Chapter 0 Algebra Preliminaries
Calculus is the reformulation of elementary mathematics through the use of a limit process. Students who
have weak elementary mathematic skills may have a difficult time learning calculus. In this chapter, we
first provide a pretest. If the result of your pretest is above 70% of the total score, your elementary
mathematics may be sufficient. Then we will review some basic algebraic formulas and skills which are
absolutely necessary for students to know before proceeding to calculus courses. After the review, we will
provide a posttest to see how much you have improved your algebraic skills. Finally, we will study the
results of the posttest.
Pretest__The following problems will be demonstrated in the review.
1. Simplify and eliminate the radical by writing the given expression with fractional exponents:
1
1

2. Simplify: 2  h 2
h
3. Factor: 6 x 2  11x  10
4. Multiply: (e2 x  e x )(e2 x  e x )
5. Simplify:
x2  2x  1
x2  x  2
6. Solve for x: x3  5x 2  14 x
7. Rationalize the numerator:
2h  2
h
8. For g ( x)  1  2 x2 , find and simplify: g ( x  h)  g ( x)
9. Find all values of x that satisfy 3  2 x  5 .
10. Rewrite f ( x)  4 x  2 as a piecewise-defined function.
(3a ) 2
4  3 a5
Algebra Review
0.1 Properties of Exponents
An expression of the form a n is called an exponential expression, where the variable a is called the base,
and n is called the exponent or power to which the base is raised. Any exponential expression with a
rational exponent can be written as a radical expression. The following table provides the properties of
exponents that are frequently used in simplifying certain expressions.
Table 0.1 Properties of Exponents
Let a and b be positive numbers and let m and n be real numbers. Then,
2. a  m 
1. a 0  1
3. a m / n  n a m 
5.
 a
n
m
1
am
4. a m  a n  a mn
am
 a m n
n
b
6. (ab)m  a mbm
m
am
a
8.    m
b
b
7. (a m )n  a mn
Example 0.1 Use the properties of exponents to simplify the following expressions. Write the
answers so that radical notions are not used and there are no variables in the denominator.
2
2
4 3
(a) (2a b) (3ab )
(b)
3
t2
 2 x6 
(c) 
3 
 5 xy 
3
(d)
Solution (a)
(2ab) 2 (3ab 4 )3  (22 a 2b 2 )[33 a 3 (b 4 )3 ]
By property: (ab)m  a mbm
 (4a 2b 2 )(27 a 3b12 )
By property: (a m )n  a mn
 4  27  a 2  a 3  b 2  b12
Group
 108a 5b14
By property: a m  a n  a mn
(3a ) 2
4  3 a5
(b) We need to eliminate the radicals.
1
1 1

t  2  3 (t  2) 2  (t  2) 2   (t  2) 2 3  (t  2)
1
3
1
3
1
6
Note: (t  2) 6  t 6  2 6
1
1
1
(c) We only need to eliminate the variables from the denominator.
3
 2 x 6   2 6 1 3   2 6 1 3   2 5 3 
 x x y   x y   x y 

3 
 5
 5

 5 xy   5
3
3
3
3
8 15 9
 2
   ( x5 )3 ( y 3 )3 
x y
125
5
(d) We need to eliminate the radical notion and variables from the denominator.
(3a) 2
32 a 2 9 2  5 3

 aa
5
4  3 a5 4  a 3 4
9 5 9 6 5 9 1
 a 2 3  a 3  3  a 3
4
4
4

Rationalizing the Denominator
There are times when it is convenient to remove a radical from the denominator of fractions. For example,
we want to rewrite the fraction 1 2 in an equivalent form, but without the radical in the denominator. We
will multiply
1
2
by
2
2
:
1
2

2
2

2
2
The procedure used to remove a radical from the denominator is called rationalizing the denominator.
To simplify a fraction which has radicals in the denominator, multiply the numerator and denominator by
the conjugate of the denominator. For example, the expressions a  b and a  b are conjugates of
each other.

Example 0.2 (a) Rationalize the denominator and simplify:
(b) Rationalize the numerator and simplify:
1
2 1
3 x
( x  0 and x  3 )
3 x
Solution (a) After multiplying the numerator and denominator by the conjugate of ( 2  1) , we need
use the formula (a  b)(a  b)  a2  b2 to simplify the denominator.
1
1
2 1
2 1



2 1
2  1 2  1 ( 2  1)  ( 2  1)
=
( 2  1)
( 2  1)

 2 1
2
2
2 1
( 2)  1
(b) In calculus, it is sometimes necessary to rationalize a numerator. The same techniques as in (a) can be
used to rationalize the numerator of a fraction.
3 x
3  x 3  x ( 3  x) ( 3  x)



3 x
3 x
3 x
(3  x)  ( 3  x )
=
( 3) 2  ( x ) 2
(3  x)
1


(3  x)  ( 3  x ) (3  x)  ( 3  x )
3 x

0.2 Operations on Fractions
a
, where a and b can be integers or algebraic expressions. If a and b
b
1
3
a
(b ≠ 0) are integers, is called a rational number. For example,
and are rational numbers. If a and
2
4
b
3 x
3x 2  2 x  1
a
b are algebraic expressions, is called a fractional expression. For example,
and
3 x
x3  5 x
b
are fractional expressions. The following table provides the operation rules for combining two (or more)
fractions into one.
Recall that a fraction has a form of
Table 0.2 Operations on Fractions
(The denominators in each case are assumed to be nonzero.)
1.
a c ac
,
 
b b
b
a c ac
 
b b
b
2.
a c ad  cb
 
b d
bd
3.
a c ac
 
b d bd
4.
a c a d ad
   
b d b c bc
Example 0.3 Carry out the indicated operations and simplify:
3
1
(a)

2
4x
6x
x
7  2x
(b)

1
2 x  4 3x  6
1
1

(c) 2  h 2
h
Solution (a) To add or subtract two fractions with different denominators, we need to find the least
common denominator (LCD). Since 4 x 2  2  2  x  x and 6x  2  3  x , the LCD is 2  2  3  x  x = 12x 2 .
Multiply the first term by
3
3
and the second term by
2x
2x
:
3
1
3 3 1 2x

 2  
2
4x
6x 4x 3 6x 2x
9
2x
9  2x



2
2
12 x 12 x
12 x 2
(b)
x
7  2x
x
7  2x

1 

1
2 x  4 3x  6
2( x  2) 3( x  2)
x
3 7  2x 2
6( x  2)

 
  1
2( x  2) 3 3( x  2) 2
6( x  2)
3x
14  4 x 6( x  2)



6( x  2) 6( x  2) 6( x  2)
3 x  (14  4 x)  6( x  2)

6( x  2)
3 x  14  4 x  6 x  12

6( x  2)
13 x  26

6( x  2)
13( x  2)

6( x  2)
13

6
Factor out the common factors 2 and 3.
The LCD is 6( x  2) .
Simplify.
By properties: a  c  a  c , a  c  a  c
b b
b
b b
b
Apply distributive property for numerator.
Simplify the numerator.
Factor out the common factor 13 from numerator
Cancel ( x  2) from numerator and denominator.
(c) This type of expression occurs in calculus.
1
1
1 2 1 2h

  
2h 2  2h 2 2 2h
h
h
2  (2  h) 2  2  h
h
2  (2  h)
2  (2  h) 2  (2  h)



h
h
h
h
h
1
1

h
 
2  (2  h)
2  (2  h) h 2(2  h)
Find the common denominator
for the fractions in the numerator.
Simplify.
a
1
 a b  a
b
b
An alternate way to solve the problem is by multiplying the given expression by
2(2  h)
, where
2(2  h)
2(2  h) is the LCD for the fractions in the numerator.
1
1
1
1


2  h 2  2  h 2  2  (2  h)
h
h
2  (2  h)
2  (2  h)
h
1



h  2  (2  h) h  2  (2  h) 2(2  h)
Which way is better?

0.3 Operations on Polynomials
Adding and Subtracting Polynomials
An expression of the form an x n  an 1 x n 1    a1 x  a0 is called a Polynomial, where an , an 1 … a1 ,
a0 are real numbers and an  0 . The degree of a polynomial is the highest power of the variable. Thus,
3x  1 , 5x2  2 x  1 , and x 4  3x 2  1are polynomials of degree 1, 2 and 4, respectively. Polynomials can
be added by combining similar (or like) terms. For example, 2x 2 and 3x 2 are similar (because they have
the same power of x), hence, 2 x2  (3x2 )   x2 . But, 2x 2 and 3x3 are not similar (because they have
different powers of x), therefore, they cannot be combined. To subtract one polynomial from another, add
the opposite of each term of the polynomial you are subtracting.
Example 0.4 Take the indicated operations for the following:
(a) (2a2b  3b2  3a)  (a2b  2b2  5)
(b) (6x2  3x  7)  2( x2  5x  1)
Solution
(a)
(2a 2b  3b 2  3a)  ( a 2b  2b 2  5)
Given
 2a 2b  3b 2  3a  a 2b  2b 2  5
2a 2b and a 2b are similar, 3b 2 and 2b 2 are similar.
 (2a 2b  a 2b)  (3b 2  2b 2 )  3a  5
Group the similar terms together.
 3a b  b  3a  5
Simplify.
(6 x 2  3 x  7)  2( x 2  5 x  1)
Given
 6 x  3x  7  2 x  10 x  2
Distribute -2 to the second pair of parentheses.
 (6 x  2 x )  (3x  10 x)  (7  2)
Group the similar terms together
 4x2  7 x  5
Simplify
2
2
(b)
2
2
2
2

Multiplication of Polynomials
The product of two polynomials is the polynomial obtained by multiplying each term of one polynomial
by each term of the other polynomial and then combining like terms.
Example 0.5 Multiply:
(a) ( x2  3x  1)  (3x  2)
(b) (2 x  3)  (4 x  5)
Solution (a) Multiplication of polynomials can be carried out by the distributive properties.
( x 2  3x  1)  (3 x  2)
 ( x 2  3x  1)  (3 x)  ( x 2  3 x  1)   2 
 (3 x3  9 x 2  3x)  (2 x 2  6 x  2)
 3x3  9 x 2  3x  2 x 2  6 x  2
 3x3  7 x 2  9 x  2
(b) A binomial is a polynomial that has two terms. Here, (2 x  5) and (4 x  5) are two binomials. It is
frequently necessary to find the product of two binomials. The product can be found by using a method
called FOIL, which is based upon the Distributive Property. The letters of FOIL stand for First, Outer,
Inner, and Last.
First
Outer
(2x  3)  (4 x  5)  8x2  10 x  12 x  15  8x2  2x  15
Inner
Last
Factoring Polynomials
A polynomial of degree greater than 0 is said to be prime relative to a given set of numbers if all of its
coefficients are from that set of numbers and it can not be written as a product of two polynomials,
excluding 1 and itself, have coefficients from that set of numbers. For example, to the set of integer,
x 2  3 is a prime polynomial, but x 2  9 is not prime, because x2  ( x  3)( x  3) . A polynomial is said to
be factored completely relative to a given set of number if it can be written as a product of prime
polynomials relative to that set of numbers. The set of numbers most frequently used to factoring
polynomials is the set of integers. Here, we will review some of the basic factoring techniques for
polynomials with integer coefficients.

Table 0.3
Basic Factoring Techniques
Technique
Example
Common Factors
3x2 y  6xy  3xy( x  1)
Grouping
ab  4a  2b  8  (ab  4a)  (2b  8)
 a(b  4)  2(b  4)
 (b  4)( a  2)
Difference of Squares: a2  b2  (a  b)(a  b)
x2  25  x2  52  ( x  5)( x  5)
Difference of Cubes: a  b  (a  b)(a  ab  b )
3
Sum of Cubes:
3
2
2
a3  b3  (a  b)(a2  ab  b2 )
x3  27  x3  33
 ( x  3)( x 2  3 x  32 )
 ( x  3)( x 2  3 x  9)
x2  8x  15  ( x  3)( x  5)
This will be explained in example 0.6(c).
Trial and Error for the form ax2  bx  c
Example 0.6 Factor each expression completely:
(a)
(b)
(c)
(d)
(e)
a3  4a  a 2b  4b
( x  2)3  64
x2  8x  15
6 x 2  11x  10
x2  4 x  3
Solution
(a)
a 3  4a  a 2b  4b  (a 3  a 2b)  (4a  4b)
Grouping
 a (a  b)  4(a  b)
Factor out the common factors a 2 and -4
 (a  b)(a 2  4)
Factor out the common factor (a  b)
2
 (a  b)(a  2 )
2
2
 (a  b)(a  2)(a  2)
Difference of squares
(b)
Rewrite as the form: a3  b3
Apply: a3  b3  (a  b)(a2  ab  b2 )
Apply the Distributive Property
( x2  10) is a prime polynomial.
( x  2)3  64  ( x  2)3  43
 [( x  2)  4]  [( x  2)2  4( x  2)  42 ]
 ( x  6)( x 2  4 x  2  4 x  8  16)
 ( x  6)( x 2  10)
(c) Polynomials of the form ax2  bx  c , where a, b, and c are real numbers with a  0 , are called
quadratic. The term ax 2 is the leading term, bx is the linear term, and c is the constant term. The
Trial and Error method is often used to factor a quadratic. For example, if x2  8x  15 can be factored,
we look for factors of the form (kx  s) and (ux  v) , such that
x 2  8 x  15  (kx  s)(ux  v)
 kux 2  (kv  su) x  sv
Setting corresponding coefficients equal gives
ku  1
kv  su  8
and
sv  15
These equations suggest a way to find k, s, u, and v. Since the coefficient of x 2 is 1, ku  1 . Therefore,
both k and u equal 1. Since sv  15 , there are only four pair of factors possible:
sv  3  5 , sv  (3)  (5) , sv  1  15 , and sv  (1)  (15) .
Now we need to solve kv  su  8 to find out s and v. Since k and u are 1, v  s  8 . Therefore choose
the pair of factors that have a sum of -8: -3 and -5.
This is the final answer: x2  8x  15  ( x  3)( x  5)
We also can put kv  su  8 as a cross product form:
k
The factors of the
coefficient of x 2
s
u
(+)
kv
Then
(+)
The factors of the
constant term: c
v
+ us = -8
1
3
1
5
5 +
3
The coefficient of the linear term
= 8
The result of the cross product is not the coefficient of the linear term. Try another pair.
(+)
( x  3) .
1
-3
The first line suggests
1
-5
The second line suggests
(-5 ) + (- 3) = -8
( x  5) .
The result of this cross product is the coefficient of the linear term. Therefore,
x2  8x  15  ( x  3)( x  5) . You can use the same way to check that sv  1 15 and sv  (1)  (15)
don’t give the right answer.
(d) For 6 x 2  11x  10 , the coefficient of x 2 is 6. The possible factors of 6 are: 2  3 , (2)  (3) , 1  6 and
(1)  (6) . The possible factors of the constant term -10 are: (2)  5 , 2  (5) , (1) 10 and 1  (10) . By
doing all of possible cross products, only the following cross product gives the right coefficient of the
linear term,
(+)
3
-2
This line suggests (3x  2) .
2
5
This line suggests (2 x  5) .
15 + (- 4) = 11
Therefore, the answer is 6x2  11x  10  (3x  2)(2 x  5) .
(e) For x 2  4 x  3 , since 1 is the coefficient of x 2 , we need factors of -3 whose sum is 4. The only
possible factors of -3 are (-1)(3) and (1)(-3). Neither pair of factors has sum 4. Therefore
x 2  4 x  3 cannot be factored over the set of integers and it is called irreducible over the integers.

The method of trial and error illustrated in the above can be long and tedious, particularly if the
coefficients of the polynomial are large and have many prime factors. But for simple cases, it is often
possible to arrive at the correct choice very quickly.
0.4 Rational Expressions
As we mentioned in section 0.1, a fractional expression is a quotient of two algebraic expressions. As a
p
special case, if p and q are polynomials, the algebraic expression
(q  0) is called a rational expression.
q
The operations on fractions we reviewed in section 0.2 are also used for rational expressions. In algebra,
as in arithmetic, we say that a fraction is reduced or simplified to lowest terms when the numerator and
denominator contain no common factors (other than 1 and -1). The factoring techniques we reviewed in
the previous section are used to simplify rational expressions.
Example 0.7 Reduce the following expressions to lowest terms:
(a)
6 x3  9 x 2
12 x 2  18 x
Solution
(a)
(b)
x2  2x  1
x2  x  2
(c)
x3  1 x 2  8 x  9

x9
x 2  81
6 x3  9 x 2 3x 2 (2 x  3)

12 x 2  18 x 6 x(2 x  3)
3  x  x  (2 x  3) x


2  3  x  (2 x  3) 2
(b)
(c)
Factor out the common factors from
numerator and denominator
Simplify.
x 2  2 x  1 ( x  1)( x  1) x  1


x 2  x  2 ( x  1)( x  2) x  2
Factor, then simplify.
x3  1 x 2  8 x  9 x 3  1 x 2  81
a c a d
  


 2
2
b
d b c
x9
x  81
x  9 x  8x  9
2
( x  1)( x  x  1) ( x  9)( x  9)


( x  9)
( x  1)( x  9)
 x2  x  1

Example 0.8 For what values is the following rational expression undefined?
x2  2x  1
x2  x  2
Solution Rational expressions name a real number for each real number replacement of the variable
or variables. Thus all properties of real numbers apply to rational expressions. Since a real number
divided by zero is undefined, the values that make a rational expression undefined are the values that
make the denominator of the expression zero. In this case,
x2  x  2  ( x  1)( x  2)
Thus when x equals -1 or 2, the rational expression is undefined.

0.5 Difference Quotient
Recall that a function is a rule that takes certain number as inputs and assigns to each input one and only
one number as output. For example, the function f ( x)  x2 takes each real number as inputs and square
each input as an output. It is helpful to write the function f ( x)  x2 as f ( )  ( )2 , which means
whatever quantity goes in the parentheses on the left hand side, must also be placed in the parentheses on
the right.
Example 0.9 Given that g ( x)  1  2x2 , find and simplify the following expressions:
(a) g (5)
(b) g (2b)
(c) g ( x  h)
Solution (a) The number 5 is the input for the given function, we need find the output g (5) .
g ( )  1  2( )2
Plug in 5 in both sides..
g (5)  1  2  5  1  50  49
2
(b) The 2b is the input, we need find the output g (2b) .
g ( )  1  2( ) 2
g (2b)  1  2(2b) 2
Plug in 2b in both sides.
 1  2  2  b  1  8b
2
2
2
Simplify.
(c) The ( x  h) is the input, we need find the out put g ( x  h) .
g ( )  1  2( ) 2
g ( x  h)  1  2( x  h) 2
Plug in ( x  h) in both sides.
 1  2( x  2 xh  h )
Apply (a  b)2  a2  2ab  b2
 1  2 x 2  4 xh  2h 2
Distributive property
2
2
Note: g ( x  h)  g ( x)  h and also g ( x  h)  g ( x)  g (h)
In this case,
g ( x)  h  1  2 x2  h
and
g ( x)  g (h)  (1  2 x 2 )  (1  2h 2 )
 2  2 x 2  2h 2
 2(1  x 2  h2 )

A special fractional expression of a function, called the difference quotient, plays an important role in
calculus. For now, we will concentrate on the algebraic techniques used in calculating and simplifying
such expressions. You will learn the meaning and applications later when you learn calculus.
Table 0.4 Definition of The Difference Quotient
If f is any function and h is any nonzero number, then the difference quotient of f is
the expression
f ( x  h)  f ( x )
h

Example 0.10
Given that g ( x)  1  2 x2 , find and simplify
g ( x  h)  g ( x )
.
h
Solution From the example 0.9 (c), we have already found g ( x  h)  1  2x2  4xh  2h2 , so
g ( x  h)  g ( x) [1  2 x 2  4 xh  2h 2 ]  (1  2 x 2 )

h
h
1  2 x 2  4 xh  2h 2  1  2 x 2

h
2
4 xh  2h
h(4 x  2h)


h
h
 4 x  2 h

Example 0.11 Compute the difference quotient of f ( x)  1  x when x = 1. Rationalize the
numerator if needed.

Solution By the definition, the difference quotient when x = 1 is the expression
f (1  h)  f (1)
.
h
Recall that
f ( )  1 ( )
f (1)  1  (1)  2
f (1  h)  1  (1  h)  2  h
Therefore
f (1  h)  f (1)
2h  2

h
h
2h  2 2h  2


h
2h  2
(2  h)  2
h


h( 2  h  2) h( 2  h  2)
1

2h  2
Rationalize the numerator.
Simplify.


Example 0.12
Compute the difference quotient of G(t ) 

1
when t = 2.
t
Solution By the definition, the difference quotient when t = 2 is the expression
Since G(
)
1
(
)
1
1
and G( 2 ) 
. Therefore,
(2  h)
(2)
1
1

G (2  h)  G (2) 2  h 2

h
h
, G(2  h) 
G(2  h)  G(2)
.
h
The right hand side is the expression in Example 0.3 (c) that we have already simplified.

0.6 Solving Linear Equations
A linear equation in one variable is an equation that can be written in the form ax  b  0 , where a and b
are real numbers and a  0 . A solution (root) set of an equation is the set of real numbers which make
the equation true. To solve an equation is to find all its solutions.
Two equations are equivalent if they have the same solution set. To solve equations, we use the idea that
adding or subtracting the same quantity on both sides of the equation and multiplying or dividing both
sides of an equation by the same nonzero quantity produces an equivalent equation to generate a sequence
of equivalent equations until we obtain one whose solution is obvious.
Example 0.13 Solving linear equations:
(a) 2 x  5  3
(b)
x 2x

 11
3 5
(c)
9x  6
7

x 2  3x x  3
Solution
2x  5  3
2x  5  5  3  5
2x  8
1
1
 2x   8
2
2
x4
(a)
Given.
Add 5 to both sides.
Simplify.
Multiply both sides by
1
2
.
Simplify.
Thus the solution is 4. You can check this answer by substituting 4 for x in 2 x  5  3 .
Check:
24 5  85  3
Thus, the answer is right.
(b)
x 2x

 11
3 5
 x 2x 
15      15  11
3 5 
x
2x
15   15 
 15  11
3
5
5 x  6 x  15  11
11x  15  11
11x 15 11

11
11
x  15
Given.
The LCD is 15. Multiply both sides by 15.
By the distributive property
Simplify.
Add similar terms.
Divide both sides by 11.
Simplify.
Thus the solution is 15. You can check this answer by substitute 15 for x in
x 2x

 11 .
3 5
Check:
15 2 15

 5  2  3  5  6  11
3
5
The answer is right.
(c) When we multiply both sides of an equation by an expression containing a variable, we sometimes
produce an equation that is not equivalent to the original equation. In this case, to eliminate the
denominator, we need multiply both sides by x( x  3) :
9x  6
7

2
x  3x x  3
9x  6
7

x( x  3) x  3
9x  6
7
x( x  3)
 x( x  3)
x( x  3)
x3
9x  6  7x
2x  6  0
2 x  6
x  3
Since the denominator is 0 when x  3 , the number -3 does not satisfy the original equation. Therefore,
we say the equation has no solution. It is necessary and important to check the answers, particularly when
you multiply an equation by an expression containing a variable.
An alternate way to solve above equation is by cross-multiplication. The idea is if
both sides by bd, give ad  cb .
a c
 , then multiply
b d
9x  6
7

2
x  3x x  3
(9 x  6)( x  3)  7( x 2  3 x)
9 x 2  6 x  27 x  18  7 x 2  21x
9 x 2  6 x  27 x  18  7 x 2  21x  0
2 x 2  12 x  18  0
x2  6 x  9  0
( x  3) 2  0
x3
Which way is better?

0.7 Solving Quadratic Equations
Solving Quadratic Equations by Factoring
A quadratic equation is an equation that can be written in the form ax 2  bx  c  0 , where a, b, and c
are real numbers and a  0 . One of the simplest techniques for solving quadratic equation involves
factoring and applying the following theorem.
Table 0.4 The Zero-Factor Theorem
If A and B are real numbers, then
A B  0
if and only if
A0
or
B0
(or both).
Example 0.14 Solve each equation by factoring.
(a) x2  2 x  3  0
(b) x3  5x 2  14 x
Solution (a)
3x 2  5 x  2  0
(3 x  1)  ( x  2)  0
(3x  1)  0
x
The solutions are
1
3
( x  2)  0
x  2
Given.
Factor.
(3x  1)( x  2)  0 if and only if (3x  1)  0 or ( x  2)  0 .
Solve for x.
1
1
and -2. You can check the answers by substituting and -2 for x in the equation
3
3
respectively.
x 3  5 x 2  14 x
(b)
x 3  5 x 2  14 x  0
x  ( x 2  5 x  14)  0
x  ( x  2)  ( x  4)  0
x0
( x  2)  0
x  2
( x  4)  0
x4
Do not divide both sides by x.
Subtract 14x from each side.
To use the Zero-Factor theorem to
solve the equation, one side must be 0.
Factor x
Factor completely.
Set each factor to zero, solve for x.
The answers are 0, -2 and 4, which means the equation has three solutions. But if you divide both sides
by x at the beginning, you would have found only two solutions: -2 and 4.

Solving Quadratic Equations by Completing the Square
The previous method requires the quadratic equation to be factored into integers. However, not every
quadratic equation, like x2  2 x  2  0 , can be factored into integers. A technique called completing the
square can solve the problem. The idea of completing the square is using the algebraic skills to rewrite
the form ax 2  bx  c  0 into a form ( x  h)2  k , which can be easily solved by taking the square root of
each side.
Example 0.15 Solve each equation by completing the square.
(a) x2  2 x  2  0
(b) 9 x 2  6 x  5  0
Solution
(a)
x2  2x  2  0
Given
x  2x  2
2
Add 2 to each side.
x 2  2 x  12  2  12
Half of the coefficient of x is 1 and 12  1 . Add 1 to each side.
( x  1) 2  3
Using (a  b)2  a2  2ab  b2 , the left side is ( x  1)2 .
( x  1) 2  3
Take the square root of each side.
( x  1)   3
( x  1)  3
 3 means
( x  1)   3
x 1 3
3 or  3 .
Set up two equations and solve for x.
x 1 3
Therefore the equation has two solutions: 1  3 and 1  3 .
(b)
9 x2  6 x  5  0
Given
9 x  6 x  5
2
x2 
Subtract each side by 5.
2
5
x
3
9
2
2
5 1
1
x  x    
3
9  3
 3
Divide each side by 9, making the coefficient of x 2 equal to 1.
2
2
2
1
4

x  
3
9


1
4

x  
3
9

1
2

1
x 
3
3

1
2

x  i
3
3

Half of
2
3
is
1
3
. Add the square of
1
3
to each side.
2
Using (a  b)2  a2  2ab  b2 , the left hand side is  x  1  .


3

Take the square root of each side.
Simplify the right hand side. Note that 1 is not a real number, it is
called an imaginary number and denoted as i ( i 2  1 ).
1 2

x  i
3 3

1 2
x  i
3 3
1  2i

3
1
2

Set up two equations.
x  i
3
3

1 2
x    i Solve for x.
3 3
1  2i
x
3
1  2i
1  2i
and
, which are called complex numbers. A complex number is
3
3
a number of the form a  ib , where a and b are real numbers. a is called the real part of the complex
number and b is called the imaginary part of the complex number. To make it possible to solve all
quadratic equations, mathematicians invented this complex number system. Example 0.15(b) has no
solutions in the real number system, but have two solutions in the complex number system.
There are two solutions:
Solving Quadratic Equations by Using the Quadratic Formula
A general formula known as the quadratic formula can be derived by applying the method of
completing the square to the standard form of a quadratic equation. This formula can be used to solve any
quadratic equation.
Table 0.4 The Quadratic Formula
The solution of the general quadratic equation ax 2  bx  c  0 , where a  0 , is
x
b  b 2  4ac
2a
Example 0.16 Solve each equation by using the quadratic formula..
(a) 4 x 2  12 x  9  0
(b) x2  4( x  1)
(c) x2  6  2 x
Solution (a) The equation 4 x 2  12 x  9  0 is in standard form ax2  ba  c  0 . Since a  4 , b  12 ,
and c  9 , we have
b  b 2  4ac 12  122  4  4  9

2a
24
12  144  144 12
3



8
8
2
3
The solution is a real number  and it is called a double root.
2
(b) The equation x2  4( x  1) is not in the standard form ax2  ba  c  0 , we need to rewrite it in
standard form before using the formula.
x
x 2  4( x  1)  4 x  4
x 2  4 x  4  0,
where a  1, b  4, and c  4.
b  b 2  4ac (4)  (4) 2  4 1  (4)

2a
2 1
4  16  16 4  2 16 4  4 2



22 2
2
2
2
x
The solutions are two real numbers: 2  2 2 and 2  2 2 .
(c) The equation x2  6  2 x is not in the standard form ax2  ba  c  0 , we need to rewrite it in standard
form before using the formula.
x2  6  2x
x2  2x  6  0
where a  1, b  2, and c  6.
b  b 2  4ac (2)  (2)2  4  1  6
x

2a
2 1
2  4  24 2  20 2  (1)  4  5



2
2
2
2  1  4  5 2  i  2  5


1 i 5
2
2
The solutions are two complex numbers: 1  i 5 and 1  i 5 .
The expression b2  4ac in the quadratic formula is called the discriminant. Its value determines the
nature of the roots of the given quadratic equation, as summarized in the following table.
Table 0.5 The Discriminant b2  4ac
For any quadratic equation ax 2  bx  c  0 , where a, b, and c are real numbers and a  0 , If
1. b2  4ac  0
The equation has two real and unequal roots.
2. b2  4ac  0
The equation has one real root (double root).
3. b2  4ac  0
The equation has two complex number (no real) roots.
0.8 Inequalities
An inequality is a statement where two expressions are not equal. For example, 7 > 4 is an inequality
which means 7 is greater than 4. The symbols > (greater than), < (less than), ≥ (greater than or equal), and
≤ (less than or equal), are used to write inequalities.
A linear inequality in one variable is an inequality that can be written in the form ax  b  c , where a, b,
and c are real numbers, and a  0 . The solution set of an inequality is the set of real numbers which
make the inequality true. Most of properties we used for solving equations are also true for solving
inequalities except for those involving multiplying and dividing each side of an inequality by a negative
number. The basic properties are summarized in the following table.
Table 0.6 Properties of Inequalities
Let a , b and c be real numbers.
1. If a  b , then
a  c  b  c and a  c  b  c .
Example:
If 5  10 , then 5  3  10  3 , or 8  13 .
If 5  10 , then 5  2  10  2 , or 3  8 .
2. If a  b and c is positive, then
ac  bc .
If a  b and c is negative, then
ac  bc.
Example:
If 5  10 , then 5  2  10  2 , or 10  20 .
If 5  10 , then 5  (2)  10  (2) , or 10  20 .
3. If a < b and b < c , then
ac.
Example:
If 5  10 , and 10  11 , then 5  11 .
Above properties hold for other inequality relations: >,  , and  .
Example 0.15 Solve each inequality:
(a) 1  3x  10
(b) 3x  1  7 x  15
(c) 5  3x  7  8
Solution (a)
1  3x  10
1  3x  1  10  1
 3x  9
1
1
3x  ( )  9  ( )
3
3
x  3
Given
Subtract both sides by 1.
Simplify.
Multiply both sides by (  1 3 ) and reverse the
inequality symbol.
The solution is a number set which includes all real numbers less than -3. The solution set can be denoted
as x < -3 or the unbounded interval (-∞, -3).
(b)
3x  1  7 x  15
4 x  1  15
 4 x  16
x4
Given
Subtract 7x from both sides.
Subtract 1 from both sides.
Multiply both sides by (  1 4 ) and reverse the
Inequality symbol.
The solution is a number set which includes all real numbers greater than or equal to -4. The solution set
can be denoted as x ≥ -4 or the unbounded interval [-4, ∞).
(c)
5  3x  7  8
5  7  3x  7  7  8  7
12  3 x  15
1
1
1
12   3x   15 
3
3
3
4 x5
Given
Add 7 to each of the three parts of the inequality.
Simplify.
Multiply each of the three parts of the inequality by
1
3
.
The solution is a number set which includes all real numbers greater than 4 and less than or equals to 5.
The solution set can be denoted as 4  x  5 or the interval (4, 5].
0.9 Absolute Value
Recall that the absolute value of a number x, denoted by |x|, is the distance from x to 0 on the real number
line. Distance is always positive or zero, so we have |x| ≥ 0 for every number x.

Table 0.6
Definition of Absolute Value
If x is a real number, then the absolute value of x is
 x if x  0
x 
 x if x  0
Solving Absolute Value Equations and Inequalities
To solve absolute value equations and inequalities, we use the following properties.
Table 0.7 Properties of Absolute Value Equations and Inequalities
If a > 0, then
1. x  a , is equivalent to
x  a or x  a
2. x  a , is equivalent to
a  x  a
3. x  a , is equivalent to
x  a or x  a
The properties above also hold for the other inequality relations:  and  .
Example 0.16 Solve:
(a) 2 x  5  7
(b) 3  2 x  5
(c) 2 x  1  1
Solution
(a) The equation 2 x  5  7 is equivalent to two equations: 2 x  5  7 and 2 x  5  7 , which can be
solved separately:
2x  5  7
2 x  12
x6
Thus, the solutions are 6 and -1.
or
2 x  5  7
2 x  2
x  1
(b) The inequality 3  2 x  5 is equivalent to the inequalities: 5  3  2 x  5 . Therefore,
5  3  2 x  5
 8  2 x  2
4  x  1
The solution is the set 4  x  1 , or the interval [1, 4] .
(c) The inequality 2 x  1  1 is equivalent to the inequalities: 2 x  1  1 or 2 x  1  1 , which can be
solved separately:
2x  1  1
2x  0
x0
or
2 x  1  1
2 x  2
x  1
The solution are the sets x > 0 and x < -1, or the intervals (-∞, -1) and (0, ∞).

Absolute Value Functions
The definition of absolute value shown in Table 0.6 also defines a function which has a rule assigning
each real number x its absolute value |x|. The function is called an absolute value function and the graph is
shown in the following figure.
 Figure 0.1 Graph of the absolute value function: f ( x)  x
Note that the graph of f ( x)  x has a corner at x = 0 (origin) and has two straight lines: f ( x)  x for x ≥
0 and f ( x)   x for x < 0. This kind of function which is defined for specific intervals is called a
piecewise-defined function.
Example 0.17 For the following functions, graph the function and rewrite the function in piecewise
form.
(a) f ( x)  x  3
(b) f ( x)  4 x  2

Solution
(a) Since f ( x)  x has a corner when x = 0, to find the corner of f ( x)  x  3 , we let x  3  0 . The
solution of x  3  0 is -3, which means the corner of f ( x)  x  3 has shifted 3 units to the left of the
origin. See the figure below.
Also if x  3  0 or x  3 , x  3  x  3 and if x  3  0 or x  3 , x  3  ( x  3) , thus
 x  3 if x  3
f ( x)  x  3  
 x  3 if x  3
(b) To find the corner, we let 4 x  2  0 , then the solution x  1 2 means the corner has shifted
the right of the right.
1
2
units to
If 4 x  2  0 or x  1 2 , 4 x  2  4 x  2 and 4 x  2  0 or x  1 2 , 4 x  2  (4 x  2)  2  4 x , therefore
4 x  2 if x  1 2
f ( x)  4 x  2  
2  4 x if x  1 2

Posttest
1. Simplify the following expression. The answer should not have radical or negative exponents.
5
32a 3
a 7 b5
2. For f ( x)  3x2  2 x  1, find the difference quotient when x = 2.
3. Solve by completing the square: x  2  ( x  3)( x  2)
4. Factor: 3x  xy  2 y  6
5. Solve for x: x5  16 x  0
x4 x2
1


4
3
2
1 1 h
7. Rationalize the numerator:
h
6. Solve for x:
8. Simplify:
6 x3  12 x 2
x2  4

2 x3  4 x 2  8 x x3  8
9. Find all values of x that satisfy 4(1  x)  1  3(2 x  1) .
10. Rewrite f ( x)  1  3x as a piecewise-defined function.
Analyzing Posttest Result
The knowledge and skills in Algebra are essential for learning Calculus. The Pretest, Chapter 0, and the
Posttest are all designed to refresh and review your Algebraic skills. However, they are not supposed to
teach these skills to you.
Posttest score
(after review)
90% or above
Between 75% and
90%
75% or below
Analysis
You are most likely prepared with the essential algebraic skills.
You are some what prepared with the essential Algebraic skills. It
would be in your best interest to study on your own to improve
your Algebra in order to avoid struggling when learning Calculus.
You need to seriously reconsider taking this course. Your
Algebraic skills do not seem to be adequate to take Calculus.
Major difficulty exists. Please consult your advisor.
The numbers given are only guidelines. Please take all factors into account to make a mature and
responsible decision on whether you should take this course or not.
Answers to Pretest
1.
9 13
a
4
2.
1
2(2  h)
3. (3x  2)(2 x  5)
4. e4 x  e2 x
5.
x 1
x2
6. x = 0, x = -2, x = 4
7.
1
2h  2
8. 4 xh  2h2
9. [-1, 4]
4 x  2 if x  1 2
10. f ( x)  
2  4 x if x  1 2
Answers to Posttest
1.
2a 2
b
6. 10
2. 10  3h
7.
1
1 1 h
3. 1  5
8. 3x
4. ( x  2)(3  y)
9. x  3 or [3, ∞)
5. x  0,  2,  2i
3x  1 if x  1 3
10. f ( x)  
1  3 x if x  1 3