Transcript 24 = 2 * 2 * 2 * 3

Real Numbers and
Algebraic
Expressions
The Basics About Sets
The set {1, 3, 5, 7, 9} has five elements.
•
A set is a collection of objects whose contents can be clearly determined.
•
•
The objects in a set are called the elements of the set.
We use braces to indicate a set and commas to separate the elements of
that set.
For example,
The set of counting numbers can be represented by {1, 2, 3, … }.
The set of even counting numbers are {2, 4, 6, …}.
The set of even counting numbers is a
subset of the set of counting numbers,
since each element of the subset is
also contained in the set.
Important Subsets of the Real Numbers
Name
Description
Examples
Natural Numbers
N
{1, 2, 3, …}
These are the counting numbers
4, 7, 15
Whole Numbers
W
{0, 1, 2, 3, … }
Add 0 to the natural numbers
0, 4, 7, 15
Integers
Z
{…, -2, -1, 0, 1, 2, 3, …}
Add the negative natural
numbers to the whole numbers
-15, -7, -4, 0, 4, 7
Important Subsets of the Real Numbers
Name
Description
Examples
Rational
Numbers
Q
These numbers can be expressed as an
integer divided by a nonzero integer:
Rational numbers can be expressed as
terminating or repeating decimals.
17 
Irrational
Numbers
I
This is the set of numbers whose decimal
representations are neither terminating nor
repeating. Irrational numbers cannot be
expressed as a quotient of integers.
17
5
,5 
,3, 2
1
1
0,2,3,5,17
2
 0.4,
5
2
 0.666666...  0.6
3
2  1.414214
 3  1.73205
  3.142


2
 1.571
The Real Numbers
Rational numbers
Irrational numbers
Integers
Whole numbers
Natural numbers
The set of real numbers is formed by combining the rational numbers and
the irrational numbers.
The Real Number Line
The real number line is a graph used to represent the set of real numbers. An
arbitrary point, called the origin, is labeled 0;
Negative numbers
-4
-3
-2
Units to the left of the
origin are negative.
-1
Positive numbers
0
the
Origin
1
2
3
4
Units to the right of
the origin are
positive.
Graphing on the Number Line
Real numbers are graphed on the number line by placing a dot at the location
for each number. –3, 0, and 4 are graphed below.
-4
-3
-2
-1
0
1
2
3
4
Ordering the Real Numbers
On the real number line, the real numbers increase from left to right. The lesser
of two real numbers is the one farther to the left on a number line. The greater
of two real numbers is the one farther to the right on a number line.
-2
-1
0
1
2
3
4
5
6
Since 2 is to the left of 5 on the number line, 2 is less than 5. 2 < 5
Since 5 is to the right of 2 on the number line, 5 is greater than 2. 5 > 2
Inequality Symbols
Symbols Meaning
Example
Explanation
a<b
3<7
Because 3 < 7
7<7
Because 7 =7
b is greater than or equal to a. 7 > 3
Because 7 > 3
b>a
a is less than or equal to b.
-5 > -5
Because -5 = -5
Absolute Value
Absolute value describes the distance from 0 on a real number line. If a
represents a real number, the symbol |a| represents its absolute value, read
“the absolute value of a.”
For example, the real number line below shows that
|-3| = 3 and |5| = 5.
|–3| = 3
-3
-2
-1
The absolute value of –3 is 3
because –3 is 3 units from 0
on the number line.
|5| = 5
0
1
2
3
4
5
The absolute value of 5 is 5
because 5 is 5 units from 0
on the number line.
Definition of Absolute Value
The absolute value of x is given as follows:
{
x if x > 0
|x| =
-x if x < 0
Properties of Absolute Value
For all real number a and b,
1. |a| > 0
2. |-a| = |a|
3. a < |a|
4. |ab| = |a||b|
5.
6. |a + b| < |a| + |b| (the triangle inequality)
a |a|
= , b not equal to 0
b |b|
Example
• Find the following: |-3| and |3|.
Solution:
| -3 | = 3 and | 3 | = 3
Distance Between Two Points on
the Real Number Line
If a and b are any two points on a real number line, then the distance between
a and b is given by
|a – b| or |b – a|
Text Example
Find the distance between –5 and 3 on the real number line.
Solution Because the distance between a and b is given by |a – b|, the
distance between –5 and 3 is |-5 – 3| = |-8| = 8.
8
-5
-4
-3
-2
-1
0
1
2
We obtain the same distance if we reverse the order of subtraction:
|3 – (-5)| = |8| = 8.
3
Algebraic Expressions
A combination of variables and numbers using the operations of addition,
subtraction, multiplication, or division, as well as powers or roots, is called an
algebraic expression.
Here are some examples of algebraic expressions:
x + 6, x – 6, 6x, x/6, 3x + 5.
The Order of Operations Agreement
1.
2.
3.
4.
Perform operations within the innermost parentheses and
work outward. If the algebraic expression involves
division, treat the numerator and the denominator as if
they were each enclosed in parentheses.
Evaluate all exponential expressions.
Perform multiplication or division as they occur,
working from left to right.
Perform addition or subtraction as they occur, working
from left to right.
Example
The algebraic expression 2.35x + 179.5 describes the population of the
United States, in millions, x years after 1980. Evaluate the expression
when x = 20. Describe what the answer means in practical terms.
Solution We begin by substituting 20 for x. Because x = 20, we will be
finding the U.S. population 20 years after 1980, in the year 2000.
2.35x + 179.5
= 2.35(20) + 179.5
= 47 + 179.5
= 226.5
Replace x with 20.
Perform the multiplication.
Perform the addition.
Thus, in 2000 the population of the United States was 226.5 million.
Properties of the Real Numbers
Name
Meaning
Examples
Commutative
Property of
Addition
Two real numbers can be added
in any order.
a+b=b+a
• 13 + 7 = 7 + 13
• 13x + 7 = 7 + 13x
Commutative Two real numbers can be
Property of
multiplied in any order.
Multiplication ab = ba
• x · 6 = 6x
Associative
Property of
Addition
• 3 + ( 8 + x)
= (3 + 8) + x
= 11 + x
If 3 real numbers are added, it
makes no difference which 2 are
added first.
(a + b) + c = a + (b + c)
Properties of the Real Numbers
Name
Meaning
Examples
Associative
If 3 real numbers are multiplied,
Property of
it makes no difference which 2
Multiplication are multiplied first.
(a · b) · c = a · (b · c)
• -2(3x) = (-2·3)x = -6x
Distributive
Multiplication distributes over
Property of
addition.
Multiplication a · (b + c) = a · b + a · c
over Addition
• 5 · (3x + 7)
= 5 · 3x + 5 · 7
= 15x + 35
Identity
Property of
Addition
• 0 + 6x = 6x
Zero can be deleted from a sum.
a+0=a
0+a=a
Properties of the Real Numbers
Name
Meaning
Examples
Identity
One can be deleted from a
Property of
product.
Multiplication a · 1 = a and 1 · a = a
• 1 · 2x = 2x
Inverse
Property of
Addition
• (-6x) + 6x = 0
The sum of a real number and its
additive inverse gives 0, the
additive identity.
a + (-a) = 0 and (-a) + a = 0
Inverse
The product of a nonzero real
• 2 · 1/2 = 1
Property of
number and its multiplicative
Multiplication inverse gives 1, the multiplicative
identity.
a · 1/a = 1 and 1/a · a = 1
Definitions of Subtraction and
Division
Let a and b represent real numbers.
Subtraction: a – b = a + (-b)
We call –b the additive inverse or opposite of b.
Division: a ÷ b = a · 1/b, where b = 0
We call 1/b the multiplicative inverse or reciprocal of b. The quotient of
a and b, a ÷ b, can be written in the form a/b, where a is the numerator
and b the denominator of the fraction.
Text Example
Simplify: 6(2x – 4y) + 10(4x + 3y).
Solution
6(2x – 4y) + 10(4x + 3y)
= 6 · 2x – 6 · 4y + 10 · 4x + 10 · 3y
= 12x – 24y + 40x + 30y
= (12x + 40x) + (30y – 24y)
= 52x + 6y
Use the distributive property.
Multiply.
Group like terms.
Combine like terms.
Properties of Negatives
•
1.
2.
3.
4.
5.
6.
Let a and b represent real numbers, variables, or
algebraic expressions.
(-1)a = -a
-(-a) = a
(-a)(b) = -ab
a(-b) = -ab
-(a + b) = -a - b
-(a - b) = -a + b = b - a
Real Numbers and
Algebraic
Expressions
Exponents and
Scientific Notation
Definition of a Natural Number
Exponent
• If b is a real number and n is a natural
number,
n
b  b  b b ...b
• bn is read “the nth power of b” or “ b to the
nth power.” Thus, the nth power of b is
defined as the product of n factors of b.
Furthermore, b1 = b
The Negative Exponent Rule
• If b is any real number other than 0 and n is
a natural number, then
b
n
1
 n
b
The Zero Exponent Rule
• If b is any real number other than 0,
b0 = 1.
The Product Rule
b m · b n = b m+n
When multiplying exponential expressions
with the same base, add the exponents. Use
this sum as the exponent of the common
base.
The Power Rule (Powers to Powers)
(bm)n = bm•n
When an exponential expression is raised to a
power, multiply the exponents. Place the
product of the exponents on the base and
remove the parentheses.
The Quotient Rule
m
b
m n
n b
b
• When dividing exponential expressions
with the same nonzero base, subtract the
exponent in the denominator from the
exponent in the numerator. Use this
difference as the exponent of the common
base.
Example
• Find the quotient of 43/42
Solution:
3
4
3 2
1

4

4

4
2
4
Products to Powers
(ab)n = anbn
When a product is raised to a power, raise
each factor to the power.
Text Example
Simplify: (-2y)4.
Solution
(-2y)4 = (-2)4y4 = 16y4
Quotients to Powers
n
a
a
   n
b
b
n
• When a quotient is raised to a power, raise
the numerator to that power and divide by
the denominator to that power.
Example
• Simplify by raising the quotient (2/3)4 to the
given power.
Solution:
4
2 16
 2
   4 
3
81
 3
4
Properties of Exponents
1. b
n
1
 n
b
bm
5. n  b m n
b
0
m
n
mn
2. b  1 3. b  b  b
6. (ab)n  a n b n
m n
4. (b )  b
a n a n
7.    n
b  b
mn
Scientific Notation
The number 5.5 x 1012 is written in a form called scientific notation. A
number in scientific notation is expressed as a number greater than or equal
to 1 and less than 10 multiplied by some power of 10. It is customary to use
the multiplication symbol, x, rather than a dot in scientific notation.
Text Example
• Write each number in decimal notation:
a. 2.6 X 107
b. 1.016 X 10-8
Solution:
a. 2.6 x 107 can be expressed in decimal notation by moving the
decimal point in 2.6 seven places to the right. We need to add six zeros.
2.6 x 107 = 26,000,000.
b. 1.016 x 10-8 can be expressed in decimal notation by moving the
decimal point in 1.016 eight places to the left. We need to add seven
zeros to the right of the decimal point.
1.016 x 10-8 = 0.00000001016.
Scientific Notation
To convert from decimal notation to scientific notation, we reverse the procedure.
• Move the decimal point in the given number to obtain a number greater than or
equal to 1 and less than 10.
• The number of places the decimal point moves gives the exponent on 10; the
exponent is positive if the given number is greater than 10 and negative if the
given number is between 0 and 1.
Text Example
Write each number in scientific notation. a. 4,600,000 b. 0.00023
Solution
a. 4,600,000 = 4.6 x
b. 0.00023 = 2.3 x
10?
10?
Decimal point moves 6 places
Decimal point moves 4 places
4.6 x 106
2.3 x 10-4
Exponents and
Scientific Notation
Radicals and
Rational Exponents
Definition of the Principal Square
Root
• If a is a nonnegative real number, the
nonnegative number b such that b2 = a,
denoted by b = a, is the principal square
root of a.
Square Roots of Perfect Squares
2
a a
The Product Rule for Square Roots
• If a and b represent nonnegative real
number, then
ab  a b and a b  ab
• The square root of a product is the product
of the square roots.
Text Example
• Simplify a. 500
b. 6x3x
Solution:
a.
500  1005
b.
6x  3x  6x  3x
 100 5
 18x 2  9x 2 2
 10 5
 9x 2 2  9 x 2 2
 3x 2
The Quotient Rule for Square Roots
• If a and b represent nonnegative real
numbers and b does not equal 0, then
a
a

b
b
and
a

b
a
.
b
• The square root of the quotient is the
quotient of the square roots.
Text Example
• Simplify:
100
9
Solution:
100
100 10


9
9
3
Example
• Perform the indicated operation:
43 + 3 - 23.
Solution:
4 3 32 3 3 3
Example
• Perform the indicated operation:
24 + 26.
Solution:
24  2 6 
2 6 2 6 4 6
Definition of the Principal nth Root
of a Real Number
n
n
a  b meansthat b  a
• If n, the index, is even, then a is
nonnegative (a > 0) and b is also
nonnegative (b > 0) . If n is odd, a and b can
be any real numbers.
Finding the nth Roots of Perfect
nth Powers
If n is odd, a  a
n
n
If n is even a  a.
n
n
The Product and Quotient Rules
for nth Roots
• For all real numbers, where the indicated
roots represent real numbers,
n
a b  ab and
n
n
n
n
a n a

, b0
b
b
Definition of Rational Exponents
a1 / n  n a.
Furtherm ore,
1
1
1/ n
a
 1/ n  n , a  0
a
a
Example
• Simplify 4 1/2
Solution:
1
2
4  4 2
Definition of Rational Exponents
a
m/ n
m
n
m
( a)  a .
n
• The exponent m/n consists of two parts: the
denominator n is the root and the numerator
m is the exponent. Furthermore,
a
 m/ n

1
a
m/ n
.
Radicals and
Rational Exponents
Polynomials
The Degree of
n
ax
• If a does not equal 0, the degree of axn is n.
The degree of a nonzero constant is 0. The
constant 0 has no defined degree.
Definition of a Polynomial in x
• A polynomial in x is an algebraic
expression of the form
• anxn + an-1xn-1 + an-2xn-2 + … + a1n + a0
• where an, an-1, an-2, …, a1 and a0 are real
numbers. an = 0, and n is a non-negative
integer. The polynomial is of degree n, an is
the leading coefficient, and a0 is the
constant term.
Text Example
Perform the indicated operations and simplify:
(-9x3 + 7x2 – 5x + 3) + (13x3 + 2x2 – 8x – 6)
Solution
(-9x3 + 7x2 – 5x + 3) + (13x3 + 2x2 – 8x – 6)
= (-9x3 + 13x3) + (7x2 + 2x2) + (-5x – 8x) + (3 – 6)
= 4x3 + 9x2 – (-13x) + (-3)
= 4x3 + 9x2 + 13x – 3
Group like terms.
Combine like terms.
Multiplying Polynomials
The product of two monomials is obtained by using properties of exponents.
For example,
(-8x6)(5x3) = -8·5x6+3 = -40x9
Multiply coefficients and add exponents.
Furthermore, we can use the distributive property to multiply a monomial and
a polynomial that is not a monomial. For example,
3x4(2x3 – 7x + 3) = 3x4 · 2x3 – 3x4 · 7x + 3x4 · 3 = 6x7 – 21x5 + 9x4.
monomial
trinomial
Multiplying Polynomials when
Neither is a Monomial
• Multiply each term of one polynomial by
each term of the other polynomial. Then
combine like terms.
Using the FOIL Method to Multiply Binomials
last
first
(ax + b)(cx + d) = ax · cx + ax · d + b · cx + b ·
d
Product of
Product of
Product of
Product of
inner
outer
First terms
Outside terms Inside terms
Last terms
Text Example
Multiply: (3x + 4)(5x – 3).
Text Example
Multiply: (3x + 4)(5x – 3).
Solution
last
first
F
O
I
L
(3x + 4)(5x – 3) = 3x·5x + 3x(-3) + 4(5x) + 4(-3)
= 15x2 – 9x + 20x – 12
inner
= 15x2 + 11x – 12Combine like terms.
outer
The Product of the Sum and
Difference of Two Terms
2
(A  B)(A  B)  A  B
2
• The product of the sum and the difference
of the same two terms is the square of the
first term minus the square of the second
term.
The Square of a Binomial Sum
2
2
2
(A  B)  A  2AB  B
• The square of a binomial sum is first term
squared plus 2 times the product of the
terms plus last term squared.
The Square of a Binomial
Difference
2
2
2
(A  B)  A  2AB  B
• The square of a binomial difference is first
term squared minus 2 times the product of
the terms plus last term squared.
Special Products
Let A and B represent real numbers, variables, or algebraic expressions.
Special Product
Sum and Difference of Two Terms
Example
(A + B)(A – B) = A2 – B2
(2x + 3)(2x – 3) = (2x) 2 – 32
= 4x2 – 9
Squaring a Binomial
(A + B)2 = A2 + 2AB + B2
(A – B)2 = A2 – 2AB + B2
(y + 5) 2 = y2 + 2·y·5 + 52
= y2 + 10y + 25
(3x – 4) 2 = (3x)2 – 2·3x·4 + 42
= 9x2 – 24x + 16
Cubing a Binomial
(A + B)3 = A3 + 3A2B + 3AB2 + B3
(A – B)3 = A3 – 3A2B – 3AB2 + B3
(x + 4)3 = x3 + 3·x2·4 + 3·x·42 + 43
= x3 + 12x2 + 48x + 64
(x – 2)3 = x3 – 3·x2·2 – 3·x·22 + 23
= x3 – 6x2 – 12x + 8
Text Example
Multiply: a. (x + 4y)(3x – 5y)
b. (5x + 3y) 2
Solution
We will perform the multiplication in part (a) using the FOIL method. We will
multiply in part (b) using the formula for the square of a binomial, (A + B) 2.
a. (x + 4y)(3x – 5y)
F
O
Multiply these binomials using the FOIL method.
I
L
= (x)(3x) + (x)(-5y) + (4y)(3x) + (4y)(-5y)
= 3x2 – 5xy + 12xy – 20y2
= 3x2 + 7xy – 20y2
Combine like terms.
• (5 x + 3y) 2
= (5 x) 2 + 2(5 x)(3y) + (3y) 2
= 25x2 + 30xy + 9y2
(A + B) 2 = A2 + 2AB + B2
Example
• Multiply: (3x + 4)2.
Solution:
( 3x + 4 )2
=(3x)2 + (2)(3x) (4) + 42
=9x2 + 24x + 16
Polynomials
Factoring
Polynomials
Factoring
Factoring is the process of writing a polynomial as the product of
two or more polynomials. The factors of 6x2 – x – 2 are 2x + 1 and 3x – 2. In
this section, we will be factoring over the integers. Polynomials that cannot
be factored using integer coefficients are called irreducible over the
integers or prime.
The goal in factoring a polynomial is to use one or more factoring
techniques until each of the polynomial’s factors is prime or irreducible. In
this situation, the polynomial is said to be factored completely.
Common Factors
In any factoring problem, the first step is to look for the greatest common
factor. The greatest common factor is a n expression of the highest degree
that divides each term of the polynomial. The distributive property in the
reverse direction
ab + ac = a(b + c)
can be used to factor out the greatest common factor.
Text Example
Factor: a. 18x3 + 27x2
b. x2(x + 3) + 5(x + 3)
Solution
a. We begin by determining the greatest common factor. 9 is the greatest
integer that divides 18 and 27. Furthermore, x2 is the greatest expression that
divides x3 and x2. Thus, the greatest common factor of the two terms in the
polynomial is 9x2.
18x3 + 27x2
= 9x2(2x) + 9x2(3)
Express each term with the greatest common factor as a factor.
= 9x2(2 x + 3)
Factor out the greatest common factor.
b. In this situation, the greatest common factor is the common binomial factor
(x + 3). We factor out this common factor as follows.
x2(x + 3) + 5(x + 3) = (x + 3)(x2 + 5)
Factor out the common binomial factor.
A Strategy for Factoring ax2 + bx + c
(Assume, for the moment, that there is no greatest common factor.)
1. Find two First terms whose product is ax2:
( x + )( x + ) = ax2 + bx + c
2. Find two Last terms whose product is c:
(x + )(x + ) = ax2 + bx + c
3. By trial and error, perform steps 1 and 2 until the sum of the Outside
product and Inside product is bx:
( x + )( x + ) = ax2 + bx + c
I
O
(sum of O + I)
If no such combinations exist, the polynomial is prime.
Text Example
Factor: a. x2 + 6x + 8
b. x2 + 3x – 18
Solution
a. The factors of the first term are x and x: (x
)( x
)
To find the second term of each factor, we must find two numbers whose
product is 8 and whose sum is 6.
Factors of 8
Sum of Factors
8, 1
4, 2
-8, -1
-4, -2
9
6
-9
-6
This is the
desired sum.
From the table above, we see that 4 and 2 are the required integers. Thus,
x2 + 6x + 8 = (x + 4)( x + 2) or (x + 2)( x + 4).
Text Example cont.
Factor: a. x2 + 6x + 8
b. x2 + 3x – 18
Solution
b. We begin with x2 + 3x – 18 = (x )( x ).
To find the second term of each factor, we must find two numbers whose
product is –18 and whose sum is 3.
Factors of -8
Sum of Factors
18, -1
-18, 1
9, -2
-9, 2
6, -3
-6, 3
17
-17
7
-7
3
-3
This is the
desired sum.
From the table above, we see that 6 and –3 are the required integers. Thus,
x2 + 3x – 18 = (x + 6)(x – 3) or (x – 3)(x + 6).
Text Example
Factor: 8x2 – 10x – 3.
Solution
Step 1 Find two First terms whose Õ
product is 8x2.
8x2 – 10x – 3 Õ (8x )(x )
8x2 – 10x – 3 (4x )(2x )
Step 2 Find two Last terms whose product is –3. The possible factors are
1(-3) and –1(3).
Step 3 Try various combinations of these factors. The correct factorization
of 8x2 – 10x – 3 is the one in which the sum of the Outside and Inside products
is equal to –10x. Here is a list of possible factors.
Text Example cont.
Possible Factors of
8x2 – 10x – 3
Sum of Outside and Inside
Products (Should Equal –10x)
(8x + 1)(x – 3)
-24x + x = -23x
(8x – 3)(x + 1)
8x – 3x = 5x
(8x – 1)(x +3)
24x – x = 23x
(8x + 3)(x – 1)
-8x + 3x = -5x
(4x + 1)(2x – 3)
-12x + 2x = -10x
(4x – 3)(2x + 1)
4x – 6x = -2x
(4x – 1)(2x + 3)
12x – 2x = 10x
(4x + 3)(2x – 1)
-4x + 6x = 2x
Thus, 8x2 – 10x – 3 = (4x + 1)(2x – 3) or (2x – 3)(4x + 1).
This is the required
middle term.
The Difference of Two Squares
• If A and B are real numbers, variables, or
algebraic expressions, then
• A2 – B2 = (A + B)(A – B).
• In words: The difference of the squares of
two terms factors as the product of a sum
and the difference of those terms.
Text Example
• Factor: 81x2 - 49
Solution:
81x2 – 49 = (9x)2 – 72 = (9x + 7)(9x – 7).
Factoring Perfect Square Trinomials
Let A and B be real numbers, variables, or
algebraic expressions,
1. A2 + 2AB + B2 = (A + B)2
2. A2 – 2AB + B2 = (A – B)2
Text Example
• Factor: x2 + 6x + 9.
Solution:
x2 + 6x + 9 = x2 + 2 · x · 3 + 32 = (x + 3)2
Text Example
• Factor: 25x2 – 60x + 36 .
Solution:
25x2 – 60x + 36 = (5x)2 – 2 · 5x · 6 + 62 = (x + 3)2.
Factoring the Sum and
Difference of 2 Cubes
Type
Example
A3 + B3
= (A + B)(A2 – 2AB + B2)
x3 + 8 = x3 + 23
= (x + 2)( x2 – x·2 + 22)
= (x + 2)( x2 – 2x + 4)
A3 – B3
= (A – B)(A2 + 2AB + B2)
64x3 – 125 = (4x)3 – 53
= (4x – 5)(4x)2 + (4x)(5) + 52)
= (4x – 5)(16x2 + 20x + 25)
A Strategy for Factoring a Polynomial
1. If there is a common factor, factor out the GCF.
2. Determine the number of terms in the
polynomial and try factoring as follows:
a) If there are two terms, can the binomial be factored
by one of the special forms including difference of
two squares, sum of two cuubes, or difference of two
cubes?
b) If there are three terms, is the trinomial a perfect
square trinomial? If the trinomial is not a perfects
square trinomial, try factoring by trial and error.
c) If there are four or more terms, try factoring by
grouping.
3. Check to see if any factors with more than one
term in the factored polynomial can be factored
further. If so, factor completely.
Example
Factor: x3 – 5x2 – 4x + 20
Solution
x3 – 5x2 – 4x + 20
= (x3 – 5x2) + (-4x + 20)
= x2(x – 5) – 4(x – 5)
= (x – 5)(x2 – 4)
= (x – 5)(x + 2)(x – 2)
Group the terms with common factors.
Factor from each group.
Factor out the common binomial factor, (x – 5).
Factor completely by factoring x2 – 4 as the difference of two
squares.
Factoring
Polynomials
Rational
Expressions
Text Example
Find all the numbers that must be excluded from the domain of each
rational expression.
a
a.
x2
x
b. 2
x 1
SolutionTo determine the numbers that must be excluded from each
domain, examine the denominators.
a
a.
x2
This denominator
would equal zero
if x = 2.
x
x
b. 2

x  1 (x  1)( x  1)
This denominator
would equal zero
if x = -1.
This denominator
would equal zero
if x = 1.
Simplifying Rational Expressions
1. Factor the numerator and denominator
completely.
2. Divide both the numerator and
denominator by the common factors.
Example
• Simplify:
Solution:
x 4
4x  8
2
x  4 ( x  2)(x  2) x  2


4x  8
4( x  2)
4
2
Multiplying Rational Expressions
1. Factoring all numerators and
denominators completely.
2. Dividing both the numerator and
denominator by common factors.
3. Multiply the remaining factors in the
numerator and multiply the remaining
factors in the denominator.
Example
2
2
2
x

3
x
x
1
• Multiply and simplify:
 2
2
2x  x  3 x  2x
Solution:
2 x 2  3x
x2 1
 2

2
2x  x  3 x  2x
x(2 x  3)
( x  1)(x  1)


(2 x  3)(x  1)
x( x  2)
x 1
x2
Example
• Divide and simplify:
3x  6 x 3x  x

2
6 x  24 x  2
2
2
Solution:
3x 2  6 x 3x 2  x


2
6 x  24 x  2
3x 2  6 x x  2
 2
2
6 x  24 3 x  x
3x( x  2)
x2


6( x  2)(x  2) 3 x( x  1)
1
1
1


6 x 1 6x  6
Example
• Add:
2x
3

3x  1 3x  1
Solution:
2x
3
2x  3


3x  1 3x  1 3x  1
Finding the Least Common
Denominator
1. Factor each denominator completely.
2. List the factors of the first denominator.
3. Add to the list in step 2 any factors of the
second denominator that do not appear in
the list.
4. Form the product of each different factor
from the list in step 3. This product is the
least common denominator.
Adding and Subtracting Rational
Expressions That Have Different
Denominators with Shared Factors
1. Find the least common denominator.
2. Write all rational expressions in terms of the least
common denominator. To do so, multiply both the
numerator and the denominator of each rational
expression by any factor(s) needed to convert the
denominator into the least common denominator.
3. Add or subtract the numerators, placing the resulting
expression over the least common denominator.
4. If necessary, simplify the resulting rational
expression.
Example
• Subtract:
4
2

2
5a  5 a 5 a  5
Solution:
4
2


2
5a  5a 5a  5
4
2


5a (a  1) 5(a  1)
4
2
a

 
5a (a  1) 5(a  1) a
4
2a


5a (a  1) 5(a  1)a
4  2a
5a (a  1)
Rational
Expressions
Linear Equations
Terms Involving Equations
3x - 1 = 2
Left Side
Right Side
An equation consists of two algebraic expressions joined by an equal sign.
3x – 1 = 2
3x = 3
x=1
1 is a solution or root of the equation
Definition of a Linear Equation
• A linear equation in one variable x is an
equation that can be written in the form
• ax + b = 0
• where a and b are real numbers and a = 0.
Generating Equivalent Equations
An equation can be transformed into an equivalent equation by one or more of the following
operations.
Example
1. Simplify an expression by
removing grouping symbols and
combining like terms.
3(x - 6) = 6x - x
3x - 18 = 5x
2. Add (or subtract) the same
real number or variable
expression on both sides of the
equation.
3x - 18 = 5x
3x - 18 - 3x = 5x - 3x
-18 = 2x
3. Multiply (or divide) on both
sides of the equation by the same
nonzero quantity.
-18 = 2x
-9 = x
4. Interchange the two sides of
the equation.
-9 = x
x = -9
Subtract 3x from both
sides of the equation.
Divide both sides of the
equation by 2.
Solving a Linear Equation
• Simplify the algebraic expression on each
side.
• Collect all the variable terms on one side
and all the constant terms on the other side.
• Isolate the variable and solve.
• Check the proposed solution in the original
equation.
Text Example
Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2).
Solution
Step 1
Simplify the algebraic expression on each side.
2(x - 3) – 17 = 13 – 3(x + 2)
2x – 6 – 17 = 13 – 3x – 6
2x – 23 = - 3x + 7
This is the given equation.
Use the distributive property.
Combine like terms.
Text Example
Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2).
Solution
Step 2
Collect variable terms on one side and constant terms on
the other side. We will collect variable terms on the left by adding 3x to
both sides. We will collect the numbers on the right by adding 23 to both
sides.
2x – 23 + 3x = - 3x + 7 + 3x
5x – 23 = 7
5x – 23 + 23 = 7 + 23
5x = 30
Add 3x to both sides.
Simplify.
Add 23 to both sides.
Simplify.
Text Example
Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2).
Solution
Step 3
Isolate the variable and solve. We isolate the variable by
dividing both sides by 5.
5x = 30
5x/5 = 30/5
x=6
Divide both sides by 5
Simplify.
Text Example
Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2).
Solution
Step 4
Check the proposed solution in the original equation.
Substitute 6 for x in the original equation.
2(x - 3) - 17 = 13 - 3(x + 2)
?
2(6 - 3) - 17 = 13 - 3(6 + 2)
?
2(3) - 17 = 13 - 3(8)
?
6 – 17 = 13 – 24
-11 = -11
The solution set is {6}.
This is the original equation.
Substitute 6 for x.
Simplify inside parentheses.
Multiply.
This true statement indicates that 6 is the solution.
Types of Equations
• Identity:An equation that is true for all real
numbers.
• Conditional: An equation that is true for at
least one real number.
• Inconsistent: An equation that is not true
for any real number.
Example
Determine whether the equation 3(x - 1) = 3x + 5 is an identity, a conditional
equation, or an inconsistent equation.
Solution
To find out, solve the equation.
3(x – 1) = 3x + 5
3x – 3 = 3x + 5
-3 = 5
This equation is inconsistent.
Linear Equations
Quadratic Equations
Definition of a Quadratic Equation
• A quadratic equation in x is an equation
that can be written in the standard form
• ax2  bx  c  0
• where a, b, and c are real numbers with a
not equal to 0. A quadratic equation in x is
also called a second-degree polynomial
equation in x.
The Zero-Product Principle
If the product of two algebraic expressions is
zero, then at least one of the factors is equal
to zero.
If AB  0, then A  0 or B 0.
Solving a Quadratic Equation by
Factoring
1. If necessary, rewrite the equation in the form
ax2  bx  c  0, moving all terms to one side,
thereby obtaining zero on the other side.
2. Factor.
3. Apply the zeroproduct principle, setting each
factor equal to zero.
4. Solve the equations in step 3.
5. Check the solutions in the original equation.
Text Example
• Solve 2x2  7x  4 by factoring and then using the
zeroproduct principle.
Step 1 Move all terms to one side and obtain
zero on the other side. Subtract 4 from both sides
and write the equation in standard form.
2x2  7x  4  4  4
2x2  7x  4 0
Step 2 Factor.
2x2  7x  4 0
(2x  1)(x  4)  0
Solution cont.
• Solve 2x2  7x  4 by factoring and then
using the zeroproduct principle.
Steps 3 and 4 Set each factor equal to
zero and solve each resulting equation.
2 x  1 0
or x  4 0
2x1
x 4
x = 1/2
Steps 5 check your solution
Example
(2x + -3)(2x + 1) = 5
4x2 - 4x - 3 = 5
4x2 - 4x - 8 = 0
4(x2-x-2)=0
4(x - 2)*(x + 1) = 0
x - 2 = 0, and x + 1 = 0
So x = 2, or -1
The Square Root Method
If u is an algebraic expression and d is a
positive real number, then u2 = d has exactly
two solutions.
If u2 = d, then u = d or u = -d
Equivalently,
If u2 = d then u = d
Completing the Square
If x2 + bx is a binomial, then by adding (b/2)
2, which is the square of half the coefficient
of x, a perfect square trinomial will result.
That is,
x2 + bx + (b/2)2 = (x + b/2)2
Text Example
What term should be added to the binomial x2
+ 8x so that it becomes a perfect square
trinomial? Then write and factor the
trinomial.
The term that should be added is the square of
half the coefficient of x. The coefficient of x
is 8. Thus, (8/2)2 = 42. A perfect square
trinomial is the result.
x2 + 8x + 42 = x2 + 8x + 16 = (x + 4)2
Quadratic Equation
ax  bx  c  0
2
Quadratic Formula
 b  b  4ac
x
2a
2
x  8x  5  0
2
 (8)  (8)  4(1)(5)
x
2(1)
2
8  64  20
x
2
8  44
x
2
8  2 11
x
2
2(4  11)
x
2
x  4  11
The Discriminant and the Kinds of Solutions
to ax2 + bx +c = 0
Discriminant
b2 – 4ac
Kinds of solutions
to ax2 + bx + c = 0
b2 – 4ac > 0
Two unequal real solutions
Graph of
y = ax2 + bx + c
Two x-intercepts
b2 – 4ac = 0
One real solution
(a repeated solution)
One x-intercept
b2 – 4ac < 0
No real solution;
two complex imaginary
solutions
No x-intercepts
The Pythagorean Theorem
The sum of the squares of the lengths of the
legs of a right triangle equals the square of
the length of the hypotenuse.
If the legs have lengths a and b, and the
hypotenuse has length c, then
a 2 + b2 = c2
Quadratic Equations
Linear Inequalities
Graphs of Inequalities; Interval Notation
There are infinitely many solutions to the
inequality x > 4, namely all real numbers
that are greater than 4. Although we
cannot list all the solutions, we can make a
drawing on a number line that represents
these solutions. Such a drawing is called the
graph of the inequality.
Graphs of Inequalities; Interval Notation
•
Graphs of solutions to linear inequalities
are shown on a number line by shading all
points representing numbers that are
solutions. Parentheses indicate endpoints
that are not solutions. Square brackets
indicate endpoints that are solutions.
Text Example
Graph the solutions of
a. x < 3 b. x  1 c. 1< x  3.
Solution:
a. The solutions of x < 3 are all real numbers that are
less than 3. They are graphed on a number line by
shading all points to the left of 3. The parenthesis
at 3 indicates that 3 is not a solution, but numbers
such as 2.9999 and 2.6 are. The arrow shows that
the graph extends indefinitely to the left.
-5
-4
-3
-2
-1
0
1
2
3
Text Example cont.
Graph the solutions of
a. x < 3 b. x  1 c. 1< x  3.
Solution:
b. The solutions of x  1 are all real numbers that are
greater than or equal to 1. We shade all points to
the right of 1 and the point for 1 itself The
bracket at 1 shows that 1 is a solution for the
given inequality. The arrow shows that the graph
extends indefinitely to the right.
-5
-4
-3
-2
-1
0
1
2
3
Text Example cont.
Graph the solutions of
a. x < 3 b. x  1 c. 1< x  3.
Solution:
c. The inequality 1< x  3 is read "1 is less than x
and x is less than or equal to 3," or "x is greater
than 1 and less than or equal to 3." The solutions
of 1< x  3 are all real numbers between 1 and
3, not including 1 but including 3. The
parenthesis at 1 indicates that 1 is not a solution.
By contrast, the bracket at 3 shows that 3 is a
solution. Shading indicates the other solutions.
-5
-4
-3
-2
-1
0
1
2
3
Properties of Inequalities
Property
The Property In Words
Example
Addition and Subtraction
properties
If a < b, then a  c < b  c.
If a < b, then a  c < b  c.
If the same quantity is added to or
subtracted from both sides of an
inequality, the resulting inequality
is equivalent to the original one.
2x  3 < 7
subtract 3:
2x  3  3 < 7  3
Simplify: 2x < 4.
Positive Multiplication
and Division Properties
If a < b and c is positive,
then ac < bc.
If a < b and c is positive,
then a  c < b  c.
If we multiply or divide both sides
of an inequality by the same
positive quantity, the resulting
inequality is equivalent to the
original one.
2x < 4
Divide by 2:
2x  2 < 4  2
Simplify: x < 2
Negative Multiplication
and Division Properties
If a < b and c is negative,
then ac  bc.
If a < b and c is negative,
then a  c  b  c.
if we multiply or divide both sides
of an inequality by the same
negative quantity and reverse the
direction of the inequality symbol,
the result is an equivalent
inequality.
4x < 20
Divide by –4 and
reverse the sense of
the inequality:
4x  4  20  4
Simplify: x 5
Example
Solve and graph the solution set on a number line:
4x  5  9x  10.
Solution We will collect variable terms on the left and constant terms on
the right.
4x  5  9x  10
4x  5 – 9x  9x  10  9x
5x  5  10
5x  5  5  10  5
5x  15
-5x/5 > -15/5
This is the given inequality.
Subtract 9x from both sides.
Simplify.
Subtract 5 from both sides.
Simplify.
Divide both sides by 5 and reverse the sense
of the inequality.
Simplify.
x3
The solution set consists of all real numbers that are greater than or equal to
3, expressed in interval notation as (, 3]. The graph of the solution set is
shown as follows:
Solving an Absolute Value
Inequality
If X is an algebraic expression and c is a
positive number,
1. The solutions of |X| < c are the numbers that
satisfy c < X < c.
2. The solutions of |X| > c are the numbers that
satisfy X < c or X > c.
These rules are valid if < is replaced by  and
> is replaced by .
Text Example
Solve and graph: |x  4| < 3.
|X| < c means
Solution
c < X < c
|x  4| < 3 means 3< x  4< 3
We solve the compound inequality by adding 4 to all
three parts.
3 < x  4 < 3
3  4 < x  4  4 < 3  4
1< x<7
The solution set is all real numbers greater than 1 and
less than 7, denoted by {x| 1 < x < 7} or (1, 7). The
graph of the solution set is shown as follows:
Linear Inequalities