Download Factoring Polynomials: The Greatest Common Factor

Connexions module: m21913
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Factoring Polynomials: The
∗
Greatest Common Factor
Wade Ellis
Denny Burzynski
This work is produced by The Connexions Project and licensed under the
Creative Commons Attribution License †
Abstract
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. Factoring is an
essential skill for success in algebra and higher level mathematics courses. Therefore, we have taken
great care in developing the student's understanding of the factorization process. The technique is
consistently illustrated by displaying an empty set of parentheses and describing the thought process
used to discover the terms that are to be placed inside the parentheses. The factoring scheme for special
products is presented with both verbal and symbolic descriptions, since not all students can interpret
symbolic descriptions alone. Two techniques, the standard "trial and error" method, and the "collect
and discard" method (a method similar to the "ac" method), are presented for factoring trinomials with
leading coecients dierent from 1. Objectives of this module: understand more clearly the factorization
process, be able to determine the greatest common factor of two or more terms.
1 Overview
• Factoring Method
• Greatest Common Factor
2 Factoring Method
In the last two types of problems (Sections here1 and here2 ), we knew one of the factors and were able to
determine the other factor through division. Suppose, now, we're given the product without any factors.
Our problem is to nd the factors, if possible. This procedure and the previous two procedures are based on
the distributive property.
We will use the distributive property in reverse.
∗ Version
1.4: May 31, 2009 6:53 pm GMT-5
† http://creativecommons.org/licenses/by/3.0/
1 "Factoring Polynomials: Finding the factors of a Monomial" <http://cnx.org/content/m18870/latest/>
2 "Factoring Polynomials: Factoring a Monomial from a Polynomial" <http://cnx.org/content/m21906/latest/>
http://cnx.org/content/m21913/1.4/
Connexions module: m21913
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ab + ac = a (b + c)
| {z } | {z }
product
factors
We notice that in the product, ais common to both terms. (In fact, ais a common factor of both terms.)
Since ais common to both terms, we will factor it out and write
a(
)
Now we need to determine what to place inside the parentheses. This is the procedure of the previous
section. Divide each term of the product by the known factor a.
ac
ab
a = b and
a =c
Thus, band care the required terms of the other factor. Hence,
ab + ac = a (b + c)
When factoring a monomial from a polynomial, we seek out factors that are not only common to each
term of the polynomial, but factors that have these properties:
1. The numerical coecients are the largest common numerical coecients.
2. The variables possess the largest exponents common to all the variables.
3 Greatest Common Factor
A monomial factor that meets the above two requirements is called the greatest common factor of the
polynomial.
4 Sample Set A
Example 1
Factor 3x − 18.
The greatest common factor is 3.
3x − 18 = 3 · x − 3 · 6
Factor out 3.
3x − 18 = 3 (
)
Divide each term of the product by 3.
3x
3
= x and
−18
3
= −6
(Try to perform this division mentally.)
3x − 18 = 3 (x − 6)
Example 2
Factor 9x3 + 18x2 + 27x.
Notice that 9xis the greatest common factor.
9x3 + 18x2 + 27x = 9x · x2 + 9x · 2x + 9x · 3. Factor out 9x.
9x3 + 18x2 + 27x = 9x (
2
Mentally divide 9x into each term of the product.
)
9x + 18x + 27x = 9x x + 2x + 3
3
2
Example 3
Factor 10x2 y 3 − 20xy 4 − 35y 5 .
Notice that 5y 3 is the greatest common factor. Factor out 5y 3 .
(
10x2 y 3 − 20xy 4 − 35y 5 = 5y 3 (
)
Mentally divide 5y 3 into each term of the product and place the resulting quotients inside the
).
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Connexions module: m21913
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10x2 y 3 − 20xy 4 − 35y 5 = 5y 3 2x2 − 4xy − 7y 2
Example 4
Factor −12x5 + 8x3 − 4x2 .
We see that the greatest common factor is −4x2 .
−12x5 + 8x3 − 4x2 = −4x2 (
)
Mentally dividing −4x2 into each term of the product, we get
−12x5 + 8x3 − 4x2 = −4x2 3x3 − 2x + 1
5 Practice Set A
Exercise 1
(Solution on p. 7.)
Exercise 2
(Solution on p. 7.)
Exercise 3
(Solution on p. 7.)
Exercise 4
(Solution on p. 7.)
Factor 4x − 48.
Factor 6y 3 + 24y 2 + 36y.
Factor 10a5 b4 − 14a4 b5 − 8b6 .
Factor − 14m4 + 28m2 − 7m.
6
Consider this problem: factor Ax + Ay.Surely, Ax + Ay = A (x + y) .We know from the very beginning of
our study of algebra that letters represent single quantities. We also know that a quantity occurring within
a set of parentheses is to be considered as a single quantity. Suppose that the letter Ais representing the
quantity (a + b) .Then we have
Ax +
Ay =
A (x + y)
(a + b) x + (a + b) y = (a + b) (x + y)
When we observe the expression
(a + b) x + (a + b) y
we notice that (a + b)is common to both terms. Since it is common, we factor it out.
(a + b) (
)
As usual, we determine what to place inside the parentheses by dividing each term of the product by
(a + b) .
(a+b)x
(a+b)
=x
Thus, we get
and
(a+b)y
(a+b)
=y
(a + b) x + (a + b) y = (a + b) (x + y)
This is a forerunner of the factoring that will be done in Section 5.4.
7 Sample Set B
Example 5
Factor (x − 7) a + (x − 7) b.
Notice that (x − 7)is the greatest common factor. Factor out (x − 7) .
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Connexions module: m21913
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(x − 7) a + (x − 7) b =
Then, (x−7)a
(x−7)
=
(x − 7) a + (x − 7) b =
(x − 7)
a and (x−7)b
(x−7) = b.
(x − 7) (a + b)
Example 6
Factor 3x2 (x + 1) − 5x (x + 1).
Notice that xand (x + 1)are common to both terms. Factor them out. We'll perform this factorization by letting A = x (x + 1) . Then we have
3xA − 5A
=
A (3x − 5)
But A
=
x (x + 1) , so
3x (x + 1) − 5x (x + 1)
=
x (x + 1) (3x − 5)
2
8 Practice Set B
Exercise 5
(Solution on p. 7.)
Exercise 6
(Solution on p. 7.)
Factor (y + 4) a + (y + 4) b.
Factor 8m3 (n − 4) − 6m2 (n − 4) .
9 Exercises
For the following problems, factor the polynomials.
Exercise 7
(Solution on p. 7.)
9a + 18
Exercise 8
6a + 24
Exercise 9
(Solution on p. 7.)
8b + 12
Exercise 10
16x + 12
Exercise 11
(Solution on p. 7.)
4x − 6
Exercise 12
8x − 14
Exercise 13
(Solution on p. 7.)
21y − 28
Exercise 14
16f − 36
Exercise 15
12x2 + 18x
Exercise 16
10y 2 + 15y
http://cnx.org/content/m21913/1.4/
(Solution on p. 7.)
Connexions module: m21913
Exercise 17
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(Solution on p. 7.)
8y 2 + 18
Exercise 18
7x2 − 21
Exercise 19
(Solution on p. 7.)
3y 2 − 6
Exercise 20
2x2 − 2
Exercise 21
(Solution on p. 7.)
6y 2 − 6y
Exercise 22
ax2 − a
Exercise 23
(Solution on p. 7.)
by 2 + b
Exercise 24
7by 2 + 14b
Exercise 25
(Solution on p. 7.)
5a2 x2 + 10x
Exercise 26
24ax2 + 28a
Exercise 27
(Solution on p. 7.)
10x2 + 5x − 15
Exercise 28
12x2 − 8x − 16
Exercise 29
(Solution on p. 7.)
15y 3 − 24y + 9
Exercise 30
ax2 + ax + a
Exercise 31
(Solution on p. 7.)
by 3 + by 2 + by + b
Exercise 32
2y 2 + 6y + 4xy
Exercise 33
(Solution on p. 7.)
9x2 + 6xy + 4x
Exercise 34
30a2 b2 + 40a2 b2 + 50a2 b2
Exercise 35
(Solution on p. 7.)
13x2 y 5 c − 26x2 y 5 c − 39x2 y 5
Exercise 36
−4x2 − 12x − 8
Exercise 37
−6y 3 − 8y 2 − 14y + 10
Exercise 38
Ab + Ac
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(Solution on p. 7.)
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Exercise 39
(Solution on p. 7.)
Nx + Ny
Exercise 40
Qx + Qy
Exercise 41
(Solution on p. 7.)
Ax − Ay
Exercise 42
(x + 4) b + (x + 4) c
Exercise 43
(Solution on p. 7.)
(x − 9) a + (x − 9) b
Exercise 44
(2x + 7) a + (2x + 7) b
Exercise 45
(Solution on p. 8.)
(9a − b) w − (9a − b) x
Exercise 46
(5 − v) X + (5 − v) Y
Exercise 47
(Solution on p. 8.)
3x5 y 4 − 12x3 y 4 + 27x5 y 3 − 6x2 y 6
Exercise 48
8a3 b15 + 24a2 b14 + 48a3 b6 − 20a3 b7 + 80a4 b6 − 4a3 b7 + 4a2 b
Exercise 49
(Solution on p. 8.)
−8x3 y 2 − 3x3 y 2 + 16x4 y 3 + 2x2 y
10 Exercises for Review
Exercise 50
( here3 ) A quantity
Exercise 51
( here4 ) Solve
Exercise 52
( here5 ) Given
plus 21% more of that quantity is 26.25. What is the original quantity?
the equation 6 (t − 1) = 4 (5 − s) if s = 2.
(Solution on p. 8.)
that 4a3 is a factor of 8a3 − 12a2 ,nd the other factor.
3 "Solving Linear Equations and Inequalities: Application II - Solving Problems"
<http://cnx.org/content/m21980/latest/>
4 "Solving Linear Equations and Inequalities: Linear Equations in Two Variables"
<http://cnx.org/content/m21982/latest/>
5 "Factoring Polynomials: Factoring a Monomial from a Polynomial" <http://cnx.org/content/m21906/latest/>
http://cnx.org/content/m21913/1.4/
Connexions module: m21913
Solutions to Exercises in this Module
Solution to Exercise 1 (p. 3)
4 (x − 12)
Solution to Exercise
2 (p. 3)
6y y 2 + 4y + 6
Solution to Exercise
3 (p. 3)
2b4 5a5 − 7a4 b − 4b2
Solution to Exercise
4 (p. 3)
−7m 2m3 − 4m + 1
Solution to Exercise 5 (p. 4)
(y + 4) (a + b)
Solution to Exercise 6 (p. 4)
2m2 (n − 4) (4m − 3)
Solution to Exercise 7 (p. 4)
9 (a + 2)
Solution to Exercise 9 (p. 4)
4 (2b + 3)
Solution to Exercise 11 (p. 4)
2 (2x − 3)
Solution to Exercise 13 (p. 4)
7 (3y − 4)
Solution to Exercise 15 (p. 4)
6x (2x + 3)
Solution to
Exercise 17 (p. 4)
2 4y 2 + 9
Solutionto Exercise 19 (p. 5)
3 y2 − 2
Solution to Exercise 21 (p. 5)
6y (y − 1)
Solutionto Exercise 23 (p. 5)
b y2 + 1
Solution to Exercise 25 (p. 5)
5x a2 x + 2
Solution to Exercise
27 (p. 5)
5 2x2 + x − 3
Solution to Exercise
29 (p. 5)
3 5y 3 − 8y + 3
Solution to Exercise
31 (p. 5)
b y3 + y2 + y + 1
Solution to Exercise 33 (p. 5)
x (9x + 6y + 4)
Solution to Exercise 35 (p. 5)
13x2 y 5 (−c − 3)
Solution to Exercise 37 (p. 5)
3
2
−2 3y + 4y + 7y − 5
Solution to Exercise 39 (p. 5)
N (x + y)
Solution to Exercise 41 (p. 6)
A (x − y)
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Connexions module: m21913
Solution to Exercise 43 (p. 6)
(x − 9) (a + b)
Solution to Exercise 45 (p. 6)
(9a − b) (w − x)
Solution to Exercise 47 (p. 6)
3x2 y 3 x3 y − 4xy + 9x3 − 2y 3
Solution to Exercise 49 (p. 6)
−x2 y 11xy − 16x2 y 2 − 2
Solution to Exercise 51 (p. 6)
t=3
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