Mathematics A30
Module 1
Lesson 3
Mathematics A30
Polynomials and Rational Expressions
119
Lesson 3
Polynomials and Rational Expressions
Introduction
This lesson begins with the division of polynomials using long division and synthetic
division. There are many different strategies to help you factor polynomials and solve
problems more efficiently. This lesson teaches you a few of these strategies such as the
factor theorem and the remainder theorem.
The next topic is rational expressions. Rational expressions combine two different thought
processes. One thought process uses the properties of rational numbers. If you can recall
from past mathematics courses, rational numbers are numbers that can be written in the
form of a fraction, providing the denominator is not equal to zero.
The other thought process extends the basic principles of rational expressions to include
polynomial expressions. From the previous lesson, you have seen how these expressions
can be factored and simplified.
A graphing calculator is used in the last section of this lesson to show how to graph the
relation determined by an equation which contains a rational expression with a nonpermissible values. If you are doing some of the questions in the exercises using a
graphing calculator it may not be necessary to complete the table of values. Instead,
it will be sufficient to draw a sketch of the graph since the shape of the graph is more
important than the actual points.
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Lesson 3
Objectives
After completing this lesson, you will be able to
•
divide a polynomial by a binomial.
•
factor polynomials using the factor theorem.
•
use the remainder theorem to determine the remainder when a polynomial is
divided by (x – r).
•
simplify rational expressions involving opposites.
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Lesson 3
3.1 Division of Polynomials
Division of a polynomial by a binomial uses the same pattern ideas as doing this process
with numbers. Just as 276 is divided by 23 using long division, many polynomials can be
divided by binomials using the same procedure.
Example 1
Divide 3 y 2  10 y  8 by y  2 using long division.
quotient
Solution:
3 y2
Since
 3 y , this becomes the first term in the quotient.
y
Multiply ( y  2 ) by 3y and subtract to eliminate the 3 y 2 .
(Remember:  10 y   6 y   10 y  6 y  4 y )
4y
 4 , this becomes the next term in the quotient.
y
Multiply ( y  2 ) by (4 ) and subtract to eliminate the  4 y .
Since
•
3y 4
y  2 3 y  10 y  8
2
3 y2  6 y
4y8
4y8
0
The remainder is 0.
Check:
?
( y  2)( 3 y  4 )  3 y 2  10 y  8
?
3 y 2  4 y  6 y  8  3 y 2  10 y  8
3 y 2  10 y  8  3 y 2  10 y  8 ()
Therefore (3 y 2  10 y  8)  ( y  2)  3 y  4 .
•
3 y  4 is the quotient when 3 y 2  10 y  8 is divided by y  2 .
Polynomials may be without one or more terms. It is important in long division to have
every term represented, and it is therefore necessary to put zeros as coefficients in the
missing terms.
For example x 3  7 x  1 has the squared term missing. For division purposes, it should be
written x 3  0 x 2  7 x  1 .
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Lesson 3
Example 2
Divide 5 x 4  9 x 2  44 by x  2 .
Solution:
Express every degree of the dividend by putting zeros for the coefficients.
•
5 x 4  9 x 2  44  5 x 4  0 x 3  9 x 2  0 x  44
5 x 3  10 x 2  11 x  22
x  2 5 x  0 x 3  9 x 2  0 x  44
4
5 x 4  10 x 3
10 x 3  9 x 2
10 x 3  20 x 2
11 x 2  0 x
11 x 2  22 x
22 x  44
22 x  44
0
The quotient, when 5 x 4  9 x 2  44 is divided by x  2 , is 5 x 3  10 x 2  11 x  22 .
Some problems result in a non zero remainder. When this happens, the answer is
expressed as a mixed fraction.
Example 3
Evaluate
3x2  5x 1
using long division.
x 2
Solution:
Divide x  2 into 3 x 2  5 x  1 .
3 x  11
x  2 3x2  5x 1
3x2  6x
11 x  1
11 x  22
21
3x2  5x 1
21
 3 x  11 
.
x 2
x 2
Express the answer as a mixed fraction.
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remainder
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Lesson 3
Synthetic Division
Synthetic Division is a simpler method of dividing a polynomial by another polynomial
that is in the form x  k where the coefficient of x is 1.
The problem is set up with only the coefficients of the dividend, and the divisor is the k
term from x  k .
All the terms that are missing in the dividend must be represented with a zero.
To divide ax 3  bx 2  cx  d by x  k , use the following pattern.
k a
b
c
d
ka
a
aa
•
•
•
r
Remainder
Vertical Pattern: Add terms.
Diagonal Pattern: Multiply by k.
The pattern for higher degree polynomials is similar.
Example 4
Use synthetic division to divide x 3  8 x 2  4 x  48 by x  4 .
Solution:
•
From x  4 , k  4 .
•
The coefficients of the dividend are:
x 3  8 x 2  4 x  48
1
8
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48
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Lesson 3
Follow the pattern.
4 1
1
8
4
48
4
 16
 48
4
 12
0
Coefficients of Quotient
Zero Remainder
A written explanation of the steps of synthetic division for this question is:
•
•
•
•
•
•
•
•
•
•
Bring the 1 down and write it in the first column underneath the line.
Multiply 4 × 1 to get 4. Put this 4 underneath the  8 in the 2nd column.
Add  8 and 4. Put the answer,  4 , in the same column underneath the line.
Multiply 4 ×  4 to get  16 . Put this  16 underneath the 4 in the 3rd column.
Add 4 and  16 . Put the answer,  12 , in the same column underneath the line.
Multiply 4 ×  12 to get  48 . Put this  48 underneath the 48 in the 4th column.
Add 48 and  48 . Put the answer, 0, in the same column underneath the line.
The first three terms underneath the line are the three coefficients for the terms
in the answer.
There is no remainder because the last term is zero.
The quotient is 1 x 2  4 x  12 .
•
x 3  8 x 2  4 x  48 divided by x  4 is equal to 1 x 2  4 x  12 .
•
Therefore,
x 3  8 x 2  4 x  48
 x 2  4 x  12
x 4
Example 5
Use synthetic division to divide 2 x 4  6 x 2  3 by x  3 .
Solution:
•
From x + 3, k   3 .
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Lesson 3
Remember that k is originally from the form x  k . This means that k  3 when the
divisor is x  3 .
•
The coefficients of the dividend are:
2x 4  0x 3  6x 2  0x  3
2
0
6
0
3
Remember to put zeros for coefficients where the terms are missing.
Follow the pattern.
3 2
2
0
6
0
3
6
18
 36
108
6
12
 36
111
Coefficients of Quotient
Remainder
•
The answer is 2 x 3  6 x 2  12 x  36 with a remainder of 111.
•
2 x 4  6 x 2  3 divided by x  3 is equal to 2 x 3  6 x 2  12 x  36 
•
Therefore,
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.
x 3
2x4  6x2  3
111
 2 x 3  6 x 2  12 x  36 
.
x 3
x 3
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Lesson 3
Exercise 3.1
1.
Divide, using long division.
a)
b)
c)
d)
e)
f)
g)
h)
2.
x  3 x  9 x  5   x  1
y  3 y  9 y  5    y  5 
m  2m  9 m  18   m  2 
3
2
3
2
3
2
(b 4  6 b)  (b  3)
x 3  2 x 2  5 x  6   x  1
a 3  a 2  18   a  3 
12 t 2  11 t  2   3t  2 
x 3  3 x 2  13 x  15   x  2 
Divide each of the questions in #1, using synthetic division.
Hint for g): The coefficient of t in the divisor must be 1 to use synthetic division.
11
2
4t 2  t 
2
12 t  11 t  2
3
3
Factor out a 3.

2
3t  2
t
3
3.2 Factor Theorem
To start this section, a review of function notation is required. This concept will be
incorporated into this lesson in a very basic manner. Function notation will be studied in
more detail in a later lesson.
Function Notation
If an equation, y  2 x  1 , defines a relation which is a function, it is often written as:
f ( x)  2 x  1
This is an example of function notation and translated in words means:
A function of x is equal to 2x + 1.
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Lesson 3
To find f (3) for this relation, simply substitute 3 into the equation in place of x.
f ( x )  2( x )  1
f (3)  2(3)  1
 6 1
7
Example 1
If f ( x )  3 x  1 , find the value of
a)
b)
f (2)
c)
f (3 )
f (1)
Solution:
a)
f ( x)  3 x  1
f (2)  3(2)  1
b)
f ( x)  3 x  1
c)
f (3)  3(3)  1
 6 1
 9 1
5
8
f ( x)  3 x  1
f ( 1)  3( 1)  1
 3  1
 4
One of the strategies in problem solving is to look for patterns. A pattern can be found
when applying the principles of factoring to polynomials.
A factor of a polynomial is any term that divides evenly into the polynomial without a
remainder.
The zeros of a polynomial are any numbers that make a polynomial equal to zero.
If ab  0 , then either a  0 , b  0 or both a and b are equal to zero.
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Lesson 3
Example 2
Factor the expression x 2  x  6 and find the zeros.
Solution:
Write the expression.
Factor.
Set each factor to zero and solve.
2
x  x 6
= x  3  x  2 
x 3 0
x  3
or
x 2 0
x 2
From this example, you can see a pattern.
•
•
•
If x  3 is a factor, then f  3   0 .
If x  2 is a factor, then f 2   0 .
You can find a factor of the polynomial if a given value of x makes the polynomial
equal to zero.
This is the same with any polynomial. To determine the factors of any polynomial, find a
value x such that f  x   0 .
Factor Theorem
•
•
If f a   0 , then x  a  is a factor.
If x  a  is a factor, then f a   0 .
Example 3
Is x  4 a factor of the polynomial f  x   3 x 2  7 x  20 ?
Solution:
•
•
In the factor x  4 , a  4 .
Determine if f 4   0 .
f  x   3 x 2  7 x  20
Write the polynomial expression.
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Lesson 3
f 4   3 4 2  7 4   20
f 4   48  28  20
f 4   0
Evaluate f (4 ) .
Yes, x  4 is a factor of the polynomial f  x   3 x 2  7 x  20 because f ( 4 )  0 .
Example 4
Is x  1 a factor of x 3  6 x 2  5 x  12 ?
Solution:
•
•
In the factor x  1 , a  1 .
Determine if f 1  0 .
f  x   x 3  6 x 2  5 x  12
Write the polynomial expression.
f 1   13  6 1 2  5 1   12
f 1  1  6  5  12
f 1  12
Evaluae f (1) .
No, x  1 is not a factor of x 3  6 x 2  5 x  12 because f (1)  0 .
Example 5
Is x  5 a factor of x 3  4 x 2  x  20 ?
Solution:
•
•
In the factor x  5, a  5 .
Determine if f ( 5 )  0 .
Write the polynomial expression.
f ( x)  x 3  4 x 2  x  20
Evaluate.
f ( 5)  ( 5) 3  4( 5) 2  ( 5)  20
f ( 5)  125  100  5  20
f ( 5)  0
Yes, x + 5 is a factor of x 3  4 x 2  x  20 , because f ( 5 )  0 .
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Lesson 3
If the factor has a positive sign in it, the value that makes the polynomial equal to
zero will be negative.
•
If (x + 5) is a factor, then f ( 5 )  0 .
The factor theorem is usually used when factoring polynomials with a degree of at least
three. Once the factor theorem has determined one of the factors of the polynomial, that
factor can then be divided into the original polynomial to determine the other factor(s).
Example 6
Factor the polynomial, x 3  7 x  6 .
Solution:
Read the problem.
Factor the polynomial, x 3  7 x  6 .
Determine a plan.
•
Determine a factor of the expression by using the factor theorem and evaluating the
polynomial for f (1) , f  1 , f (2) , etc. until f  x   0 .
•
Divide this factor into x 3  7 x  6 to determine the other factors.
Carry out the plan.
Write the expression.
Evaluate f (1) .
•
f x   x 3  7 x  6
f 1   1 3  7 1   6
f 1  0
Therefore x  1 is a factor of x 3  7 x  6 because f (1) = 0.
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Lesson 3
Divide.
x2  x  6
x  1 x3  0 x2  7 x  6
Write 0x2 to take the place of
the missing x2 term.
x3  x2
x2  7 x
x2  x
 6x  6
 6x  6
0


Two factors of x 3  7 x  6 are  x  1 and x 2  x  6 .
Now factor x 2  x  6 .
x2  x  6
  x  3  x  2 
Write a concluding statement.
Using the factor theorem, x  1 is one factor.
From the quotient, x 2  x  6 , ( x  3), and ( x  2) are also factors.
Therefore, x 3  7 x  6   x  1 x  3  x  2  .
•
•
•
Activity 3.2
•
Provide a check for Example 6.
•
Multiply the factors ( x  1)( x  3)( x  2) together to see if the resulting polynomial is
equal to x 3  7 x  6 .
•
Use the distributive property. First, multiply two of the factors. Then multiply this
polynomial by the last factor.
Example 7
What value does c have to be if x  3 is a factor of 2 x 3  7 x 2  cx  12 ?
Solution:
•
If x  3 is a factor of 2 x 3  7 x 2  cx  12 , then f (3) = 0.
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Lesson 3
Write the original expression.
2 x 3  7 x 2  cx  12
Evaluate for f (3 )  0 .
2(3)3  7(3) 2  c(3)  12  0
54  63  3 c  12  0
 9  3 c  12  0
3 c  21  0
3c = 21
c=7
Therefore c must equal 7 if x  3 is a factor of 2 x 3  7 x 2  cx  12 .
Exercise 3.2
1.
If f a   0 , state one factor of f(x).
a)
b)
c)
d)
e)
2.
a
a
a
a
a
3
 5
7
 2
8
Determine a factor of each of the following polynomials by finding a value a such
that f ( a )  0 .
a)
b)
c)
f ( x)  3 x 2  7 x  10
d)
f ( y)  y3  3 y 2  9 y  27
f ( x)  2 x 3  7 x 2  x  6
f ( x)  2 x 3  7 x 2  x  10
3.
Use the factors that were found in question #2 to factor the polynomial completely.
Use long division or synthetic division.
4.
Determine the value of k in each of the following questions.
a)
b)
c)
x 3  5 x 2  8 x  k ; if x  2 is a factor.
3 a 3  7 a 2  ka  8 ; if a 1 is a factor.
3
2
if y  3 is a factor.
y  ky  y  3 ;
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Lesson 3
3.3 Remainder Theorem
If a polynomial, f ( x ) , is divided by x  a ,
the remainder will be f (a ) .
Example 1
Find the remainder when f ( x)  x 2  6 x  5 is divided by x  2 .
Solution:
•
•
In the divisor x  2 , a  2 .
It is then necessary to find f(2).
Write the expression.
f ( x)  x 2  6 x  5
Substitute f(2).
f ( 2 )  ( 2 ) 2  6( x )  5
 4  6( 2 )  5
Evaluate.
 4  12  5
 3
This is the remainder.
Check using long division .
x4
x  2 x2  6 x  5
x2  2x
 4x  5
 4x  8
 3  Remainder
Therefore, the remainder when x 2  6 x  5 is divided by x  2 is  3 .
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Lesson 3
Example 2
Find the remainder when f ( x)  x 3  2 x 2  6 is divided by x  3 .
Solution:
•
•
In the divisor x  3, a  3 .
It is then necessary to find f (3 ) .
Write the expression.
f ( x)  x 3  2 x 2  6
Substitute f (3 ) .
f ( 3)  ( 3) 3  2( 3) 2  6
Evaluate.
 27  18  6
 39
Remainder
Check using long division.
x 2  5 x  15
x  3 x3  2x2  0x  6
x3  3x2
 5x2  0x
 5 x 2  15 x
15 x  6
15 x  45
 39

Remainder

The remainder of x 3  2 x 2  6   x  3  is  39 .
Example 3
What value does c have to be if, when f(x) = 3 x 3  4 x 2  cx  6 is divided by
x  2 , the remainder is 42?
Solution:
•
•
When 3 x 3  4 x 2  cx  6 is divided by x  2 the remainder is 42.
This means f(2) = 42.
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Lesson 3
Write the expression.
Evaluate for f(2) = 42.
f ( x)  3 x 3  4 x 2  cx  6
42 = 3(2)3  4(2) 2  c(2)  6
42 = 24  16  2 c  6
42 = 2 c  2
40 = 2c
20 = c
Therefore c must equal 20 if the remainder is 42 when 3 x 3  4 x 2  cx  6 is divided by
x 2.
Exercise 3.3
1.
Determine the remainder when the polynomial x 3  2 x 2  4 x  5 is divided by:
a)
c)
e)
2.
x 1
x 2
x 3
x3  2x  6
x4  2x3  x2  3x  7
8x2  4
x 3  5 x 2  12
Determine the value of k in each of the following questions.
a)
b)
c)
4.
b)
d)
f)
Determine the remainder when each of the following polynomials is divided by
x 2.
a)
b)
c)
d)
3.
x 1
x2
x 3
When x 3  4 x 2  x  k is divided by x  1 , there is a remainder of 4.
When x 3  kx 2  5 x  1 is divided by x  1 , there is a remainder of 1.
When 2 x 3  5 x 2  2 x  k is divided by x + 1, there is a remainder of  11 .
Determine the remainder in each of the following questions.
a)
b)
c)
d)
e)
2 x  7 x  3 x  10   ( x  1)
n  n  7 n  2   (3  n )
 3 a  9 a  8   (2  a)
 c  3 c  c  1  (c  2)
 c  3 c  c  1  (c  2)
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Lesson 3
3.4 Simplifying Rational Expressions
A rational number is defined as any number in the form
integers and b is not equal to zero.
Q=
a
where a and b are
b
a
where a, b  I ; b  0
b
Some examples of rational numbers are:
4
•
7
7
22
•
8
•
9.762
13 .2 59
•
A rational expression is any algebraic expression that can be written as
the quotient of two polynomials.
polynomial
Rational expression =
polynomial
Some examples of rational expressions are:
•
3x
x+ 2
•
2
y 5y 6
y2
•
1
9a  8
3
Mathematics A30
140
Lesson 3
Non-Permissible Values
In the same way that a rational number cannot have a denominator that is equal to zero,
a rational expression also cannot have the denominator equal to zero.
Any number that makes the value of the denominator equal to zero is non-permissible.
Another name for these values are excluded values.
Since the denominator of a rational expression is in the form of a polynomial, the
polynomial must first be factored to determine its zeros. Any values that make any of the
factors in the denominator equal to zero are non-permissible.
Example 1
State the non-permissible values in the rational expression
a 3
.
2 a 2  18 a
Solution:
Factor the denominator.
Set each factor equal to zero.
2 a 2  18 a
= 2 a( a  9)
2a  0
a 0
a 9  0
a 9
The non-permissible values for this rational expression are a  0, 9 .
The domain of this rational expression is all real numbers except a  0, 9 .
a  , a  0 , 9
Example 2
State the non-permissible values for the rational expression
4
.
x 2
Solution:
•
The denominator has only one factor, x  2 .
Set the factor in the denominator equal to zero.
Mathematics A30
141
x 2 0
x 2
Lesson 3
The non-permissible value for this rational expression is x  2 .
The domain of this rational expression is all real numbers except x  2 .
x  , x  2
4
4
or y 
shows what happens to the value of the
x 2
x 2
expression for values of x which are near to the excluded value, x = 2.
A graph of the equation f ( x ) 
The following table shows the ordered pairs that satisfy this equation. Values for x were
picked and substituted into the equation to calculate the f ( x ) values. Calculate out a few
x values to verify the f ( x ) values in the chart below.
x
f(x)
1

4
3
0
1
2
4
1
1
2
2
2
8
undefined
1
2
8
3
4
4
5
2
4
3
Using the ordered pairs from the table above, the graph of this equation is:
f(x)
x
asymptote (x = 2)
Mathematics A30
142
Lesson 3
What happens when x is equal to 2?
The vertical line, x = 2 , is called an asymptote. There is no point on this line that will
satisfy the function. Both of the curved lines will go increasingly closer to this line, but will
never touch it. Asymptotes will be explored further in Math B30. Just know, a nonpermissible value in a rational expression determines an asymptote when graphing.
A graphing calculator can be used to graph this function.
Use the following key stroke pattern on the TI-83 Plus graphing calculator.
CLEAR
Y=
4  ( X,T,,n - 2 )
ENTER
GRAPH
On the calculator screen the asymptote at x  2 may appear as a solid vertical
line. Choose the TRACE function key and follow the curved lines to see what
happens as the value of x gets close to 2.
Example 3
Find the non-permissible values for the rational expression,
6
.
x 4
2
Solution:
Factor the denominator.
Set each factor to zero.
x2  4
= ( x  2)( x  2)
x 2 0
x 2
x2 0
x  2
The non-permissible values for the rational expression,
Mathematics A30
143
6
are x  2 ,  2 .
x 4
2
Lesson 3
When graphing example three, first create a table of values.
x
4
3
f (x )
1
2
6
5
2
8
3
1
2
2
und.
1
1
2
1
3
3
7
2
0

3
2
1
1
2
1
2
3
2
2
3
7
und.
1
2
8
3
3
4
6
5
1
2
Take note where the
asymptotes lie. These are the
values where the graph cannot
exist. Are these values your
non-permissible values?
( x  2) asymptotes ( x  2)
Use the following key stroke pattern on your graphing calculator to show
the asymptotes.
CLEAR
Mathematics A30
Y = 6  ( X,T,,n  2 - 4 )
144
ENTER
GRAPH
Lesson 3
Simplifying Rational Expressions
It is very important when simplifying rational expressions to always state the nonpermissible values.
The non-permissible values are determined before the expression is simplified or
reduced.
Rational expressions can be simplified or reduced by removing factors that are common to
both the numerator and the denominator.
Only common factors can be removed, not common terms. The following example will
illustrate this concept.
6 x  1 and 6 x  1
are common factors.
3 y  7 and 3 y  2
are not common factors.
A rational expression is simplified if its numerator and denominator have no factors in
common except 1 or  1 .
Example 4
x2  3x
State the non-permissible values and simplify the expression, 2
.
x 9
Solution:
x2  3x
x2  9
Write the original expression.
Factor the numerator and denominator.
State the non-permissible values.
=
x( x  3 )
, x   3, 3
 x  3  x  3 
Simplify by eliminating the common factor x  3 .
=
=
Mathematics A30
145
xx  3 
, x  3 , 3
 x  3  x  3 
x
, x  3 , 3
x 3
Lesson 3
The factors x and x  3 are different factors, therefore the expression is in
it's simplest form.
If you compare the values of the original expression,
x2  3x
, with the
x2  9
x
, you will notice that for any
x 3
permissible value x, the values of the expressions are the same.
values of the reduced expression
It may happen when simplifying rational expressions, that one factor is the “opposite” of
the other. The following example illustrates this.
x 5
5x
In this case multiply both the numerator and the denominator by  1 to make the factors
common.
x 5
5x
Distribute only one of the (1) factors.
=
 1( x  5)
 1(5  x)
=
( 1) x  5 
x  5 
=
1
1
= 1
Example 5
m 3 n  9 mn
Simplify
, m  0 , 3, n  0 .
3 mn  m 2 n
Solution:
m 3 n  9 mn
, m  0, 3, n  0
3 mn  m 2 n
Write the original expression.
Mathematics A30
146
Lesson 3
Factor the numerator and denominator.
=
Factor again and multiply by  1 .
=
mn m 2  9 
mn 3  m 
 1mn m  3 m  3 
 1mn 3  m 
Distribute the (  1 ) in the denominator.
=
Simplify by eliminating the common factors.
 1( mn )( m  3)( m  3)
( mn )( m  3)
=
 1mn m  3 m  3 
mn m  3 
=
 1( m  3 )
1
=  m  3 ,
m  0 , 3, n  0
You may have noticed that b  a  (1)( a  b) . Using this fact instead of multiplying
the numerator and denominator by (1) makes it easier to create common factors.
Example 6
Use factoring to simplify
4 b
.
b  9 b  20
2
Solution:
4 b
b  9 b  20
Write the original expression.
2
Factor the numerator and denominator.
State the non-permissible values.
=
b  5, 4
 1b  4 
, b  5, 4
b  5 b  4 
 1b  4 
, b  5, 4
=
b  5 b  4 
1
, b  5, 4
=
b5
Replace 4  b by (1)( b  4 ) .
=
Simplify by eliminating common factors.
Mathematics A30
4  b ,
b  5 b  4 
147
Lesson 3
Exercise 3.4
1.
2.
State the non-permissible values of the following rational expressions.
a.
a4
2 a( a  1)
d.
4 x 2  36
3x2  8x  3
b.
m 7
m  5m  6
e.
x 1
x3 1
c.
2t 2  3t  5
t2 9
f.
1
4 x  12 x  7 x 2  3 x  2
4
3
State the domain for each of the following rational equations. Complete the table of
values and graph the equation on the graph paper provided. Be sure to include the
asymptote line.
a.
x
f(x)
2
f ( x) 
 1 .4
 1 .2
x
x 1
 1 .1
1
 0 .9
 0 .8
 0 .6
 0 .4
 0 .2
0
1
y
5
–1
1
x
–5
Mathematics A30
148
Lesson 3
x 2.6
b.
f ( x) 
2.7
2.8
2
Draw the graph for x values near x = 3 only.
x 9
2
2.9
2.95
3
3.05
3.1
3.2
3.3
3.4
f(x)
y
x
3.
State the keystroke pattern on the graphing calculator for questions 2a and 2b.
4.
Simplify each of the following rational expressions.
a.
5 x2 y3
 15 xy 4
e.
4 ab
16 a b  12 ab
b.
y2  3 y
y 2  7 y  12
f.
r 3  2 r 2  63 r
7r
c.
n 2  7n  8
n 2  8n
g.
b2  6b  8
b 2  16
d.
6 p 2  22 p
9 p 3  121 p
h.
3  2a
4a  4a  3
Mathematics A30
149
2
2
Lesson 3
Mathematics A30
150
Lesson 3
Answers to Exercises
Exercise 3.1
1.
a.
x2  4 x  5
x  1 x3  3x2  9 x  5
x3  x2
 4 x2  9x
 4 x2  4 x
 5x  5
 5x  5
0
( x 3  3 x 2  9 x  5)  ( x  1)  x 2  4 x  5
b.
y2  2 y  1
y  5 y3  3 y2  9 y  5
y3  5 y2
 2 y2  9 y
 2 y 2  10 y
y5
y5
0
( y 3  3 y 2  9 y  5)  ( y  5)  y 2  2 y  1
c.
Mathematics A30
m2 9
151
Lesson 3
d.
b 3  3 b 2  9 b  33
b  3 b4  0b3  0b2  6b  0
b4  3b3
 3b3  0b2
 3b3  9b2
9b2  6b
9 b 2  27 b
 33 b  0
 33 b  99
99
b
4

 6 b  b  3   b 3  3 b 2  9 b  33 
e.
x2  x  6
f.
a 2  2a  6
a  3 a 3  a 2  0 a  18
remainder
99
b3
a 3  3a 2
2a 2  0a
2a 2  6a
6 a  18
6 a  18
0
( a 3  a 2  18 )  ( a  3)  a 2  2 a  6
g.
Mathematics A30
4t  1
152
Lesson 3
x 2  x  15
x  2 x  3 x  13 x  15
3
h.
2
x3  2x2
 x 2  13 x
 x2  2x
 15 x  15
 15 x  30
 15
x
2.
a.
b.
c.

 3 x 2  13 x  15   x  2   x 2  x  15 
3
x2  4 x  5
y2  2 y  1
2
2
1
2
1
m
remainder
3
0
9
0
18
 18
9
0
15
x 2

 2 m 2  9 m  18  m  2   1m 2  0 x  9
m2 9
d.
b 3  3 b 2  9 b  33 
e.
1
x
3
99
b3
1
2
1
5
1
6
6
1
1
6
0

 2 x 2  5 x  6   x  1  1 x 2  1 x  6
f.
a 2  2a  6
g.
4t  1
h.
2
1
3
2
 13
15
 2  30
1
1
 15  15
remainder
x 3  3 x 2  13 x  15   x  2   1 x 2  1 x  15  x 152
 x 2  x  15 
Mathematics A30
153
15
x2
Lesson 3
Exercise 3.2
1.
a.
b.
c.
d.
e.
x 3
x 5
x 7
x2
x 8
2.
a.
x  1 since f  1  0
f  1   3  1   7  1   10
 3  7  10
f  1  0
x  3 since f  3   0
x  1 since f 1  0
y  3 since f 3   0
2
b.
c.
d.
f 3   3   3 3   9 3   27
 27  27  27  27
f 3   0
3
3.
a.
b.
c.
d.
4.
a.
b.
2
x  13 x  10 
x  3 2 x 2  x  2 
x  12 x 2  9 x  10   x  12 x  5 x  2 
 y  3  y 2  6 y  9    y  3  y  3 2
k  12
f  2   0
3
2
0   2   5  2   8  2   k
0  8  20  16  k
12  k
k  18
f  1   0
0  3  1   7  1   k  1   8
3
2
0  3  7  k  8
0  18  k
 18  k
Mathematics A30
154
Lesson 3
c.
k 3
f 3   0
0  3   k 3   3  3
3
2
0  27  9 k
27  9 k
3k
Exercise 3.3
1.
a.
b.
c.
d.
f (1)  2  R
3
2
f  1    1   2  1   4  1   5
 1  2  4  5
f  1  2
remainder
f 1  8  R
f  2   19  R
f 2   13  R
f 2   2   2 2   4 2   5
 8 8 8 5
f 2   13
remainder
3
2.
2
e.
f.
f  3   52  R
f 3   26  R
a.
b.
f 2   10
f 2   9
f 2   2   2 2   2   3 2   7
 16  16  4  6  7
f 2   9
f 2   28
f 2   0
remainder
4
c.
d.
3.
a.
3
2
f 1   1   4 1   1  k ; f 1   4
3
2
4  1  4  1  k
4 6k
2  k
Mathematics A30
155
Lesson 3
b.
f (2 )  (1) 3  k (1) 2  5(1)  1; f (2 )  1
 1  1  k  5 1
15k
k 4
c.
f ( 1)  2(1) 3  5( 1) 2  2(1)  k; f (1)  11
   11  2  5  2  k
 11  5  k
 16  k
4.
Mathematics A30
a.
2 x 3  7 x 2  3 x  10
f ( 1)  2( 1) 3  7( 1) 2  3( 1)  10
f ( 1)  2  7  3  10
f ( 1)  18
remainder
b.
n 3  n 2  7n  2
f (3 )  (3 ) 3  (3 ) 2  7 (3 )  2
 41
remainder
c.
 3a 3  9a 2  8
f ( 2)  3( 2) 3  9( 2) 2  8
 3( 8 )  9(4 )  8
 4
remainder
d.
 c4  3c2  c  1
f (2)  (2) 4  3(2) 2  2  1
 16  12  2  1
 5
remainder
156
3  n  n  3
 1( n  3)
Lesson 3
Exercise 3.4
1.
e.
 c4  3c2  c  1
f ( 2)  ( 2) 4  3( 2) 2  ( 2)  1
 16  12  2  1
 1
remainder
a.
b.
c.
e.
0, 1
2, 3
3,  3
1
 ,3
3
1
f.
 1,  2 ,
a.
Domain is x  
d.
2.
1
1
,
2
2
x  1
y
5
–1
1
x
–5
Mathematics A30
157
Lesson 3
Domain is x  , x  3,  3
b.
y
x
2
4.
a.
x
3y
b.
y
y4
c.
n 1
n
d.
Mathematics A30
3
4
6 p 2  22 p
9 p 3  121 p
2 p(3 p  11 )

p(9 p 2  121 )
2 p(3 p  11 )

p(3 p  11 )( 3 p  11 )
2

3 p  11
158
Lesson 3
e.
f.
g.
h.
Mathematics A30
1
4a  3
r 3  2 r 2  63 r
7r
2
r ( r  2 r  63 )

7r
r ( r  7 )( r  9 )

 1( r  7 )
r(r  9)

1
 r ( r  9 )
b2
b4
3  2a
4a  4a  3
3  2a

(2 a  3)( 2 a  1)
 1(2 a  3)

(2 a  3)( 2 a  1)
1

2a  1
2
159
Lesson 3
Mathematics A30
Module 1
Assignment 3
Mathematics A30
161
Assignment 3
Staple here to the upper left
corner of your assignment
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please complete the following
procedures:
Print your name and address, with postal code. This address
sheet will be used when mailing back your corrected assignment.
1. Write your name and address and the
course name and assignment number
in the upper right corner of the first
page of each assignment.
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2. Number all the pages and place them
in order.
3. Complete the required information
details on this address sheet.
4. Staple this address sheet to the
appropriately numbered assignment.
Use one address sheet for each
assignment.
5. Staple the appropriately numbered
Assignment Submission Sheet to the
upper left corner, on top of this address
sheet.
Name
8404
Course Number
03
Assignment Number
Street Address or P.O. Box
Mathematics A30
Course Title
City/Town, Province
Postal Code:
Country
Distance-Learning Teacher’s Name
Mark Assigned:
Assignment 3
Values
(40)
A.
Multiple Choice: Select the correct answer for each of the following and
place a () beside it.
1.
The non-permissible values of the expression
____
____
____
____
2.
Mathematics A30
a.
b.
c.
d.
x
x
x
x
2x  3 
has vertical asymptotes at ***.
x2  4
4
 3
 3, x  2 , x  2
 2, x  2
A vertical line passing through 3,  4  has the equation ***.
____
____
____
____
4.
1,  1
0 , 1,  1
0, 1
0,  1
The graph of the equation y 
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____
____
____
3.
a.
b.
c.
d.
1 x 1

are ***.
x x 1
a.
b.
c.
d.
x3
x  4
y3
y  4
When simplified,
____
a.
____
b.
____
c.
____
d.
2x  4
is equal to ***.
2x  6
x4
x 6
2

3
4
6
x2
x 3
165
Assignment 3
5.
If f ( x)  x 3  7 x 2  5 x  30 , then x  2 is a factor and ***.
____
____
____
____
6.
9.

Mathematics A30
x  27 is a factor
the remainder of f ( x )  ( x  1) is  27
x  1 is a factor
x  1 is a factor

a.
b.
c.
d.

x  1  2 x  3  equals
2 x  3   x  1 equals
2 x2  x  3
2 x2  x  3
x  1 and 2 x  3 are factors of 2 x 2  x  3
2 x  3   2 x 2  x  3 is x  1



In x 3  3 x  1   x  1  , the remainder is ***.
____
____
____
a.
b.
c.
____
d.
x 1
1
1
1
x 1
A factor of the polynomial f  x   x 4  x 3  2 x 2  x  18 is ***.
____
____
____
____
10.
a.
b.
c.
d.
If 2 x 2  x  3   x  1   2 x  3 , then ***.
____
____
____
____
8.
x  2 is a factor
f (0 )  2
f (2)  0
f ( 2 )  0
For f ( x)  x 3  7 x 2  5 x  30 , f (1)  27 and ***.
____
____
____
____
7.
a.
b.
c.
d.
a.
b.
c.
d.
x+2
x 1
x+3
x2
x  1 is a factor of ***.
____
____
____
a.
b.
c.
____
d.
x2  6 x  7
2 x2  x  1
x4  x3  3 x2  x  4
x 4  x3  3x 2  x  4
166
Assignment 3
11.
When a polynomial f x  is divided by x  5  the remainder is  1 .
The value of f  5  is ***.
____
____
____
____
12.
15.
Mathematics A30
a.
b.
c.
d.
f 1 
f  1
2
1
If 3 y 2  10 y  8  ( y  2)( 3 y  4 ) , then (3 y 2  10 y  8)  (3 y  4 ) is equal
to ***.
____
____
____
____
14.
1
1
5
5
The remainder when f  x   2 x 3  x 2  x  1 is divided by x  1 is ***.
____
____
____
____
13.
a.
b.
c.
d.
a.
b.
c.
d.
y2
y2
0
y
When 3 x 2  5 x  21 is divided by x  2 , the following is true:
____
a.
____
b.
____
____
c.
d.
3x + 11 is a factor of 3 x 2  5 x  21 .
3 x 2  5 x  21
 (3 x  11 )  1 .
x 2
the remainder is 1.
x  2 is a factor of 3 x 2  5 x  21 .
If the remainder and quotient are  5 and x 2  x  3 when
x 3  x 2  x  1 is divided by x  2 , then
____
____
____
a.
b.
c.
____
d.
x  2 and x 2  x  3 are factors.
x 2  x  3  ( x 3  x 2  x  1)  x  2 .
the value of the cubic polynomial at x  2 is 5.
5
( x 3  x 2  x  1)  ( x  2)  x 2  x  3 
.
x2
167
Assignment 3
16.
The zeros of the polynomial ( x  1)( x 2  2 x  15 ) are ***.
____
____
____
____
17.
a.
b.
c.
d.
0
f(a)
f (a)
 f (a )
a.
b.
c.
d.
2
 28
 14
14
The reduced form of
____
____
____
____
Mathematics A30
1
0
1
2
When x 2  7 x  k is divided by x  3 the remainder is 2. The value of k
must be ***.
____
____
____
____
20.
a.
b.
c.
d.
If a polynomial f(x) is divided by x  a , the remainder is ***.
____
____
____
____
19.
1,  5 , 3
 1, 5 ,  3
1,  1, 15
1,  1,  15
If x  1 is a factor of x 3  3 x 2  x  c , then c must be ***.
____
____
____
____
18.
a.
b.
c.
d.
a.
b.
c.
d.
( x 2  1)( x  3)
is ***.
1 x
x( x  3)
 x( x  2)
( x  1)( x  3)
 ( x  1)( x  3)
168
Assignment 3
Answer Part B and Part C in the space provided. Evaluation of your
solution to each problem will be based on the following:
• A correct mathematical method for solving the problem is shown.
• The final answer is accurate and a check of the answer is shown where
asked for by the question.
• The solution is written in a style that is clear, logical, well organized,
uses proper terms, and states a conclusion.
(5)
B.
1.
Show use of long division to find the quotient and the remainder.
8 x3  4 x  1
2x 1
Mathematics A30
169
Assignment 3
(5)
2.
Show that x  2 is a factor of 2 x 3  x 2  7 x  6 and find the remaining
linear factors of the polynomial.
(5)
3.
Is 3 x  2 a factor of 6 x 3  x 2  5 x  2 ? Justify your answer.
Mathematics A30
170
Assignment 3
(5)
4.
Show use of synthetic division to determine f 2  in
f x   2 x 3  x 2  7 x  6 .
(5)
5.
Show use of synthetic division to divide x 3  9 by x  3 .
Mathematics A30
171
Assignment 3
(5)
6.
(5)
7.
Mathematics A30
2 n  32 n 3
Reduce
to lowest terms.
20 n 3  3 n 2  2 n
Determine the non-permissible values of the expression
1
.
3
2
2 x  3 x  11 x  6
172
Assignment 3
( 4)
C.
( 7)
1.
Determine the non-permissible value of the equation
5
f ( x)  2
.
x  x  12
2.
Make a table of values and sketch the graph of f ( x ) 
5
.
x  x  12
2
x
f(x)
Mathematics A30
173
Assignment 3
( 4)
Mathematics A30
3.
In a paragraph, describe what happens on the graph in question
2 near the non-permissible values.
174
Assignment 3
(10)
4.
(STUDENT JOURNAL).
Write a summary of this lesson that would be suitable as a review for
an examination.
(100 )
Mathematics A30
175
Assignment 3