x 3 − 5x 2  1  −3x 3  7x − 6 − 3x 3  6x 2 − 7x
 2x 3 − 5x 2  1 − 3x 3  7x − 6 − 3x 3 − 6x 2  7x
 2 − 3 − 3x 3  −5 − 6x 2  7  7x  1 − 6
 −4x 3 − 11x 2  14x − 5
y 3  1 − y 3 − 1  y 2 − 2y  1  y 3  1 − y 3 − 1 − y 2 − 2y  1
 y 3  1 − y 3  1 − y 2  2y − 1
 1 − 1y 3 − y 2  2y  1  1 − 1
 −y 2  2y  1
y 3  1 − y 3 − 1 − y 2 − 2y  1  y 3  1 − y 3 − 1  y 2 − 2y  1
 y 3  1 − y 3  1  y 2  2y  1
 1 − 1y 3  y 2  2y  1  1  1
 y 2  2y  3
Recall that if a  b  c, then a  c − b, so the polynomial we seek is
2x 2  4x  7 − x 3  3x 2 − 3x  2  2x 2  4x  7 − x 3 − 3x 2  3x − 2
 −x 3  2 − 3x 2  4  3x  7 − 2
 −x 3 − x 2  7x  5
3x − 25x  7
8.
−2
3x
5x 15x
−10x
2
7 21x
−14
 15x 2  21x − 10x − 14
 15x 2  11x − 14
−9x 2  4x − 5
9.
−5
x
−9x
4
2
−9x
4x
3
−45x 2 2
−20
 −9x 3 − 45x 2  4x − 20
3x − 2x 2 − 2x  2
10.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
−2x
x2
3x 3x
−6x
3
2
2
−2 −2x 2 4x
6x
−4
 3x 3 − 6x 2 − 2x 2  6x  4x − 4
 3x 3 − 8x 2  10x − 4
4x  5 2  4x  54x  5
11.
5
4x
4x 16x
2
5 20x
20x
25
 16x 2  20x  20x  25
 16x 2  40x  25
12. We could square the first term using a  b 2  a 2  2ab  b 2 , but notice the following:
3x  7 2 3x − 7
 3x  73x  73x − 7
Now use difference of squares on the term in brackets:
 3x  79x 2 − 49
9x 2
−49
3x 27x 3 −147x
7 63x 2 −343
 27x 3  63x 2 − 147x − 343
13. We will use FOIL here because of the multiple operations:
2x  1x − 3  5x − 3
 2xx  2x−3  1x  1−3  5x − 15
 2x 2 − 6x  x − 3  5x − 15
 2x 2 − 18
14.
2x − 1 3
 2x − 1 2 2x − 1
 4x 2 − 4x  12x − 1
2x
4x 2 8x 3
−1
−4x 2
−4x −8x 2 4x
1
2x
−1
 8x 3 − 12x 2  6x − 1
15. Multiply out the expression:
x9 − 2x 2  x18 − 36x  4x 2   4x 3 − 36x 2  18x
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
Therefore, the degree of the polynomial is 3, the leading coefficient is 4, and the constant
term is 0.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
x 2 − 6x  10  1 x − 15 
85
4
2x  3
2
42x  3
x3 − 1  x2  x  1
x−1
x4 − 1  x3  x2  x  1
x−1
x5 − 1  x4  x3  x2  x  1
x−1
x n − 1  x n−1  x n−2    x  1
x−1
x2  1  1  x  2
x−1
x−1
x 3 − 6x 2  12x − 8  x 2 − 4x  4
x−2
x 4 − 2x 3  4x 2 − 2x − 7  x 2  1 −
10
x 2 − 2x  3
x 2 − 2x  3
6x 3 − 4x  2x3x 2 − 2
x − 22x − 3x − 2  x − 22x − 3
NOTE: you could multiply out the expression first and then factor the resulting trinomial.
Use difference of cubes:
y 3 − 27z 3  y 3 − 3z 3  y − 3zy 2  3yz  9z 2 
27. Factor the common factor, then use sum of cubes:
2x 3  16  2x 3  8  2x  2x 2 − 2x  4
28. Factor by grouping, then use difference of cubes:
x 4 − 2x 3 − 8x  16
 x 3 x − 2 − 8x − 2
 x 3 − 8x − 2
 x − 2x 2  2x  4x − 2
 x − 2 2 x 2  2x  4
29. a 3 − 27a  aa 2 − 27 We cannot factor this any further (using only integers).
30. Factor by grouping, then use difference of squares:
4x 3  4x 2 − 9x − 9
 4x 2 x  1 − 9x  1
 4x 2 − 9x  1
 2x  32x − 3x  1
31. We need two numbers whose product is 10 and whose sum is 11; the numbers are 10 and 1:
x  10x  1
32. We need two numbers whose product is 6 and whose sum is −5; the numbers are −2 and −3
y − 2y − 3
33. We need two numbers whose product is −2 and whose sum is 1; the numbers are −1 and 2:
x  2x − 1
34. We need two numbers whose product is −20 and whose sum is 8; the numbers are 10 and
−2:
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1C
x  10x − 2
35. Common factor first:
y 3 − 2y 2 − 3y
 yy 2 − 2y − 3
We need two numbers whose product is −3 and whose sum is −2; the numbers are −3 and 1:
 yy − 3y  1
36. We need two numbers whose product is 2−15  −30 and whose sum is 1; the numbers are
6 and -5:
 2x 2  6x − 5 − 15
 2xx  3 − 5x  3
 2x − 5x  3
37. We need two numbers whose product is 38  24 and whose sum is 10; the numbers are 6
and 4:
 3x 2  6x  4x  8
 3xx  2  4x  2
 3x  4x  2
38. We need two numbers whose product is 152  30 and whose sum is −11; the numbers are
−5 and −6:
 15x 2 − 5x − 6x  2
 5x3x − 1 − 23x − 1
 5x − 23x − 1
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
Chapter 1D - Rational Expressions
Definition of a Rational Expression
A rational expression is the quotient of two polynomials. (Recall: A function px is a polynomial in x
of degree n, if there are constants a 0 , a 1 , , a n , with a n ≠ 0 such that px  a 0  a 1 x    a n x n .)
p
More formally, a rational expression is an expression of the form q , where p and q are polynomials,
and qx cannot be the zero polynomial. The denominator of a rational expression can be a constant
polynomial though. For example, rx  x is a polynomial of degree 1, and it is also a rational
expression, for rx  x  x .
1
3
4x
−
7
Example 1:
, 2 , x − 28 , and x2  5 are examples of rational expressions.
6x − 16 x − 9
2
x  25
Example 2:
x 2  12
is not a rational expression.
x7
A rational expression involves a quotient, and since division by 0 is not defined for real numbers, there
are sometimes restrictions on the variable x. In particular, the restrictions occur wherever the
denominator is zero.
Example 3:
For the rational expression 4x − 7 , the values of x which make the denominator zero
6x − 16
can be found by solving the equation
6x − 16  0
6x  16
x  16  8
6
3
8
Therefore, the variable x cannot equal
in the rational expression 4x − 7 . We also say that the
3
6x − 16
domain of this expression is all x not equal to 8 .
3
Example 4:
For the rational expression 2 , the values of x which make the denominator zero
x−9
can be found by solving the equation
x−9  0
x9
Therefore, the variable x cannot equal 9 in the rational expression 2 . The domain of this
x−9
expression is all x not equal to 9.
For the rational expression x2  5 , there are no values of x which make the
x  25
denominator zero since x 2  25 is always positive. Therefore, there are no restrictions on the variable x.
The domain of this expression is all x.
Example 5:
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
2
For the rational expression x 2 − 9 , the values of x which make the denominator zero
x −1
can be found by solving the equation
x2 − 1  0
Example 6:
x2  1
x  1
2
Therefore, the variable x cannot equal 1 in the rational expression x 2 − 9 . The domain of this
x −1
expression is all x not equal to 1.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
Simplifying Rational Expressions
To simplify a rational expression means to reduce it to lowest terms. From working with fractions, you
may recall that simplifying is done by cancelling common factors. Therefore, the key to simplifying
rational expressions (and to most problems involving rational expressions) is to factor the polynomials
whenever possible.
Example 1:
Solution:
cancel the x’s:
4
Write 8x 6 in reduced form.
12x
Factor the numbers and cancel common factors. Use properties of exponents to help
8x 4  4  2  x 4
12x 6
4  3  x4  x2
 22
3x
Example 2:
Write
y 3  5y 2  6y
in reduced form.
y2 − 4
Solution:
yy 2  5y  6
y 3  5y 2  6y

2
y −4
y2 − 4
yy  2y  3

y  2y − 2
yy  3
y 2  3y


; y ≠ 2
y−2
y−2
Question:
Why is y ≠ 2?
Answer:
In order for our answer to be equivalent to the original fraction, the variable must have
the same restrictions. Since y cannot equal 2 in the original expression (the denominator would then
be zero), we must restrict the domain of our answer in order for these fractions to be equivalent.
Example 3:
Write
4x − x 3 in reduced form.
x −x−2
2
Solution:
2
4x − x 3  x4 − x 
2
x −x−2
x −x−2
x2  x2 − x

x − 2x  1
2
At this point, it doesn’t look like anything will cancel. However, if we factor −1 from the last term in
the numerator, we obtain the following:
−x2  xx − 2

x − 2x  1
2
−xx  2

 −x − 2x ; x ≠ 2 or − 1
x1
x1
Question:
Answer:
©
What property of real numbers tells us that −x2  x  −xx  2 ?
The commutative property of addition.
x  2  2  x.
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
Operations with Rational Expressions
The operations of addition, subtraction, multiplication, and division with rational expressions follow
the same rules that are used with common fractions. The key, as with simplifying rational expressions,
is to factor the polynomials.
I. Multiplication of Rational Expressions
Recall with fractions a and c that a  c  ac . Also recall when mulitplying fractions that you
b
d
b d
bd
may cross-cancel; that is, reduce common factors of any numerator with any denominator.
Example 1:
2  3
9 10
 2  3
33 25
 1  1
3 5
 1
15
The same is true of multiplying rational expressions. As before, the key is to factor the polynomials
first.
2
Example 2:
Simplify 5  x − 1
x − 1 25x − 50
Solution:
First factor the polynomials. Don’t forget to cancel common factors!
5  x 2 − 1  5  x  1x − 1
x−1
x − 1 25x − 50
25x − 2
 1  x1
1 5x − 2
 x  1 ; x ≠ 1 or 2
5x − 10
Before looking at the next example, recall our strategies for factoring:
1. Factor out any common factors
2. If more than 3 terms, try factoring by grouping
3. Recognize special products
4. Factor trinomials using product/sum strategies:
a) x 2  bx  c: factors into x  mx  n where
mn  c and m  n  b
b) ax 2  bx  c: split bx into mx  nx, where
mn  ac and m  n  b, then factor by grouping.
2
3
2
 2x  x  5
Example 3:
Simplify 2x2  x − 6  x − 3x
2
x3 − 8
x  4x − 5
4x − 6x
Solution:
2x 2  x − 6  x 3 − 3x 2  2x  x  5
x3 − 8
x 2  4x − 5
4x 2 − 6x
2x − 3x  2 xx − 2x − 1
x5



2x2x − 3
x  5x − 1
x − 2x 2  2x  4
x2
x2


; x ≠ −5, 1, 0, 3 , 2
2
2x 2  2x  4
2x 2  4x  8
Example 4:
©
Simplify 2x2 − 4
x −4
: Pre-Calculus
©
Solution:
: Pre-Calculus - Chapter 1D
2x − 2
2x − 4 
 2 , x ≠ 2.
x2
x − 2x  2
x2 − 4
II. Division of Rational Expressions
a/b
Recall with fractions a and c that a  c 
 a  dc . In other words, dividing by a fraction
b
d
b
d
b
c/d
is the same as multiplying by its reciprocal. The same is true for rational expressions. As before, the
key to making the work easier is to factor the polynomials first.
x 2  2x − 15  x 2  7x  10
Example 5:
x2
2
Solution:
Begin by writing the second term as x  7x  10 :
1
x 2  2x − 15  x 2  7x  10  x 2  2x − 15 
1
x2
1
x2
x 2  7x  10
x  5x − 3
1


x2
x  5x  2
 x − 3 2  2 x − 3 ; x ≠ −5, − 2
x  4x  4
x  2
Notice that the x  2 terms cannot be cancelled since both are in the denominator.
Multiplication and Division can also appear together. Only take the reciprocal of the fractions you are
dividing.
y 2  3y − 28
3y  21
y 2 − 7y  12
Example 6:


2
2
4y  24
y  3y − 18
y  12y  36
Solution:
We only take the reciprocal of the second fraction:
y 2  3y − 28
3y  21
y 2 − 7y  12


2
2
4y  24
y  3y − 18
y  12y  36
y 2 − 7y  12 y 2  12y  36 3y  21
 2
 2

4y  24
y  3y − 18
y  3y − 28

3y  7
y − 3y − 4
y  6 2


y  6y − 3 y  7y − 4 4y  6
 3; y ≠ −7, − 6, 3, 4
Question:
Explain why the values −7, −6, 3, and 4 are excluded in the previous example.
Answer:
The easiest way to understand why these values have been excluded is to write
y 2  3y − 28
3y  21
y 2 − 7y  12
 2

as a fraction.
2
4y  24
y  3y − 18
y  12y  36
y 2 − 7y  12
y 2  3y − 18 3y  21
y 2  3y − 28 4y  24
y 2  12y  36
y ≠ −6 follows since 4y  24 cannot equal 0.
We also get y ≠ 3 because the term
y 2 − 7y  12
y 2 − 7y  12

y − 6y  3
y 2  3y − 18
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
The next observation is that the denominator of
y 2 − 7y  12
y 2  3y − 18
y 2  3y − 28
y 2  12y  36
y 2  3y − 28
that is, 2
cannot equal zero. Factoring the numerator and denominator of this rational
y  12y  36
expression we have
y 2  3y − 28  y  7y − 4
The numerator is zero if y  −7 or if y  4. This accounts for excluding the numbers −7 and 4. The
denominator of
y 2  3y − 28
y 2  12y  36
factors into y  6 2 which means we have to exclude −6, but we have already done that.
What values of x are not allowed in the rational expression x − 1  x − 3 ?
x−4
x−6
x
−
3
is zero at x  3 and we cannot
Answer:
3, 4, 6 (The value 3 is not allowed because
x−4
divide by 0. x  4 is not allowed because we have the term x − 4 in the denominator. Similarly x  6 is
not allowed because the term x − 6 is the denominator of x − 1 .)
x−6
Question:
III. Addition and Subtraction of Rational Expressions
Recall that addition and subtraction is done by first finding a common denominator. The least common
denominator (LCD) of several fractions is the product of all prime factors of the denominators. A
factor only occurs more than once in the LCD if it occurs more than once in any one fraction. Once
you have the LCD, convert each fraction to an equivalent fraction with the LCD, then add or subtract
the numerators.
1 − 1
Example 7:
x
x−1
Solution:
The LCD here is xx − 1. We convert each fraction to an equivalent one with this
x
denominator: That is, 1x  x − 1 and 1 
x−1
xx − 1
xx − 1
1x
1 − 1  1x − 1 −
x
x−1
xx − 1
x − 1x
x
 x−1 −
xx − 1
xx − 1
x − 1 − x
−1


xx − 1
xx − 1
Example 8:
Solution:
©
y−4
y 2  3y − 28
− 2
y6
y  12y  36
We must factor the denominators first to find the LCD:
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
y−4
y 2  3y − 28
−
y6
y  6 2
The LCD is y  6 2 . We now convert each fraction to an equivalent fraction using this denominator:
y − 4y  6
y−4

y6
y  6 2
y 2  3y − 28
y 2  3y − 28

2
y  6
y  6 2
Next subtract the two expressions
y − 4y  6
y 2  3y − 28
y 2  3y − 28
y−4
− 2

−
2
y6
y  12y  36
y  6 2
y  6
y 2  2y − 24
y 2  3y − 28

−
2
y  6
y  6 2
y 2  2y − 24 − y 2  3y − 28

y  6 2
y 2  2y − 24 − y 2 − 3y  28

y  6 2
−y  4

y  6 2
Note that the minus sign in the numerator was distributed across the expression
−y 2  3y − 28  −y 2 − 3y  28.
Question:
Answer:
©
What is the common denominator in
x − 7x  1x 2 − 4
: Pre-Calculus
x−5
− 21 ?
x − 7x  1
x −4
©
: Pre-Calculus - Chapter 1D
Compound Fractions
A compound fraction (also called a complex fraction) is an expression which contains nested fractions.
In general, there is one ”main fraction” which will have one or more fractions in the numerator and/or
denominator. There are two strategies which may be used to simplify compound fractions. First,
simplify both the numerator and the denominator individually, then divide the numerator by the
denominator by multiplying with the reciprocal of the denominator.
5
6 −
y
2y  1
Example 1:
6 4
y
Solution:
First simplify the numerator using the common denominator y2y  1:
Example
5y
62y  1
5
6 −
−
y
y2y  1
2y  1y
2y  1

6 4
6 4
y
y

12y  6 − 5y
y2y  1
6 4
y
7y  6
y2y  1

6 4
y
Now simplify the denominator using the common denominator y (remember that 4  4 :
1
7y

6
5
6 −
y
y2y  1
2y  1

6 4
6  4y
y
y
1y
7y  6
y2y  1

6  4y
y
7y  6
6  4y
We now divide the two fractions
and
y :
y2y  1
5
6 −
y
2y  1
7y  6
6  4y


y
6 4
y2y  1
y
7y  6
y


y2y  1 6  4y
7y  6

; y ≠ −3 , −1 , 0 .
2
2
2y  16  4y
As you can see, this is often a long, tedious process. A second technique is to multiply the
numerator and the denominator of the ”main fraction” by the LCD of all the smaller fractions and
simplify the result.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
5
6 −
y
2y  1
Example 2:
6 4
y
Solution:
The LCD of all the smaller fractions is y2y  1. We multiply the numerator and
denominator by this LCD:
5
5
6 −
6 −
y2y  1
y
y
2y  1
2y  1

6 4
6  4 y2y  1
y
y




6 y2y  1 −
5
y2y  1
y
2y  1
6 y2y  1  4y2y  1
y
62y  1 − 5y
62y  1  4y2y  1
12y  6 − 5y
2y  16  4y
7y  6
2y  16  4y
The following example is a common one in many calculus classes:
2x  h  1 −1 − 2x  1 −1
Example 3:
Simplify the expression
.
h
First rewrite the expression without negative exponents. (Recall x −1  1x )
2
− 2
2x  h  1 −1 − 2x  1 −1
x1
x

h

1

h
h
Method I Solution: Now simplify the numerator using the common denominator x  h  1x  1:
2x  h  1
2x  1
−
2x  h  1 −1 − 2x  1 −1
x  1x  h  1
x  h  1x  1

h
h
2x  2 − 2x  2h  2
x  h  1x  1

h
2x  2 − 2x − 2h − 2
x  h  1x  1

h
−2h
x  h  1x  1

h
h
The denominator is h 
which is already simplified. We now divide the two fractions
1
−2h
h
and :
1
x  h  1x  1
2x  h  1 −1 − 2x  1 −1
−2h

 1
h
x  h  1x  1 h
−2

; h≠0
x  h  1x  1
Solution:
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
Method II Solution: The LCD of all the smaller fractions is x  h  1x  1. We mulitply the
numerator and denominator by this LCD:
2
− 2
x  h  1x  1
x1
xh1
hx  h  1x  1




©
2
2
x  h  1x  1 −
x  h  1x  1
x1
xh1
hx  h  1x  1
2x  1 − 2x  h  1
hx  h  1x  1
2x  2 − 2x − 2h − 2
hx  h  1x  1
−2h
−2

; h≠0
hx  h  1x  1
x  h  1x  1
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
Exercises for Chapter 1D - Rational Expressions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
For problems 1-5, determine if the given expression is a rational expression. If it is a
rational expression, state any restrictions on the variable.
3x 2 − 6x  7
5x − 10
5x  10
x8
x 2  4x  6
5x  7
4
2
x−4
For problems 6-8, reduce the rational expression if possible.
x3y
x2y2
x 2  4x − 12
5x − 10
x 3 − 4x
x2  x − 2
For problems 9-20, perform the indicated operations and simplify.
x3  8
x3 − 8 
2
2
x − 4 x  2x  4
z 2 − 5z − 24  9z 2 − 12z  4  z 2 − z − 6
9z − 12
z2 − 9
z 2 − 10z  16
3  6x  9
x−1
x 2 − 9x  8
y 2  8y − 20
y3  1
y2 − y  1


y 2  11y  10 y 3 − 8
y 2  2y  4
3
y − 27
4y
2y − 6
 2
 2
3
y − 5y  6
y − 2y
2y
x 
2
x−3
3x  4
1 − 3
x3
x−2
1  1 − 10
x3
x−3
x2 − 9
7
8x
−
4
2x  1
2x − 1
3 − 2  x3
x
x−1
x2 − 1
2
y − 7y  12
y 2 − 25

(HINT: Simplify first)
y 2  3y − 18
y 2 − 8y  15
y−7
− 5
y1
y 2  2y  1
For problems 21-25, simplify the expression completely.
3x  y
4
21.
xy
2
©
: Pre-Calculus
©
22.
23.
24.
25.
©
: Pre-Calculus - Chapter 1D
5  x
x−1
x1
2 1
x2 − 1
1
1  1
a
b
1 − 1
x
2
x−2
x  h −2  x −2
h
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
Answers to Exercises for Chapter 1D - Rational
Expressions
1.
2.
This is a rational function and there are no restrictions on the variable x.
Is a rational function
Restriction:
5x  10 ≠ 0
5x ≠ −10
x ≠ −2
3.
4.
5.
Not a rational function because of the radical in the numerator.
This is a rational function and there are no restrictions on the variable x.
Is a rational function.
Restriction:
x−4 ≠ 0
x≠4
6.
x2  x  y
x3y
 2
 xy ; x ≠ 0 and y ≠ 0
2 2
x y
x yy
7.
x 2  4x − 12  x  6x − 2  x  6 ; x ≠ 2
5
5x − 10
5x − 2
8.
x 3 − 4x  xx  2x − 2  xx − 2 ; x ≠ −2 and x ≠ 1
x−1
x  2x − 1
x x−2
2
9.
x − 2x 2  2x  4 x  2x 2 − 2x  4
x3 − 8 
x3  8


 x 2 − 2x  4; x ≠ 2, −2
2
2
x  2x − 2
x − 4 x  2x  4
x 2  2x  4
10.
z 2 − 5z − 24  9z 2 − 12z  4  z 2 − z − 6
9z − 12
z2 − 9
z 2 − 10z  16
2
z − 8z  3
3z − 2
z − 3z  2



33z − 4
z  3z − 3 z − 2z − 8
2
z2
 1 3z − 2
z ≠ −3, 3, 8, −2, 4
3
3
z − 23z − 4
11.
2
x − 8x − 1
3  6x  9
 3  x − 9x  8  3 
 x − 8 ; x ≠ 1, 8
2
x−1
x−1
2x  3
6x  9
x−1
32x  3
x − 9x  8
12.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
y 2  8y − 20
y3  1
y2 − y  1
 3
 2
2
y  11y  10 y − 8
y  2y  4
y  1y 2 − y  1
y  10y − 2
y 2  2y  4



y  10y  1 y − 2y 2  2y  4
y2 − y  1
 1; y ≠ −10, −1, 2
13.
4y
2y − 6
y 3 − 27
 2
 2
y − 5y  6
y − 2y
2y 3
y − 3y 2  3y  9
yy − 2
4y



3
y − 2y − 3 2y − 3
2y
2
y  3y  9

; y≠2
yy − 3
14. LCD: x − 33x  4
2
3x 2  4x  2x − 6
x3x  4
2x − 3
x 
2



 3x  6x − 6
x−3
3x  4
x − 33x  4
x − 33x  4
3x  4x − 3
x − 33x  4
15. LCD: x − 2x  3
1x  3
3x − 2
x  3 − 3x − 6
−2x  9
1 − 3 
−


x3
x−2
x − 2x  3
x  3x − 2
x − 2x  3
x − 2x  3
16. LCD: x  3x − 3
1x − 3
1x  3
10

−
x  3x − 3
x − 3x  3
x  3x − 3
x − 3  x  3 − 10

x  3x − 3
2x − 10

x  3x − 3

17.
7
− 8x  4
2x  1
2x − 1
1
LCD: 2x  12x − 1
72x − 1
8x2x  1
42x − 12x  1

−

2x  12x − 1
2x − 12x  1
2x − 12x  1
14x − 7 − 16x 2  8x  16x 2 − 4

2x  12x − 1
6x − 11

2x  12x − 1
18.
3 − 2  x3
x
x−1
x2 − 1
x3
 3 − 2x 
x−1
x  1x − 1
LCD: xx  1x − 1
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D
3xx  1
2x  1x − 1
x  3x
−

xx  1x − 1
x − 1xx  1
x  1x − 1x
2
2
2
3x  3x − 2x − 2  x  3x

xx − 1x  1
2
 2x  6x  2
xx − 1x  1

19.
y 2 − 25
y 2 − 7y  12
 2
2
y  3y − 18
y − 8y  15
y − 4y − 3
y  5y − 5


y  6y − 3
y − 5y − 3
y−4
y5


; y≠5
y6
y−3
So LCD is y  6y − 3
y − 4y − 3
y  5y  6

y  6y − 3
y − 3y  6
y 2 − 7y  12  y 2  11y  30

y  6y − 3
2
2y  4y  42

; y≠5
y  6y − 3

20.
y−7
− 5
y1
y  2y  1
y−7

− 5
y1
y  1 2
2
So LCD  y  1 2
5y  1
y−7
−
2

1
y
y  1 2
y − 7 − 5y  5

y  1 2
−4y − 12

y  1 2

21.
3x  y
4
3x  y
4

 3
xy
2
2x

y
4
2
22.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1D




5  x
x1
x−1
2 1
x2 − 1
5  x
x  1x − 1
x1
x−1
2
 1 x  1x − 1
x  1x − 1
5x − 1  xx  1
2  x  1x − 1
5x − 5  x 2  x
2  x2 − 1
2
x  6x − 5 ; x ≠ 1, −1
x2  1
23.
1ab
 ab ; a ≠ 0; b ≠ 0
ba
1  1 ab
a
b
24.




1 − 1
x
2
x−2
1 − 1 2x
x
2
x − 2 2x
1
2−x
2xx − 2
−x − 2
2xx − 2
−1
2x
25.
x  h −2 − x −2
h







©
: Pre-Calculus
1
− 12 x  h 2 x 2 
2
x
x  h
hx  h 2 x 2 
x 2 − x  h 2
hx 2 x  h 2
x 2 − x 2  2xh  h 2 
hx 2 x  h 2
x 2 − x 2 − 2xh − h 2
hx 2 x  h 2
−2xh − h 2
hx 2 x  h 2
h−2x − h
hx 2 x  h 2
−2x − h ; h ≠ 0
2
x x  h 2
©
: Pre-Calculus - Chapter 1E
Chapter 1E - Complex Numbers
Definition of a Complex Number
Not everything can be done using real numbers. For example, there is no real number x such that
x 2  −1. To handle this, mathematicians (such as William R. Hamilton 1805-1865) expanded the real
number system by introducing the imaginary unit i  −1 . Putting real and imaginary numbers
together creates the Complex Number System.
Definition
A complex number is a number that can be written in the form a  bi,
where a and b are real numbers. a is referred to as the real part and b is referred to as
the imaginary part. This is called the standard form of the complex number.
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are
equal.
Example 1:
Solution:
Write the number 24 − −64 in its standard form a  bi.
We begin by simplifying the radicals, noting that −64  64  −1 :
24 − −64

4  6 − 64  −1
 2 6 − 8 −1
 2 6 − 8i
The real part is 2 6 and the imaginary part is −8.
Example 2:
Solution:
Find real numbers a and b such that 2a − 1 − 4i  −7 
Find a by setting the real parts equal:
2a − 1  −7
b i
3
2a  −6
a  −3
Then find b by setting the imaginary parts equal:
−4  b
3
− 12  b
We now define the absolute value of a complex number.
If z  a  bi is a complex number in standard form, the absolute value of z, |z|, is defined as
|z|  |a  bi|  a 2  b 2 .
If z 1 and z 2 are two complex numbers, then the distance between these two numbers is
defined to be (We’ll see how to subtract complex numbers soon.)
|z 1 − z 2 |
If z is a real number, then the absolute value of z is the same as its real number absolute value.
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1E
Example 3:
Compute the absolute values of the following complex numbers.
2 2  −3 2 

|2 − 3i| 

|5|  |5  0i| 

|−7|  |−7  0i| 


|7  11i| 
|16 − 5i| 

|−5| 
49 
52  02 
5 2  5.
−7 2  0 2 
7 2  11 2 
16 2  5 2 
13 .
49  7.
49  121 
281 .
170 .
−5 2  0 2  5.
Some properties of the absolute value of a complex number:
1.
2.
For z  a  bi a complex number, the absolute value of z represents the distance from the
origin 0, 0 to the point with coordinates a, b. If you have not learned about the Cartesian
coordinate system for the Euclidean plane yet, then ignore this for now.
Let z 1 and z 2 be any two complex numbers, then
|z 1 z 2 |  |z 1 ||z 2 | .
3.
Let z 1 and z 2 be any two complex numbers, then
z 1  |z 1 | .
z2
|z 2 |
4.
Let z 1 and z 2 be any two complex numbers, then
|z 1  z 2 | ≤ |z 1 |  |z 2 | .
This inequality is called the triangle inequality.
Example 4:
Compute the absolute value of the following compex numbers.
2 − 5i, 7  3i, −11 − 4i, and 3i.
Solution:
|2 − 5i| 
4  25 
29 ≈ 5. 385 2
|7  3i| 
49  9 
58 ≈ 7. 615 8
|−11 − 4i| 
|3i| 
121  16 
137 ≈ 11. 705
9 3
Example 5:
Show that the absolute value of the product of 3 − 2i and −4  i equals the product of
their absolute values.
Solution:
|3 − 2i−4  i|  |−10  11i| 
|3 − 2i||−4  i| 
©
: Pre-Calculus
100  121 
9  4 16  1 
13 17 
221
13 ∘ 17 
221 .
©
: Pre-Calculus - Chapter 1E
Example 6:
Show that the absolute value of the sum of −8  i and 4  3i is less than or equal to the
sum of their absolute values.
Solution:
|−8  i  4  3i|  |−4  4i| 
|−8  i|  |4  3i| 
We now need to show that 32 ≤
32  65 .
16  16 
64  1  16  9 
32
65  25 
65  5 .
65  5, but this is obvious since 32  65 and therefore
We now define the conjugate of a complex number.
The conjugate of the complex number z  a  bi is defined to be a − bi, where we assume
that z is in standard form.
The conjugate of a complex number z is commonly denoted by writing a line above z. That is, z̄ is used
to denote the conjugate of z.
Example 7:
Compute the conjugate of the following complex numbers:
1 − i, 3i, −8, and −3 − 6i.
Solution:




The conjugate of 1 − i  1 − i  1  i.
The conjugate of 3i  3i  −3i.
The conjugate of −8  −8  −8.
The conjugate of −3 − 6i  −3 − 6i  −3  6i.
Some Properties of the Conjugate of a Complex Number
In the following z, z 1 , and z 2 represent arbitrary complex numbers.
1. z  z  |z| 2 , a  bia − bi  a 2  b 2  0i  a 2  b 2  |a  bi| 2 .
2. If x is a real number, then x  x.
3. z̄  z (The conjugate of the conjugate of z equals z.)
4. z 1  z 2   z 1  z 2 (The conjugate of a sum is the sum of the conjugates.)
5. z 1  z 2  z 1  z 2 (The conjugate of a product is the product of the conjugates.)
z 1  z 1 (The conjugate of a quotient is the quotient of the conjugates.)
6.
z2
z2
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1E
Exercises for Chapter 1E - Complex Numbers
1.
Write each of the following in standard form:
−2  −16 , 5 − −50 , 2i 2 − 4i, − 9 − 3
4
4
2. Find real numbers a and b such that a  bi  14i − 3
3. Find real numbers a and b such that 2a − 5 − 6i  11 − 3bi
4. What is the absolute value and conjugate of z  5?
5. What is the absolute value and conjugate of z  −7i?
6. If the conjugate of the complex number z equals −2  7i, what does z equal?
7. What is the absolute value and conjugate of z  −3 − 5i?
8. If the real part of z equals −5 and the absolute value of z  5, what must z equal?
9. If the imaginary part of z equals −13, what does the imaginary part of z̄ equal?
10. If the real part of z equals −23, what does the real part of z̄ equal?
11. If the real and imaginary parts of a complex number are equal, and the real part equals −5,
what is the complex number?
12. What are the real and imaginary parts of the complex number 2 − 5i?
©
: Pre-Calculus
©
: Pre-Calculus - Chapter 1E
Answers to Exercises for Chapter 1E - Complex Numbers
1.
2.
3.
−2  −16  −2  4i
5 − −50  5 − 5 2 i
2i 2 − 4i  −2 − 4i
−9 − 3  −3  3i
4
4
4
2
a  −3
b  14
2a − 5 − 6i  11 − 3bi
This equation of complex numbers implies the following two real equations.
2a − 5  11
− 6  −3b
4.
5.
6.
7.
Thus, 2a  16 or a  8, and b  −6  2.
−3
5̄  5
|5|  5
−7i  7i
|−7i|  7
Since z̄  −2  7i, and z̄  z, we have z  z̄  −2  7i  −2 − 7i.
−3 − 5i  −3  5i
|−3 − 5i|  9  25  34
Suppose z  a  bi. We know that a  −5. Thus, 5  |z|  −5 2  b 2 .Squaring both
sides of this equality, we have 25  25  b 2 , which implies that b  0. Thus, z  −5.
9. 13
10. −23
11. −5 − 5i
12. Real part of 2 − 5i is 2, and the imaginary part of 2 − 5i is −5.
8.
©
: Pre-Calculus