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Section 6.1
Rational Expressions
OBJECTIVES
A
Find the numbers that
make a rational
expression undefined.
OBJECTIVES
B
Write an equivalent
fraction with the
indicated denominator.
OBJECTIVES
C
Write a fraction in the
standard forms.
OBJECTIVES
D
Reduce a fraction to
lowest terms.
DEFINITION
Rational Expressions
If P and Q are polynomials:
P
(Q  0)
Q
DEFINITION
Undefined Rational
Expressions
The variables in a rational
expression may not be
replaced by values that will
make the denominator zero.
DEFINITION
Fundamental Property of
Fractions
If P, Q, and K are polynomials
P =P • K
Q Q•K
PROCEDURE
Reducing Fractions
1. Write numerator and
denominator in factored
form.
2. Find the GCF.
PROCEDURE
Reducing Fractions
3. Replace the quotient of the
common factors by 1.
4. Rewrite in lowest terms.
DEFINITION
Quotient of Additive Inverses
a – b = –1
b–a
Chapter 6
Section 6.1A,B
Practice Test
Exercise #1
Find the undefined value(s) for
x –2
a.
3x + 4
is undefined when:
3x + 4 = 0
3x = – 4
4
x =–
3
Write the fraction with the indicated
denominator.
2x 2
?
b.
=

4
9y
36 y 7
 36y 7 = 9y 4 • 4y 3 


2x 2 • 4y 3
9y 4 • 4y 3
=
8x 2 y 3
36y 7
Chapter 6
Section 6.1C
Practice Test
Exercise #2
Write in standard form
–5
a. –
y
=
5
y
Write in standard form
x–y
b. –
5
x+y
–
=
5
y–x
=
5
Chapter 6
Section 6.1D
Practice Test
Exercise #4
Reduce to lowest terms.
y2 – x 2
x3 – y3
=
=

Factor out – 1

–1 x 2 –y 2 Difference of Squares
x 3 –y 3
Difference of Cubes
–  x –y   x +y 
 x –y   x 2 + xy + y 2 
Reduce to lowest terms.
=
=
–  x –y   x +y 
 x –y   x 2 + xy + y 2 
–  x +y 
x 2 + xy + y 2
Section 6.2
Multiplication and
Division of Rational
Expressions
OBJECTIVES
A
Multiply rational
expressions.
OBJECTIVES
B
Divide rational
expressions.
OBJECTIVES
C
Use multiplication and
division together.
DEFINITION
Multiplication of Rational
Expressions
a • c =a•c
b d b•d
(b  0, d  0)
PROCEDURE
To Multiply Rational
Expressions
1. Factor the numerators and
denominators completely.
2. Simplify each expression.
PROCEDURE
To Multiply Rational
Expressions
3. Multiply remaining factors.
4. The final product should
be in lowest terms.
DEFINITION
Division of Real Numbers
a c a • d
÷ =
b d b c
(b, d, and c  0)
Chapter 6
Section 6.2B
Practice Test
Exercise #6
Perform the indicated operations.
2– x
x 3 – 8 x 3 + 27
÷

x +3
x +5
x +5
 x + 3

x 2 – 3x + 9
–  x – 2
x +5
x 3 + 27
=


3
x +3
x +5
x – 8
 x – 2

x 2 + 2x + 4


Perform the indicated operations.
–  x – 2
x +5
=


x +3
 x – 2 x 2 + 2x + 4

–  x 2 – 3x + 9
=
x 2 + 2x + 4

 x + 3

x 2 – 3x + 9
x +5

Chapter 6
Section 6.2C
Practice Test
Exercise #7
Perform the indicated operations.
2– x
x 3 – 8 x 3 + 27
÷

x +3
x +5
x +5
 x + 3

x 2 – 3x + 9
–  x – 2
x +5
x 3 + 27
=


3
x +3
x +5
x – 8
 x – 2

x 2 + 2x + 4


Perform the indicated operations.
–  x – 2
x +5
=


x +3
 x – 2 x 2 + 2x + 4

–  x 2 – 3x + 9
=
x 2 + 2x + 4

 x + 3

x 2 – 3x + 9
x +5

Section 6.3
Addition and
Subtraction of
Rational Expressions
OBJECTIVES
A
Add or subtract rational
expressions with the
same denominator.
OBJECTIVES
B
Add or subtract rational
expressions with
different denominators.
PROCEDURE
Finding the LCD of Two or
More Rational Expressions
1. Factor denominators.
Place factors in columns.
(Not necessary to factor
monomials).
PROCEDURE
Finding the LCD of Two or
More Rational Expressions
2. Select the factor with the
greatest exponent from
each column.
PROCEDURE
Finding the LCD of Two or
More Rational Expressions
3. The product of all the
factors obtained is the LCD.
PROCEDURE
To Add or Subtract Fractions
with Different Denominators.
1. Find the LCD.
2. Write all fractions as
equivalent ones with LCD
as denominator.
PROCEDURE
To Add or Subtract Fractions
with Different Denominators.
3. Add numerators.
4. Simplify.
Chapter 6
Section 6.3B
Practice Test
Exercise #9a
Perform the indicated operations.
a.
x +1
x2 + x – 2
+
x +4
x2 – 1
Perform the indicated operations.
a.
x +1
x2 + x – 2
+
 x +1
= 
 x + 2  x – 1
x +4
x2 – 1
+
 x + 4
 x +1 x – 1
LCD =  x + 2  x – 1 x +1
Perform the indicated operations.
LCD =  x + 2  x – 1 x +1
=
 x +1  x + 1
 x + 4  x + 2
+
 x + 2  x – 1  x + 1  x +1 x – 1  x + 2 
= 
 x 2 + 2 x +1 +  x 2 + 6 x + 8 
 x + 2  x – 1 x +1
x 2 + 2x + 1 + x 2 + 6x + 8
= 
 x + 2  x – 1 x +1
Perform the indicated operations.
x 2 + x 2 + 2 x + 6 x +1+ 8
= 
 x + 2  x – 1 x +1
2x 2 + 8x + 9
= 
 x + 2  x – 1 x +1
Section 6.4
Complex Fractions
OBJECTIVES
A
Write a complex fraction
as a simple fraction in
reduced form.
PROCEDURE
Simplifying Complex Fractions
METHOD 1
Multiply the numerator and
denominator of the complex
fraction by the LCD of all
simple fractions.
PROCEDURE
Simplifying Complex Fractions
METHOD 2
Perform operations indicated in
numerator and denominator.
Then divide numerator by
denominator.
Chapter 6
Section 6.4A
Practice Test
Exercise #10
Simplify.
x+
x–
1
x2
1
x3
Multiply by LCD
Simplify.

x +
= 

x –

=

1  3
 •x
2
x4 + x
x 
=
4
1  3
x
–1
•x

x3 

x x 3 +1



x 2 +1 x 2 – 1
Simplify.
=
=


x x 3 +1

2

x +1 x – 1


x x 2  x +1


=
 
2
x 2 +1  x  1


x  x +1 x 2 – x +1

x 2 +1  x +1 x – 1
Section 6.5
Division of
Polynomials and
Synthetic Division
OBJECTIVES
A
Divide a polynomial by a
monomial.
OBJECTIVES
B
Use long division to
divide one polynomial
by another.
OBJECTIVES
C
Completely factor a
polynomial when one of
the factors is known.
OBJECTIVES
D
Use synthetic division to
divide one polynomial
by a binomial.
OBJECTIVES
E
Use the remainder
theorem to verify that a
number is a solution of a
given equation.
RULE
Dividing a Polynomial by a
Monomial
Divide each term in the
polynomial by the monomial.
DEFINITION
The Remainder Theorem
If P(x) is divided by x – k ,
then the remainder is P(k).
DEFINITION
The Factor Theorem
When P(x) has a factor (x –k) ,
it means that P(k) = 0.
Chapter 6
Section 6.5B
Practice Test
Exercise #13
Divide.


2x 3 – 6 – 4 x ÷  2 + 2 x 
Write in descending order.
 2x 3 +0x 2 – 4 x – 6  ÷  2x + 2 
Use 0x 2 for missing term.
Divide.
x 2 –x –1
2x+ 2 2x 3 + 0x 2 – 4x – 6
2x 3 + 2x 2
– 2x 2 – 4x
– 2x 2 – 2x
– 2x – 6
– 2x – 2
– 4 Remainder
Chapter 6
Section 6.5C
Practice Test
Exercise #14
Factor 2x 3 + 3x 2 – 23x – 12 if x – 3 is one of its factors.
2x 2 + 9x + 4
x –3
2x 3 + 3x 2 – 23x – 12
2x 3 – 6x 2
9x 2 – 23x
2
9x – 27x
4x – 12
4x – 12
0
Factor 2x 3 + 3x 2 – 23x – 12 if x – 3 is one of its factors.
Factor 2x 2 + 9x + 4
=  2x + 1 x + 4 
Factors of 2x 3 + 3x 2 – 23x – 12 are :
 x – 3  2x + 1 x + 4 
Chapter 6
Section 6.5E
Practice Test
Exercise #16
Use synthetic division to show that – 1 is a solution of
x 4 – 4x 3 – 7x 2 + 22x + 24 = 0
–1
1 –4
–1
–7
+5
22 24
+2 –24
1 –5
–2
24
(0)
–1 is a solution of the equation
since the remainder R=0.
Section 6.6
Equations Involving
Rational Expressions
OBJECTIVES
A
Solve equations
involving rational
expressions.
OBJECTIVES
B
Solve applications using
proportions.
PROCEDURE
Solving Equations Containing
Rational Expressions
1. Factor denominators and
multiply both sides of the
equation by the LCD.
PROCEDURE
Solving Equations Containing
Rational Expressions
2. Write the result in reduced
form. Use the distributive
property to remove
parentheses.
PROCEDURE
Solving Equations Containing
Rational Expressions
3. Determine whether the
equation is linear or
quadratic and solve
accordingly.
PROCEDURE
Solving Equations Containing
Rational Expressions
4. Check that the proposed
solution satisfies the
equation. If not, discard it
as an extraneous solution.
DEFINITION
Property of Proportions
a
c
If = (where b, d  0),
b d
then a • d = b • c
A proportion is true if the
cross products are equal.
Chapter 6
Section 6.6A
Practice Test
Exercise #18
Solve: 18x – 2 – 3x – 1 = 1
1
–
2
x =
,
x2
18 • x 2
x2
1
–
1
x =
x
3 • x2
–
= 1 • x2
x
Solve: 18x – 2 – 3x – 1 = 1
x
2
2
18 • x
3 •x
–
= 1 • x2
x 1
x2
18 – 3x = x 2
O
0 = x 2 + 3x – 18
F
0 = (x + 6)(x – 3)
F
x + 6 = 0 or x – 3 = 0
x = –6
or
x =3
Chapter 6
Section 6.6B
Practice Test
Exercise #19
A recipe for curried shrimp that normally serves
four was once served to 200 guests at a wedding
reception. One of the ingredients in the recipe is
1 1 cups of chicken broth.
2
a. How much chicken broth was
required to make the recipe for
200 people?
a. People
4
200
=
=
Broth
1
x
1
2
3
•
4x = 200
2
4x = 300
300
x=
4
= 75 cups
b. If a medium-sized can of chicken broth
contains 2 cups of broth, how many
cans are necessary?
75 cups
2 cups
1
= 37
2
 38 cans
Section 6.7
Applications:
Problem Solving
OBJECTIVES
A
Solve integer problems.
OBJECTIVES
B
Solve work problems.
OBJECTIVES
C
Solve distance
problems.
OBJECTIVES
D
Solve for a specified
variable.
PROCEDURE:
RSTUV Method for Solving
Word Problems
Read
Select
Think
Use
Verify
Chapter 6
Section 6.7B
Practice Test
Exercise #21
Jack can mow the lawn in 4 hours and Jill can
mow it in 3. How long would it take them
to mow the lawn if they work together?
let x = Time working together (hr)
Time worked
= amount done
Time working alone
x
Jack does of the work
4
x
Jill does of the work
3
Jack can mow the lawn in 4 hours and Jill can
mow it in 3. How long would it take them
to mow the lawn if they work together?
Together they do 1 full job
x x
+ =1
4 3
3x + 4x = 12
7x = 12
Jack can mow the lawn in 4 hours and Jill can
mow it in 3. How long would it take them
to mow the lawn if they work together?
12
x=
7
5
or 1
7
5
It takes 1 hours if they work together.
7
Section 6.8
Variation
OBJECTIVES
A
Direct variation.
OBJECTIVES
B
Inverse variation.
OBJECTIVES
C
Joint variation.
OBJECTIVES
D
Solve applications
involving direct, inverse,
and joint variation.
DEFINITION
Direct Variation
y varies directly as x if
there is a constant k:
y = kx
DEFINITION
Inverse Variation
y varies inversely as x if
there is a constant k:
k
y =
x
DEFINITION
Joint Variation
z varies jointly with x and y if
there is a constant k:
z = kxy
Chapter 6
Section 6.8A
Practice Test
Exercise #24
C is directly proportional to m.
a. Write an equation of variation
with k as the constant.
1
b. Find k when C = 12 and m = .
3
C is directly proportional to m.
a. Write an equation of variation
with k as the constant.
Direct Variation
y = kx
C = km
C is directly proportional to m.
1
b. Find k when C = 12 and m = .
3
C = km
1
12 = k
3
k = 36