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Mathematics A30
Module 1
Lesson 2
Mathematics A30
Factoring Polynomials
69
Lesson 2
Factoring Polynomials
Introduction
Factoring polynomials is a very important concept in Algebra. The first section will be a
review of factoring. This factoring will be expanded to include the factoring of special
polynomials.
In past courses polynomials have been linear or quadratic. In this course cubic
polynomials will be introduced with the concept of factoring sums and differences of cubes.
The concepts that are developed in this lesson will be the foundation for higher learning
and problem solving in the real world.
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Lesson 2
Objectives
After completing this lesson, you will be able to
•
expand polynomials.
•
factor trinomials.
•
factor the difference of squares.
•
factor the difference of squares of special polynomials.
•
factor the sum and difference of cubes.
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Lesson 2
2.1 Review of Factoring
The process of determining polynomials whose product equals a given polynomial is called
factoring.
A review of some of the concepts that have already been taught in earlier mathematics
courses is necessary at the beginning of this lesson.
In order to understand factoring well, it is necessary to know how to expand (or multiply)
polynomials and to recognize the fact that expanding and factoring are inverse operations,
the same way that adding and subtracting are inverse operations.
The following illustration shows expanding and factoring with numbers.
Expanding (multiplying)
3  7  21
Factoring
3
 7

21
3 and 7 are factors of 21.
Expanding Polynomials
Expanding polynomials may involve one or many of the following multiplication processes.
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Lesson 2
Multiplying a Polynomial by a Monomial
The distributive property is used to multiply a polynomial by a monomial.
The Distributive Property
The product of a and (b  c) is given by:
a(b  c)  ab  ac or (b  c)a  ba  ca .
The product of a and (b  c) is given by:
a(b  c)  ab  ac or (b  c)a  ba  ca .
Example 1


Expand  2 x 5 xy  x 2  1 .
Solution:

Write the expression.
 2 x 5 xy  x 2  1

Apply the distributive property.
= –
2
x
(
5
x
yx
–+
1
)
Multiply.
=  2 x(5 xy)  2 x( x 2 )  2 x(1)
Simplify.
=  10 x 2 y + 2 x 3  2 x
2
Multiplying Two Binomials
The distributive property is used twice to multiply two binomials together.
Distributive Property
( a  b)( c  d )  ( a  b)c  ( a  b)d
 ac  bc  ad  bd
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( a  b) is distributed.
c and d are distributed.
Lesson 2
The distributive property is often remembered as the FOIL method.
Multiply the: 



F
O
I
L
First terms of the binomials.
Outside terms of the binomials.
Inside terms of the binomials.
Last terms of the binomials.
Example 2
Multiply x  3m m  x  .
Solution:
Write the expression.
x  3m m  x 
Apply the FOIL Method.
O
F
x  3m m  x 
I
L
Simplify.
F
O
I
L
 xm  x 2  3 m 2  3 mx
=  2 mx  x 2 + 3 m 2
Squaring a Binomial
•
•
This is when a binomial expression is multiplied by itself.
Here is an example of the distributive property. The FOIL method could also be
used.
Example 3
Expand  y  2 2 .
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Lesson 2
Solution:
 y  2 2
Write the expression.
Expand.
=  y  2  y  2
Use the distributive property.
= y y  2   2 y  2 
Simplify.
= y2  2 y  2 y  4
= y2  4 y  4
When a binomial is squared, a pattern shows how the answer can be expressed in
a simple form.
(a + b)2
=
•
•
•
a2 + 2ab + b2
Square the first term.
Square the last term.
The middle term is 2 times the first and last terms.
Factoring Polynomials
Remember that factoring polynomials is the opposite of multiplying or expanding
polynomials.
Greatest Common Factor
Polynomials are factored using the reverse of the distributive property. Whenever you are
asked to factor a polynomial, the first question should always be:
"Is there a common factor in all of the terms?".
The greatest common factor (GCF) of two numbers can be found by factoring each of the
numbers, finding all the factors that are common to both and multiplying these common
factors together.
The concept can be shown first with numbers.
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Lesson 2
The greatest common factor of 18 and 42 can be determined by factoring each number
separately.
18 = 2 × 3 × 3
42 = 2 × 3 × 7
The greatest common factor of 18 and 42 is 2  3 = 6 .
The same is true when finding the greatest common factor of algebraic terms in a
polynomial.
Example 4
Factor 15 x 2 y  30 xy  35 xy 2 .
Solution:
Factor each algebraic term.
15 x 2 y = 3  5  x  x  y
30 xy = 2  3  5  x  y
35 xy 2 = 5  7  x  y  y
Determine the common factors.
•
•
•
The common factors are 5, x and y.
Multiply these common factors together.
The GCF is 5xy.
Write the original expression.
15 x 2 y  30 xy  35 xy 2
Factor the common factor 5xy from
each term.
= 5 xy3 x   5 xy6   5 xy(7 y)
Use the distributive property.
= 5 xy3 x  6  7 y 
Check by expanding.
5 xy3 x  6  7 y 
= 15 x 2 y  30 xy  35 xy 2
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()
Lesson 2
Example 5
Factor 30 m 2  40 m 2 n  50 m 2 n 2 .
Solution:
Determine the GCF of the terms.
•
The GCF is 10 m 2 .
Write the original expression.
30 m 2  40 m 2 n  50 m 2 n 2

Factor the common factor 10m 2 from
each term.
= 10 m 2 3   10 m 2 4 n   10 m 2 5 n 2
Use the distributive property.
= 10 m 2 3  4 n  5 n 2



Check the answer by expanding.
Factoring by Grouping
Sometimes the GCF in all the terms of a polynomial is 1. In this case it is necessary to
group terms together that do have a common factor. In most cases this type of factoring
occurs when there are four (4) terms in the polynomial. Two terms are grouped together.
When factoring by grouping, the steps are:
•
Group terms together that have a common factor.
•
Factor a common factor from each group.
•
The terms inside the two sets of brackets must be the same. This will be one factor
of the polynomial.
•
The two common factors from each group are combined to make the other factor of
the polynomial.
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Lesson 2
Example 6
Factor a 2  ab  ac  bc .
Solution:
Write the original expression
a 2 + ab + ac + bc .


Group terms with a common factor.
= a 2 + ab + ac + bc 
Factor a common factor from each group.
= aa + b  + c a + b 
Write as two factors.
= a + c a + b 
a + c a + b 
Check by expanding. (FOIL)
= a 2 + ab + ac + bc ()
Another possible way of grouping the above terms is ( a 2  ac)  ( ab  bc) . Does this
give the same factors? Try it.
Example 7
Factor 4 x 2  7 yz  2 xy + 14 xz .
Solution:
Write the original expression.
4 x 2  7 yz  2 xy + 14 xz
4 x

Group terms with a common factor.
Factor a common factor from each group.
=
=
 2 xy   7 yz + 14 xz 
2 x2 x  y  + 7 z  y + 2 x 
Write as two factors.
=
2 x + 7 z 2 x
 y
2 x + 7 z 2 x
Check by expanding. (FOIL)
=
•
2
 y
4 x  2 xy  14 xz  7 zy ()
2
Sometimes you may need to try different combinations of grouping in order to have
the same factor inside both brackets.
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Lesson 2
Difference of Squares
A difference of squares may occur when you are asked to factor a polynomial that is a
binomial.
There is a special way of factoring an expression where each of the two terms is a perfect
square and there is a negative sign between the two terms. When this pattern occurs it is
called factoring a difference of squares.
Some examples of perfect squares are:
25
a2
16b2c2
5×5
a×a
4bc × 4bc
Some examples of binomials which are a difference of perfect squares are:
x
2
 9
y
2
4 a 2  25 b 2
 81
To factor a difference of squares, follow these steps:
•
Find the principal (positive) square root of the first term.
•
Find the principal (positive) square root of the last term.
•
Multiply the sum of these two square roots by the difference of these two square
roots.
Difference of Squares
In general:
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2
a  b = a + b a  b
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Lesson 2
Example 8
Factor y 2  81 .
Solution:

Find the principle square root of the first term.
2
y =y

Find the principle square root of the last term.
81 = 9
Write the original expression.
y
Write the product of the sum
and the difference of the roots.
=
2
 81
(y+9
)(y–9
)
su
m
o
fth
e
sq
u
a
rero
o
ts
d
iffe
re
n
c
eo
fth
e
sq
u
a
rero
o
ts
Example 9
Factor 64 a 2  9 b 2 .
Solution:
•
•
The square root of the first term is 8a.
The square root of the last term is 3b.
Write the original expression.
Write in factored form.
64 a 2  9 b 2
= 8 a + 3b8 a  3b
8 a +
Check by expanding.
3b8 a  3b
= 64 a 2  24 ab + 24 ab  9 b 2
= 64 a 2  9 b 2 ()
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Lesson 2
Exercise 2.1
1.
Expand the following polynomials.
a)
b)
c)
d)
e)
f)
g)
h)
i)
j)
k)
2.
r  7 2
k  10 2
2 a  3 2
(3 x  5 y) 2
Factor the following polynomials.
a)
b)
c)
d)
e)
f)
g)
3.
7 a 6 a  5 
 3 xy8 x  4 y  2 xy  1
4 m 2 6 m  3 n 2  9 
x  5 x  8 
2 y  12 y  1
3m  4 n 2m  n 
 p  11  p  9 
3 a 2  6 ab
m2 9
n 2  2 n  np  2 p
9x2 1
x 2  121
25 x 2  36 y 2
5 a 2  80 b 2
h)
i)
j)
k)
l)
m)
n)
34 u 9  24 u 7  18 u 5
21 w 4  14 w 2  35
2 nr  3 p 2  6 pr  pn
k 2  5 k  mk  5 m
5 x  5 y  x 2  xy
3 p 2 r 2  12 p 2 r  5 pr 2  20 pr
2 x 2 y  10 xy  6 x 2  30 x
What is the first question that you ask yourself when you are factoring polynomials?
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Lesson 2
2.2 Factoring Trinomials
Perfect Trinomial Square
When two identical binomials are multiplied together, they produce a perfect trinomial
square.
To identify a perfect trinomial square, the following conditions must be true.
•
The first and last terms are both perfect squares such that the principle square root
of either term can be found exactly.
•
The middle term is equal to twice the product of the square roots of the first and
last term.
A Perfect Trinomial Square has the form a 2 + 2ab + b 2 or a 2  2ab + b 2 .
In general:
a 2 + 2ab + b 2
=
=
a 2  2ab + b 2
=
=
a + ba + b
a + b 2
a
a
 ba  b
 b
2
Example 1
Factor the perfect trinomial square, y 2  10 y  25 .
Solution:
•
•
•
The square root of the first term is y.
The square root of the term is 5.
Two times the product last 5 and y is 10y.
Write the expression.
y2 + 10y + 25
Factor.
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Lesson 2
Example 2
Factor 9 x 2  48 x + 64 .
Solution:
•
•
The square root of the first term is 3x.
The square root of the last term is 8.
23 x 8  = 48 x
Multiply twice the product of the square roots.
This is the middle term; therefore, this is a perfect trinomial square.
Write the original expression.
9 x 2  48 x  64
Factor.
 3 x  8 3 x  8 


Use the negative signs in the answer because the middle term of the original expression is
negative.
Factored polynomials can always be checked by expanding.
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Lesson 2
Example 3
Factor 25 a 2 + 40 ab + 16 b 2 .
Solution:
Write the original expression.
25 a 2 + 40 ab + 16 b 2
 5 a  4 b5 a  4 b
Factor.
25 a 2 is the square of 5a.
25 a 2  40 ab  16 b 2  5 a  4 b5 a  4 b
The middle term is 25 a 4 b .
16 b 2 is the square of 4b.
The most important thing is to be able to identify a perfect trinomial square. Once
you have done this, then use the pattern for factoring.
Factoring Trinomials
Trinomials can be factored by using a specific pattern.
When the coefficient of the x2 term is equal to 1, use the following pattern.
•
Find two numbers such that:
•
their sum is the coefficient of the 2nd term.
•
their product is the last term.
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Lesson 2
Example 4
Factor x 2  6 x  8.
Solution:
Find two numbers whose sum is 6 and whose product is 8.
•
The product of 2 and 4 is 8 (the last term of the trinomial).
•
The sum of 2 and 4 is 6 (the coefficient of the middle term of the trinomial).
Factor.
sum
x 2  6 x  8   x  4  x  2 
product
Example 5
Factor y 2 + 13 y + 30 .
Solution:
Find two numbers whose product is 30 and whose sum is 13.
•
The two numbers are 3 and 10.
Write the original expression.
2
y + 13 y + 30
=  y + 3  y + 10 
Factor.
Trinomials can also be in the form x 2  bx  c where the middle term is negative. When
looking for factors, use the same pattern of factoring the product, or last term, but
remember that when the middle term is negative and the last term is positive, the two
factors will both be negative.
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Lesson 2
Example 6
Factor g 2  21 g + 80 .
Solution:
Find two numbers whose product is 80 and whose sum is  21 .
•
The two numbers are  5 and  16 .
Write the original expression.
g
2
 21 g + 80
=  g  5  g  16 
Factor.
g
 5  g  16 
= g  16 g  5 g  80
= g 2  21 g  80 ()
Check by expanding.
2
If the last term of the trinomial is negative, then the factors of the trinomial will have
different signs.
Example 7
Factor a 2  4 ab  32 b 2 .
Solution:
Find two numbers whose product is  32 and whose sum is  4 .
•
The two numbers are 4 and  8 .
Write the original expression.
a 2  4 ab  32 b 2
Remember to factor the b2
in the last term.
= a  8 ba  4 b
Factor.
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Lesson 2
There are many different ways to factor a trinomial where the coefficient of the x2 term is
not equal to 1.
One method of factoring these trinomials is to use trial and error.
•
This can be done by finding the factors of the first term, finding the factors of the
last term and writing down all the possible combinations. Expand each of the
expressions to see which set of factors gives you the correct middle term.
Trinomials of this form can also be factored using decomposition.
•
A pattern is used to factor trinomials where the x2 term is not equal to 1. Factoring
by grouping is one of the steps.
The steps are:
•
Multiply the coefficients of the first and last terms.
•
Find two numbers whose product is this number and whose sum is the coefficient of
the middle term.
•
These two numbers now become the middle two terms.
•
Group the first two terms together and the last two terms together.
•
Remove a common factor from each.
•
Factor by grouping.
Example 8
Factor 15 y 2  7 y  4 by the trial and error method.
Solution:
•
Pairs of factors of 15 are:
(15, 1)
(1, 15)
(5, 3)
(3, 5)
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Pairs of factors of  4 are:
(4 , 1)
(1,  4 )
(4,  1 )
(  1 , 4)
(2,  2 )
(  2 , 2)
Lesson 2
Create a table of possible solutions.
Middle term
11y
 59 y
 11 y
59y
28y
 28 y
7 y
 17 y
7y
17y
4y
4y
(15 y  4 )(1 y  1)
(15 y  1)(1 y  4 )
(15 y  4 )(1 y  1)
(15 y  1)(1 y  4 )
(15 y  2)(1 y  2)
(15 y  2)(1 y  2)
(5 y  4 )( 3 y  1)
(5 y  1)( 3 y  4 )
5 y  4 3 y  1 
(5 y  1)( 3 y  4 )
(5 y  2)( 3 y  2)
(5 y  2)( 3 y  2)
correct middle term
Therefore, 15 y 2  7 y  4  (5 y  4 )( 3 y  1) .
Example 9
Factor 15 y 2  7 y  4 by decomposition.
Solution:
Multiply the first and last terms.
15  ( 4 )  60
Find two numbers whose product is  60 and whose sum is 7 (coefficient of middle term).
•
These two numbers are 12 and  5 .
Rewrite the expression with these two numbers as coefficients for the middle terms.
= 15y2+7y–4
15y2+12y–5y–4
Group terms together and factor.
= 3 y(5 y  4 )  1(5 y  4 )
= (3 y  1)( 5 y  4 )
Check by expanding.
(3 y  1)( 5 y  4 )
= 15 y 2  12 y  5 y  4
= 15 y 2  7 y  4 ()
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Lesson 2
Example 10
Factor 6 m 2+ mn  15 n 2 by decomposition.
Solution:
6   15  =  90
Multiply the first and last terms.
Find two numbers whose product is  90 and whose sum is 1.
•
These two numbers are  9 and 10 .
Rewrite the expression with these two numbers as coefficients for the middle terms.
6 m 2 + mn  15 n 2
=
Group terms together and factor.
6 m 2  9 mn + 10 mn  15 n 2
=
Write in factored form.
=
3m +
3 m (2 m  3 n )  5 n (2 m  3 n )
5n 2m  3n 
The greatest common factor can also be determined from mathematical expressions. The
greatest common factor of two expressions can be found by factoring each of the
expressions, finding all the factors that are common to both and multiplying these
common factors together.
Example 11
Find the greatest common factor from the expressions 6 c 3  6 c 2  36 c and 3 c2  27 c  66 .
Solution
Factor each expression. See if you can get the listed factors for each expression by using
the decomposition method!
6 c 3  6 c 2  36 c  6 c ( c 2  c  6 )
 3  2  c  ( c  2)  ( c  3)
3 c 2  27 c  66  3  ( c  11 )  ( c  2)
The greatest common factor of 6 c 3  6 c 2  36 c and 3 c2  27 c  66 is 3  ( c  2)  3( c  2) .
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Lesson 2
Exercise 2.2
1.
Determine which of the following are perfect trinomial squares. State the reason
why if they are not.
a)
b)
c)
d)
e)
f)
g)
h)
2.
a 2  8 a  16
x 2  28 x  196
1  2a  a 2
100  10 w  w 2
4 w 2  36 wx  81 x 2
m 2  16 m  64
9k 2  6k  1
a 2  24 a  169
In the following trinomials, determine the two numbers needed to factor the
expression by first finding the sum of the middle term and the product of the last
term.
a)
b)
c)
d)
e)
x  13 x  42
p 2  70 p  600
32  12 m  m 2
c 2  15 c  250
15  2w  w 2
2
Sum
________________
________________
________________
________________
________________
Product
________________
________________
________________
________________
________________
Numbers
______________
______________
______________
______________
______________
3.
Factor 2 y 2  8 y  5 . What is your conclusion?
4.
Factor the following polynomials. (Remember to first look for a common factor.)
a)
b)
c)
d)
e)
f)
x2  6x  9
5 n 2  15 n  20
x2  3x  2
3 y2  7 y  6
g)
h)
i)
j)
k)
4 a 2  12 ab  9 b 2
2
y  y  20
m 2  12 m  36
6 n 2  n  15
6 a 2 x 2  19 ax 2  10 x 2
6 a 4  19 a 3  10 a 2
y4  2 y2  1
l)
4 y 4  37 y 2  9
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Lesson 2
2.3 Factoring Special Polynomials
Difference of Squares of Special Polynomials
The two processes of grouping and factoring a difference of squares will be incorporated
into this section.
In Section 2.1, when grouping within a polynomial, pairs of terms were grouped together.
It is also possible to group together three terms that form a special polynomial. This
special polynomial will usually be a trinomial square. It is therefore necessary to be able
to recognize a perfect trinomial square at a glance. This is the first step.
The second step involves factoring a difference of squares.
Example 1
Factor x 2  4 x  4  25 .
Solution:
Write the original expression.
x 2  4 x  4  25


Group together the terms that form
a special polynomial.
= x 2  4 x  4  25
Factor the perfect trinomial square.
=  x + 2 2  25
Factor as a difference of squares.
= ( x + 2) + 5 ( x + 2)  5 
= x + 2 + 5 x + 2  5 
Simplify.
= ( x  7)( x  3)
Mathematics A30
Perfect Trinomial Square
94
Lesson 2
Example 2
Factor x 2  6 x  9   x  4 2 .
Solution:
x 2  6 x  9   x  4 2
Write the original expression.
Group together terms that form
a special polynomial.
=
Factor the perfect trinomial square.
=
x
2

 6 x  9   x  4 2
Perfect Trinomial Square
Factor as a difference of squares.
 x  3 2   x  4 2
=
( x  3)  ( x  4)( x  3)  ( x  4)
Simplify.
=
=
=
x  3  x  4 x  3  x  4 
2 x  7 (1)
2x 7
Mental Math Calculations
•
Multiply 33  27 mentally.
•
Think of the question as:
30  3  30  3 
•
Hint: Square the first terms, square the last terms and
subtract.
•
Now try:
42  58
55  65
26  34
71  69
Exercise 2.3
1.
Factor the following special polynomials.
a)
x 2  y2  2 y  1
b)
c)
d)
e)
f)
x 2  2 xy  y 2  z 2
a 2  10 a  25  b 2
9 m 2  6 mn  n 2  36
25  x 2  4 xy  4 y 2
t 2  4 t  4  81 s 2
Mathematics A30
95
Lesson 2
2.4 Sum and Difference of Cubes
The "cube" of a number is a number b is b  b  b  b 3 .
•
•
4  4  4  4  64
3
xxx  x
3
The cube root of a number is a factor of the number that when multiplied by itself three
times, the result is the original number. The cube root of a term is a factor of the term that
can be multiplied by itself three times and the result is the original term.
•
•
•
The cube root of 8 is 2 because 8  2  2  2 .
The cube root of a 3 is a because a 3  a  a  a .
The cube root of 64 t 3 is 4 t because 64 t 3  4 t  4 t  4 t .
The volume of a cube visually shows the concept of cubes and cube roots.
Volume  V  x  x  x  x 3
Length of a side  3 V  x
3
is the symbol for cube root.
A sum or a difference of two cubed terms forms a special polynomial.
Example
•
Sum of Cubes:
•
Difference of Cubes:
3
3
x + y
3
3
x y

a 3  27

8 m 3  125
These two special polynomials can be factored using a special pattern.
Sum of Cubes
Difference of Cubes

x 3  y3   x + y  x 2  xy  y 2
Mathematics A30


x 3  y3   x  y  x 2  xy  y 2
96

Lesson 2
The volume of a cube with sides equal to x cm can illustrate the process that determines
the rule or pattern for factoring a difference of cubes.
•
•
•
Start with the larger cube whose volume is x 3 cm 3 .
Remove a smaller cube whose volume is y3 cm 3 .
The remaining volume will be x 3  y 3 .
y
(x – y )
y
B
x
C
y
y
A
y
x
x
The portion of the cube that is remaining can be divided into three sections.
1)
The volume of section A would be:
x  x  x  y  cm 3
= x2 x  y 
2)
The volume of section B would be:
x  y  x  y  cm 3
= xyx  y 
3)
The volume of section C would be:
y  y  x  y  cm 3
= y2 x  y 
The total volume would be:
3
2
2
3
x  y  x  x  y   xy  x  y   y  x  y 
  x  y  x 2  xy  y 2

Mathematics A30

97
Lesson 2
Example 1
Factor a 3  8b 3 .
Solution:
•
•
•
•
The cube root of a 3 is a.
x=a
The cube root of 8b 3 is 2b.
y = 2b

x 3  y3   x  y  x 2  xy  y 2
Write the formula for a difference of cubes.

a  2 b a


 2 ab  4 b 
a 3  8 b 3  a  2 b  a 2  2 ab  2 b 2
Substitute the values, x = a, y = 2b.

2
2
Remember to
square the
entire term.
Example 2
Factor 125 p3  1 .
Solution:
•
•
•
•
The cube root of 125 p3 is 5p.
x 5p
The cube root of 1 is 1.
y 1

3
2
3
2
x  y   x  y  x  xy  y
Write the formula for a sum of cubes.

 5 p  1  25 p

 5 p  1
125 p3  1  5 p  1  5 p 2  5 p  12
Substitute the values, x = 5p, y = 1.
Mathematics A30

98
2
Lesson 2
Example 3
Factor 64 m 3  729 n 3 .
Solution:
•
•
•
•
The cube root of 64 m 3 is 4m.
x  4m
The cube root of 729 n 3 is 9n.
y  9n

3
2
3
2
x  y   x  y  x  xy  y
Write the formula for a difference of cubes.
Substitute the values.




64 m 3  729 n 3  4 m  9 n  4 m 2  36 mn  9 n 2

 4 m  9 n  16 m 2  36 mn  81 n 2
Exercise 2.4
1.
Determine the cube root of the following terms. If the term is not a perfect cube,
state that.
a)
b)
c)
d)
e)
f)
2.
8 x3
125 m 3
42 y 3
27 a 6
9
r
1 000 b 8
Use the distributive property to show that each of the statements is true.
a)
b)
a  ba 2  ab  b 2   a3  b3
 x  y x 2  xy  y2   x 3  y3
Mathematics A30
99
Lesson 2
3.
Factor completely.
a)
b)
c)
d)
e)
x 3  64
a 3  8b 3
4 t 3  108
729 s 3  125 t 3
1 000 m 3 n 3  p 3
f)
x 6  y6
g)
1 27 b 3

64
8
Mathematics A30
100
Lesson 2
Answers to Exercises
Exercise 2.1
Mathematics A30
1.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
k.
 42 a 2  35 a
 24 x 2 y  12 xy 2  6 x 2 y 2  3 xy
24 m 3  12 m 2 n 2  36 m 2
x 2  3 x  40
4 y2 1
6 m 2  11 mn  4 n 2
p 2  20 p  99
r 2  14 r  49
k 2  20 k  100
4 a 2  12 a  9
9 x 2  30 xy  25 y 2
2.
a.
b.
c.
d.
e.
f.
g.
3 a( a  2 b)
(m  3)( m  3)
n ( n  2)  p( n  2)  ( n  2)( n  p)
(3 x  1)( 3 x  1)
( x  11)( x  11 )
(5 x  6 y)( 5 x  6 y)
5( a  4 b)( a  4 b)
h.
34 u 9  24 u 7  18 u 5
 2u 5 (17 u 4  12 u 2  9)
i.
7(3w 4  2w 2  5)
j.
2 nr  6 pr  3 p 2  pn
= 2 r ( n  3 p)  p(3 p  n )
= (2 r  p)( 3 p  n )
k.
k 2  5 k  mk  5 m
= k ( k  5)  m ( k  5)
= ( k  m )( k  5)
l.
5 x  5 y  x 2  xy
= 5( x  y)  x( x  y)
= (5  x )( x  y)
101
Lesson 2
Exercise 2.2
m.
pr (3 pr  12 p  5 r  20 )
= pr 3 p(r  4)  5(r  4)
= pr (3 p  5)( r  4 )
n.
2 x( xy  5 y  3 x  15 )
= 2 xxy  3 x  5 y  15 
= 2 xx( y  3)  5( y  3)
= 2 x( x  5)( y  3)
3.
“Is there a common factor in all of the terms?”
1.
a.
b.
c.
d.
e.
f.
g.
h.
yes
yes
yes
no
yes
no
yes
no
a.
b.
c.
d.
e.
Sum
13
 70
 12
 15
2
2.
3.
2(10 )( w )  20 w  this should be the middle term.
The last term cannot be a negative.
2(13)(a) = 26a  this should be the middle term.
Product
42
600
32
 250
 15
Numbers
6, 7
 60 ,  10
 8,  4
 25 , 10
5, 3
Factoring by decomposition, the first step is:
2  5  10
•
•
Find two numbers whose product is 10 and whose sum is
8 .
•
This is impossible.
This expression cannot be factored further. It is therefore prime.
4.
Mathematics A30
a.
b.
c.
d.
e.
( x  3) 2
5(n 2  3n  4)  5(n  4)( n  1)
( x  2)( x  1)
(3 y  2)( y  3)
(2 a  3 b)( 2 a  3 b)
102
Lesson 2
Exercise 2.3
1.
f.
( y  5)( y  4 )
g.
h.
i.
j.
k.
l.
(m  6) 2
(2 n  3)( 3 n  5)
x 2 (2 a  5)( 3 a  2)
a 2 (2 a  5)( 3 a  2)
( y 2  1)( y 2  1)  ( y  1) 2 ( y  1) 2
(4 y 2  1)( y 2  9)  (2 y  1)( 2 y  1)( y  3)( y  3)
a.
x 2  ( y 2  2 y  1)
 x 2  ( y  1) 2
 ( x  y  1)( x  y  1)
b.
( x 2  2 xy  y 2 )  z 2
 ( x  y) 2  z 2
 ( x  y  z )( x  y  z )
d.
e.
f.
( a 2  10 a  25 )  b 2
 ( a  5) 2  b 2
 ( a  5  b)( a  5  b)
(3 m  n  6)( 3 m  n  6)
(5  x  2 y)( 5  x  2 y)
(t  2  9 s)( t  2  9 s)
a.
b.
c.
d.
e.
2x
5m
3
42 , 42 is not a perfect cube.
3a 2
r3
f.
10 3 b 8 , b 8 is not a perfect cube.
a.
b.
( a  b)( a 2  ab  b 2 )
 a 3  a 2 b  ab 2  ba 2  ab 2  b 3
 a 3  b3
( x  y)( x 2  xy  y 2 )
 x 3  x 2 y  xy 2  yx 2  xy 2  y 3
 x3  y3
a.
b.
c.
( x  4)( x 2  4 x  16 )
( a  2b)( a 2  2 ab  4 b 2 )
4(t 3  27 )  4(t  3)( t 2  3t  9)
c.
Exercise 2.4
1.
2.
3.
Mathematics A30
103
Lesson 2
d.
e.
f.
(9 s  5t )(81 s 2  45 st  25 t 2 )
(10 mn  p)(100 m 2 n 2  10 mnp  p 2 )
( x 2  y 2 )( x 4  x 2 y 2  y 4 )
 ( x  y)( x  y)( x 4  x 2 y 2  y 4 )
g.
Mathematics A30
2
 1 3 b  1 3 b 9 b






4
 4 2  16 8
104



Lesson 2
Mathematics A30
Module 1
Assignment 2
Mathematics A30
105
Assignment 2
Staple here to the upper left
corner of your assignment
Before submitting your assignment,
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.
Student Number
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
02
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 2
Values
(40)
A.
Multiple Choice: Select the correct answer for each of the following and
place a () beside it.
1.

____
____
____
____
2.
a.
b.
c.
d.
5 xy  2 x 2  3 y 2
6 xy  2 x 2  6 y
6 xy  2 x 2
 3 y 2  xy
2
a.
b.
c.
d.
4a2
4a2
4a2
4a2
 49
 49
 14 a  49
 28 a  49
The greatest common factor of each of the terms in the
expression, 12 a 2 b 3  16 ab 3  8 a 3 b is ***.
____
____
____
____
Mathematics A30
6 a 2  15 ax  6 a 2 x 2
5 a  8 ax  6 ax 2
 6 a 2  15 ax  6 a 2 x 2
6 a 2  5 x  2 ax 2
The expanded form of 2 a  7  is ***.
____
____
____
____
4.
a.
b.
c.
d.
The expanded form of 2 x  y 3 y  x  is ***.
____
____
____
____
3.

The expanded form of  3 a 2 a  5 x  2 ax 2 is ***.
a.
b.
c.
d.
16 a 3 b 3
4 ab
4 a 3 b3
48 a 3 b 3
109
Assignment 2
5.
To factor 7 x  7 xc  2 b  2 bc , a correct grouping of the terms is
***.
____
____
____
____
6.
8.
____
____
a.
b.
____
c.
____
d.
2 x  12 x  a 
4 xx  a   a4 x  a 
2 x  a 2
2
 12 x  a 
2
a.
b.
c.
d.
0
 21 ab
42 ab
 42 ab
One of the factors of 2 x 2 + 11 x  6 is ***.
a.
b.
c.
d.
x 6
x 6
2x 1
2x  6
Another way of writing 8 x 2  3 xy  5 y 2 is ***.
____
____
____
____
Mathematics A30
1
3 xa
3 xa  1
xa  1
The middle term of the expanded form of 7 a  3 b  is ***.
____
____
____
____
10.
a.
b.
c.
d.
One factored form of  4 x 2  4 ax  a 2 is ***.
____
____
____
____
9.
 7 xc  2bc  7 x  2b
 7 xc  2b  7 x  2bc
7 x  2bc   7 xc  2b
7 xc  2bc  7 x  2b
One of the factors of 9 x 2 a 2  1 is ***.
____
____
____
____
7.
a.
b.
c.
d.
a.
b.
c.
d.
8 x 2  18 xy  15 xy  5 y 2
8 x 2  4 xy  xy  5 y 2
8 x 2  8 xy  3 xy  5 y 2
8 x 2  2 xy  xy  5 y 2
110
Assignment 2
11.
12.
Another way of writing x 2  6 x  11 is ***.
____
____
____
a.
b.
c.
 x  9 2  2
x  10 x  1
____
d.
x  3 
14.
a.
b.
c.
d.
____
____
____
a.
b.
c.
____
d.
2
2
9 x5
27 x 2
27 x 6
9x2
5ay 2
25 ay 2
5ay 3
125 a 9 y 18
A factor of  x  1   2 3 is ***.
3
a.
b.
c.
d.
x 1
x 3
x2  4 x 1
x2  5
A factor of 8 x 3  27 y 6 is ***.
____
____
____
____
Mathematics A30

 6x  2 1
The cube root of 125 a 3 y 6 is ***.
____
____
____
____
15.
2
The cube of 3 x 2 is ***.
____
____
____
____
13.
x
a.
b.
c.
d.
4 x 2  6 xy 2  9 y 2
2x  3 y2
2x  3 y
4 x 2  6 xy 2  9 y 4
111
Assignment 2
16.
A factor of (2 x)3  ( x  2)3 is ***.
____
____
____
____
17.
 3x2  3x 1
 3x2  3x 1
1
1
a.
b.
c.
d.
1, 3, 1
2, 3, 1
2, 3,  1
2 x 2 , 3x,  1
a.
b.
c.
d.
1 + x3
1 + x + x2
1  x  x2
1+x
One of the factors of 12 x 2  8 x  15 is ***.
____
____
____
____
Mathematics A30
x3
x2
x3
x3
One of the factors of x 2  x 5 is ***.
____
____
____
____
20.
a.
b.
c.
d.
If ax2 + 3x + c = 2x2 + bx  1 , then a, b, c are equal to ***.
____
____
____
____
19.
3x  2
4 x 2  2 x( x  2)  ( x  2) 2
x2
x 2
The expanded form of ( x  1)3 is ***.
____
____
____
____
18.
a.
b.
c.
d.
a.
b.
c.
d.
6x 5
2x  3
2x + 3
3x + 5
112
Assignment 2
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.
B.
1.
Completely factor each of the following:
(2)
a.
a 3  216
(2)
b.
8 a 3 b3  c6
(2)
c.
16 x 4  y 2
Mathematics A30
113
Assignment 2
(2)
d.
2 y3  5 y2  3 y
(2)
e.
x  3 2  2 x  1 2
(2)
f.
36 x 4 r  42 x 3 r 2  18 x 2 r 3
Mathematics A30
114
Assignment 2
2.
Completely factor each of the trinomials and show a check
of the factors by multiplying.
(3)
a.
12x 2  2x  14
(3)
b.
x 2  35 xy  300 y 2
(3)
c.
6 x 2  30 x  900
(3)
d.
8 x 2  24 x  1 440
Mathematics A30
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Assignment 2
(3)
e.
2 x 4  10 x 3  72 x 2
(3)
f.
16 x 2  34 xy  15 y 2
3.
Find the greatest common factor from each of the following
expressions.
(3)
a.
48 y 2 ( y  2)( y  3) 2 ; 32 y 3 ( y  3)
(3)
b.
6 x 2  6 x; 4 x 2  24 x  20
Mathematics A30
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Assignment 2
(4)
(6)
c.
C.
16 ab 2  a 5 b 2 ; a 3 b  8 b
1.
R
r
Cross section of pipe
The formula for the area of a circle is A  r 2 , where r represents the
radius of the circle.
Show that the cross sectional area of the pipe shown in the diagram is
 times the product of the sum of the two radii and the difference of
the two radii.
Mathematics A30
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Assignment 2
(6)
2.
The volume of a sphere of radius R is given by
V 
4
R 3
3
A hollow spherical ball is made so that the outside radius is R and the
inside radius is r. Write an expression for the volume of material used
for the ball using the thickness T of the shell in the expression.
(8)
3.
(STUDENT JOURNAL)
For this lesson, write no more than a one page summary of the
material in this lesson. This summary is to be written in such a way
that it will be useful to you for studying for examinations.
(100 )
Mathematics A30
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Assignment 2