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Algebra
Module A33
Factoring - 2
Copyright
This publication © The Northern
Alberta Institute of Technology
2002. All Rights Reserved.
LAST REVISED November, 2008
Factoring - 2
Statement of Prerequisite Skills
Complete all previous TLM modules before beginning this module.
Required Supporting Materials
Access to the World Wide Web.
Internet Explorer 5.5 or greater.
Macromedia Flash Player.
Rationale
Why is it important for you to learn this material?
Factoring is the reverse of finding a product. Some factors are so common it is useful to
recognize them by sight. This module will continue presenting strategies for approaching
common factoring problems.
Learning Outcome
When you complete this module you will be able to…
Factor a variety of algebraic expressions.
Learning Objectives
1.
2.
3.
4.
Identify the factors of x2 + qx + r.
Identify the factors of ax2 + bx + c.
Identify the factors of a3 + b3.
Identify the factors of a3 − b3.
Connection Activity
Recall from the products module that (x + 9)(x − 4) = x2 + 5x − 36. Knowing this it is
easy to recognize the factors of x2 + 5x − 36. They are (x + 9) and (x − 4). Can you think
of the factors of other common products that you have encountered?
Module A33 – Factoring - 2
1
OBJECTIVE ONE
When you complete this objective you will be able to…
Å When the x2 coefficient = 1
Identify the factors of x2 + qx + r.
Exploration Activity
In this expression r may or may not be a perfect square. We can write the factors of the
expression x2 + qx + r as follows:
x2 + qx + r = (x + a)(x + b)
If the factors x + a and x + b are multiplied using the FOIL method we arrive at the
following result:
x2 + qx + r = (x + a)(x + b)
= x2 + bx + ax + ab
= x2 + (a + b)x + ab
This implies that the two integers a and b have a product equal to r. It also implies that
the sum of these two integers a + b is equal to q.
EXAMPLE 1
Factor the following.
x2 + 8x + 15 = ____________
SOLUTION:
Step 1: Find a pair of integers whose product is 15 and whose sum is 8
(3)(5) =15 and 3 + 5 = 8
Step 2: Writing the factors you get:
= (x + 3)(x + 5)
Step 3: Checking the answer using FOIL you get:
= x2 + 5x + 3x + 15
= x2 + 8x + 15
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Module A33 – Factoring - 2
EXAMPLE 2
Factor the following.
x2 − 9x + 20 = ____________
SOLUTION:
Step 1: Find a pair of integers whose product is 20 and whose sum is −9
(−4)(−5) = 20 and −4 −5 = −9
Step 2: Writing the factors you get:
= (x − 4)(x − 5)
Step 3:Checking the answer using FOIL you get:
= x2 − 5x − 4x − 20
= x2 − 9x + 20
EXAMPLE 3
Factor the following.
x2 − 2x − 35 = ___________
SOLUTION:
Step 1: Find a pair of integers whose product is −35 and whose sum is −2
(−7)(5) = −35 and −7 + 5 = −2
Step 2: Writing the factors you get:
= (x − 7)(x + 5)
Step 3: Checking the answer using FOIL you get:
= x2 − 7x + 5x − 35
= x2 − 2x − 35
Module A33 – Factoring - 2
3
EXAMPLE 4
Factor the following.
x2 + 5x − 36 = __________
SOLUTION:
Step 1: Find a pair of integers whose product is −36 and whose sum is 5
(9)(−4) = −36 and 9 − 4 = 5
Step 2: Writing the factors you get:
= (x + 9)(x − 4)
Step 3: Checking the answer using FOIL you get:
= x2 + 9x − 4x − 36
= x2 + 5x − 36
4
Module A33 – Factoring - 2
Experiential Activity One
Factor the following expressions completely and check your answers.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
x2 + 11x + 18
x2 − 14x + 48
x2 + 6x − 7
x2 − 5x − 24
x2 − 4x − 60
x2 + 17x + 60
x2 − 15x + 36
x2 + 8x − 33
2ax2 + 24ax + 64a
Show Me.
x4 − 13x2 + 36
Experiential Activity One Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
(x + 9)(x + 2)
(x − 8)(x − 6)
(x + 7)(x − 1)
(x − 8)(x + 3)
(x − 10)(x + 6)
(x + 12) (x + 5)
(x − 12)(x − 3)
(x + 11)(x − 3)
2a(x + 8)(x + 4)
Note: common factors are removed before factoring as a trinomial.
10. Solution:
x4 − 13x2 + 36
(x2 − 9)(x2 − 4)
(x + 3)(x − 3)(x + 2)(x − 2)
Module A33 – Factoring - 2
5
OBJECTIVE TWO
When you complete this objective you will be able to…
Identify the factors of ax2 + bx + c. Å When the lead coefficient ≠ 1
Exploration Activity
In this expression both ax2 and c may or may not be perfect squares. We can write the
factors of the expression ax2 + bx + c as follows:
ax2 + bx + c = (px + q)(rx + s)
Most textbooks show students a method of trial and error to factor these types of
trinomials. Using the trial and error method you select number combinations that may
work and then test them to see if they work. This strategy may produce the answer very
quickly, however some trinomials have many combinations and this guessing and
checking becomes very tedious. There is a strategy that will help you.
A technique similar to the one used in the last section can be used to eliminate the trial
and error process.
EXAMPLE 1
Factor the following.
3x2 + 8x + 4 = ____________
SOLUTION:
From the general case above we determine a = 3, b = 8 and c = 4.
•
Determine the product a⋅ c. From the above ac = 12.
•
Now find a pair of factors of ac whose sum is equal to b. Since ac = 12 we want
to find a pair of factors whose sum is 8.
(2)(6) = 12 and 2 + 6 = 8
•
Now rewrite the original polynomial replacing 8x with 2x + 6x. The order is not
important. However, if you have one negative and one positive it is easier to
factor if you put the negative first.
3x2 + 2x + 6x + 4
•
Factor by grouping the first two terms of the expression with the last two terms
of the expression.
(3x2 + 2x) + (6x + 4)
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Module A33 – Factoring - 2
Factor these grouped expressions separately.
x(3x + 2) + 2(3x + 2)
Since (3x + 2) is a common factor from each group you can factor it from each
term.
(3x + 2)(x + 2) should be the factors of the original trinomial.
•
Check the result using FOIL
(3x + 2)(x + 2) = 3x2 + 2x + 6x + 4
= 3x2 + 8x + 4
The above technique can be used to replace the method of trial and error. The suggestion
is to first take an educated guess at the solution. If this does not produce the answer
quickly use the above method.
EXAMPLE 2
Factor the following.
4x2 − 17x + 18 = ___________
SOLUTION:
We have a = 4, b = −17 and c = 18.
•
From the above a × c = (4)(18) = 72
•
Now find a pair of factors of ac whose sum is equal to b = −17.
(−9)(−8) = 72 and (−9) + (−8) = −17
•
Now rewrite the original polynomial replacing −17x with −9x − 8x.
4x2 − 9x − 8x + 18
•
Group the terms and watch the signs.
(4x2 − 9x) − (8x − 18)
Factor these expressions separately.
x(4x − 9) − 2(4x − 9)
Since (4x − 9) is a common factor.
(4x − 9)(x − 2) should be the factors of the original trinomial.
Module A33 – Factoring - 2
7
•
Check the result using FOIL
(4x − 9)(x − 2) = 4x2 − 8x − 9x + 18
= 4x2 − 17x + 18
EXAMPLE 3
Factor the following.
24x2 − 106x − 9 = __________
SOLUTION:
We have a = 24, b = −106 and c = −9.
•
From the above a × c = (24)(−9) = −216
•
Now find a pair of factors of ac whose sum is equal to b = −106.
(2)(−108) = −216 and (2) + (−108) = −106
•
Now rewrite the original polynomial replacing −106x with 2x − 108x.
24x2 + 2x − 108x − 9
•
Group the terms and watch the signs.
(24x2 + 2x) − (108x + 9)
Factor these expressions separately.
2x(12x + 1) − 9(12x + 1)
Since (12x + 1) is a common factor.
(12x + 1)(2x − 9) should be the factors of the original trinomial.
•
Check the result using FOIL
(12x + 1)(2x − 9) = 24x2 − 108x + 2x − 9
= 24x2 − 106x − 9
•
So the factors are:
= (12x + 1)(2x − 9)
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Module A33 – Factoring - 2
EXAMPLE 4
Factor the following.
21p2 − 14r2 + 7pr = __________
SOLUTION:
We need to rearrange first to get:
= 21p2 + 7pr − 14r2
No factor out a common factor of 7 to get:
= 7(3p2 + pr − 2r2)
We can now factor the remaining trinomial. We have a = 3, b = 1 and c = −2.
•
From the above a × c = (3)(−2) = −6
•
Now find a pair of factors of ac whose sum is equal to b = 1.
(3)(−2) = −6 and (3) + (−2) = 1
•
Now rewrite the original polynomial replacing qr with 3pr − 2pr.
7(3p2 + 3pr − 2pr − 2r2)
•
Group the terms and watch the signs.
7(3p2 + 3pr) − (2pr + 2r2)
Factor these expressions separately.
7(3p(p + r) − 2r(p + r))
Since (p + r) is a common factor.
7(p + r)(3p − 2r) should be the factors of the original trinomial.
•
Check the result using FOIL
7(p + r)(3p − 2r) = 7(3p2 + 3pr − 2pr − 2r2)
= 7(3p2 + pr − 2r2)
= 21p2 + 7pr − 14r2
Module A33 – Factoring - 2
9
Experiential Activity Two
Factor the following expressions completely and check your answers.
1.
2.
3.
4.
5.
6.
7.
8.
2x2 + 7x + 5
2x2 − 15x + 7
6x2 + x − 5
6x2 − 25x − 9
6x2 + 23x + 20
16x2 − 38x − 5
18x2 − 21x − 4
36x2 − 53x + 10
Show Me.
Experiential Activity Two Answers
1.
2.
3.
4.
5.
6.
7.
8.
(2x + 5)(x + 1)
(2x − 1)(x − 7)
(6x − 5)(x + 1)
(3x + 1)(2x − 9)
(3x + 4)(2x + 5)
(8x + 1)(2x − 5)
(6x + 1)(3x − 4)
(9x − 2)(4x − 5)
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Module A33 – Factoring - 2
OBJECTIVE THREE
When you complete this objective you will be able to…
Identify the factors of a3 + b3.
Exploration Activity
This is referred to as a sum of cubes because each of the terms is a perfect cube and is
separated by a plus sign. A sum of cubes will always factor using the following
procedure:
a3 + b3= (a + b)(a2 − ab + b2)
The first factor is a binomial while the second factor is a trinomial. Therefore when
factoring a sum of cubes the following steps can be used.
• The terms in the first factor are the cube roots of the terms in the question. That
is:
3
•
•
•
•
a 3 = a and
3
b3 = b
The middle sign of the first factor is the same as the sign of the binomial to be
factored.
The first term in the second factor is the square of “a” and is always positive.
The last term in the second factor is the square of “b” and is always positive.
The middle term in the second factor is the product of “a” and “b” and always
has the opposite sign of the binomial. This trinomial CANNOT be factored
further, however the binomial factor can sometimes be factored further.
EXAMPLE 1
Factor the following: 8x3 + 27y3 = _________
SOLUTION:
Remember: a3 + b3 = (a + b)(a2 − ab + b2)
So to find the terms of the first factor we find the cube root of 8x3 and 27y3.
3
8 x 3 = 2 x and
3
27 y 3 = 3 y
The first factor is: (2x + 3y)
Now use (2x + 3y) to find the second factor, i.e. the trinomial factor.
The first term in the trinomial is (2x)2 = 4x2, the last term in the trinomial is (3y)2 = 9y2
while the middle term in the trinomial is the negative of (2x)(3y) = −6xy.
Therefore: 8x3 + 27y3 = (2x + 3y)(4x2 − 6xy + 9y2)
You should always check your answer by multiplying the factors.
Module A33 – Factoring - 2
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EXAMPLE 2
Factor the following:
27r3 + s3 = _________
SOLUTION:
Remember: a3 +b3 = (a + b)(a2 − ab + b2)
To find the terms of the first factor we find the cube root of 27r3 and s3
3
27 r 3 = 3r and
3
s3 = s
The first factor is: (3r + s)
Now use (3r + s) to find the trinomial factor.
The first term is (3r)2 = 9r2, the last term is (s)2 = s2 while the middle term is the negative
of (3r)(s) = −3rs.
Therefore: 27r3 + s3 = (3r + s)(9r2 − 3rs + s2)
Sometimes you will not immediately recognize the expression as a sum of cubes.
EXAMPLE 3
Factor the following:
4tm6 + 32tn6 = ___________
SOLUTION: Factor a 4t from each term as it is common to each term.
4t(m6 + 8n6) ← Second factor is a sum of cubes.
Remember: a3 + b3 = (a + b)(a2 − ab + b2)
To find the terms of the first factor we find the cube root of m6 and 8n6.
3
m 6 = m 2 and
3
8n 6 = 2n 2
The first factor is: (m2 + 2n2)
Now use (m2 + 2n2) to find the second factor, the trinomial part.
The first term is (m2)2 = m4, the last term is (2n2)2 = 4n4
The middle term is the negative of (m2)(2n2) = −2m2n2
Therefore: 4tm6 + 32tn6 = 4t(m2 + 2n2)(m4 − 2m2n2 + 4n4)
Don’t forget the common factor from the first step!!
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Module A33 – Factoring - 2
Experiential Activity Three
Factor the following expressions completely and check your answers.
1.
2.
3.
4.
5.
6.
7.
8.
27x3 + 125y3
x3 + 64y3
125x3 + 8y3
216x3 + 125y3
27x3 + 64
x4 + xy3
54x3 + 2y3 Show Me.
x6 + y6
Experiential Activity Three Answers
1.
2.
3.
4.
5.
6.
7.
8.
(3x + 5y) (9x2 −15xy + 25y2)
(x + 4y) (x2 − 4xy + 16y2)
(5x + 2y)(25x2 − l0xy + 4y2)
(6x + 5y)(36x2 − 30xy + 25y2)
(3x + 4)(9x2 − 12x + 16)
x(x + y)(x2 − xy + y2)
2(3x + y)(9x2 − 3xy + y2)
(x2 + y2)(x4 − x2y2 + y4)
Module A33 – Factoring - 2
13
OBJECTIVE FOUR
When you complete this objective you will be able to…
Identify the factors of a3 − b3.
Exploration Activity
The expression a3 − b3 is called a difference of cubes because each term is a perfect cube
separated by a minus sign. The procedure for factoring is the same as stated in the
previous objective except the signs change. Study the changes in the signs in the
following rule.
a3 − b3 = (a − b)(a2 + ab + b2)
Note that the first factor has the same middle sign as the binomial to be factored. The
middle term of the second factor has the opposite sign to that in the first factor.
EXAMPLE 1
Factor the following.
27x3 − 64y3 = ____________
SOLUTION:
Remember: a3 − b3 = (a − b)(a2 + ab + b2)
To find the terms of the first factor we find the cube root of 27x3 and 64y3
3
27 x 3 = 3 x and
3
64 y 3 = 4 y
The first factor is: (3x − 4y)
Now use (3x − 4y) to find the second factor.
The first term is (3x)2 = 9x2, the last term is (4y)2 = 16y2 while the middle term is the
product of (3x)(4y) = 12xy.
Therefore: 27x3 − 64y3 = (3x − 4y) (9x2 + 12xy + 16y2)
Sometimes it is necessary to factor the expression before you will recognize it as a
difference of cubes.
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Module A33 – Factoring - 2
EXAMPLE 2
Factor the following.
24ax3 − 81ay3 = ___________
SOLUTION:
You must first recognize that 3a is a common factor of both terms. When you factor 3a
from each term it becomes a difference of cubes.
3a(8x3 − 27y3)
Remember: a3 − b3 = (a − b)(a2 + ab + b2)
To find the terms of the first factor we find the cube root of 8x3 and 27y3.
3
8 x 3 = 2 x and
3
27 y 3 = 3 y
The first factor is: (2x − 3y)
Now use (2x − 3y) to find the second factor.
The first term is (2x)2 = 4x2, the last term is (3y)2 = 9y2 while the middle term is the
product of (2x)(3y) = 6xy.
Therefore: 24ax3 − 81ay3 = 3a(2x − 3y)(4x2 + 6xy + 9y2)
Sometimes it will be necessary to use factoring rules in combination with one
another.
Module A33 – Factoring - 2
15
EXAMPLE 3
Factor the following.
24x6 − 24 = ___________
SOLUTION:
First recognize that 24 is a factor of both terms. When you factor 24 from each term the
remaining terms to be factored become a difference of cubes.
24(x6 − 1)
x 6 − 1 is a difference of squares and a difference of cubes.
In this case always try to factor the difference of squares first.
(
)(
)
24 x 3 − 1 x 3 + 1
Now factor x3 − 1 and x3 + 1 as a difference and a sum of cubes
(
)
(
)
24(x − 1) x 2 + x + 1 (x + 1) x 2 − x + 1
x3 – 1 factored
x3 + 1 factored
Method of Attacking Factoring Situations:
See Appendix in Module A32 – Factoring.
Generally, the following provides an orderly process to factoring and leads to a
completely factored solution.
1. Remove all the common factors.
2. Is the expression left a binomial or a trinomial?
a. If it is a binomial (2 terms) is it
Do first if more than one applies!!
i. the difference of squares (a2 − b2)
3
3
ii. the sum of cubes (a + b )
iii. the difference of cubes (a3 − b3)
b. If it is a trinomial (3 terms) is it
i. a perfect square (first and last terms are perfect squares with the
middle term as twice the product of the square roots of the first
and last terms)
ii. factorable as ax2 + bx + c where the numbers sought have the
product of ac and the sum of b
3. Does the expression left have four or more terms − try grouping.
WHEN YOU THINK THAT YOU HAVE FINISHED, CHECK EACH FACTOR TO
SEE IF IT CAN BE FACTORED FURTHER.
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Module A33 – Factoring - 2
Experiential Activity Four
Factor the following expressions completely and check your answers.
1.
2.
3.
4.
5.
6.
7.
8.
64x3 − y3
27x3 − 8y3
125x3 − 216y3
x3 − 27y3
125x3 − 64y3
3y3 − 81
r5 − r2s3
Show Me.
p9 − q9
Experiential Activity Four Answers
1.
2.
3.
4.
5.
6.
7.
8.
(4x − y)(16x2 + 4xy + y2)
(3x − 2y) (9x2 + 6xy + 4y2)
(5x − 6y)(25x2 + 30xy + 36y2)
(x − 3y) (x2 + 3xy + 9y2)
(5x − 4y)(25x2 + 20xy + 16y2)
3(y − 3)(y2 + 3y + 9)
r2(r − s) (r2 + rs + s2)
(p − q) (p2 + pq + q2)(p6 + p3q3 + q6)
Module A33 – Factoring - 2
17
Practical Application Activity
Complete the Factoring - 2 module assignment in TLM.
Summary
Students continued to develop additional skills in factoring algebraic expressions.
18
Module A33 – Factoring - 2