2
MODULE 5. FACTORING
5a
Common Factors
The Greatest Common Factor
We begin this section with definitions of factors and divisors. Because 24 =
2 · 12, both 2 and 12 are factors of 24. However, note that 2 is also a divisor
of 24, because when you divide 24 by 2 you get 12, with a remainder of zero.
Similarly, 12 is also a divisor of 24, because when you divide 24 by 12 you get
2, with a remainder of zero.
Factors and divisors. Suppose m and n are integers. Then m is a divisor
(factor ) of n if and only if there exists another integer k so that n = m · k.
Note that the words divisor and factor are equivalent. They have the same
meaning.
You Try It!
EXAMPLE 1. List the positive divisors (factors) of 24.
Solution: First, list all possible ways that we can express 24 as a product of
two positive integers:
24 = 1 · 24
or
24 = 2 · 12
or
24 = 3 · 8
or
24 = 4 · 6
Therefore, the positive divisors (factors) of 24 are 1, 2, 3, 4, 6, 8, and 24.
!
You Try It!
EXAMPLE 2. List the positive divisors (factors) that 36 and 48 have in
common.
Solution: First, list all positive divisors (factors) of 36 and 48 separately, then
box the divisors that are in common.
Divisors of 36 are: 1 , 2 , 3 , 4 , 6 , 9, 12 , 18, 36
Divisors of 48 are: 1 , 2 , 3 , 4 , 6 , 8, 12 , 16, 24, 48
Therefore, the common positive divisors (factors) of 36 and 48 are 1, 2, 3, 4,
6, and 12.
!
3
5A. COMMON FACTORS
Greatest common divisor. The greatest common divisor (factor) of a and b
is the largest positive number that divides evenly (no remainder) both a and b.
The greatest common divisor of a and b is denoted by the symbolism gcd(a, b).
We will also use the abbreviation gcf(a, b) to represents the greatest common
factor of a and b.
Remember, greatest common divisor and greatest common factor have the
same meaning. In Example 2, we listed the common positive divisors of 36 and
48. The largest of these common divisors was 12. Hence, the greatest common
divisor (factor) of 36 and 48 is 12, written gcd(36, 48) = 12.
With smaller numbers, it is usually easy to identify the greatest common
divisor (factor).
You Try It!
EXAMPLE 3. State the greatest common divisor (factor) of each of the
following pairs of numbers: (a) 18 and 24, (b) 30 and 40, and (c) 16 and 24.
Solution: In each case, we must find the largest possible positive integer that
divides evenly into both the given numbers.
a) The largest positive integer that divides evenly into both 18 and 24 is 6.
Thus, gcd(18, 24) = 6.
b) The largest positive integer that divides evenly into both 30 and 40 is 10.
Thus, gcd(30, 40) = 10.
c) The largest positive integer that divides evenly into both 16 and 24 is 8.
Thus, gcd(16, 24) = 6.
!
With larger numbers, it is harder to identify the greatest common divisor
(factor). However, prime factorization will save the day!
You Try It!
EXAMPLE 4. Find the greatest common divisor (factor) of 360 and 756.
Solution: Prime factor 360 and 756, writing your answer in exponential form.
756
360
10
36
4
2
9
2
3
2
3
9
5
3
84
3
7
12
3
4
2
2
4
MODULE 5. FACTORING
Thus:
360 = 23 · 32 · 5
756 = 22 · 33 · 7
To find the greatest common divisor (factor), list each factor that appears in
common to the highest power that appears in common.
In this case, the factors 2 and 3 appear in common, with 22 being the highest power of 2 and 32 being the highest power of 3 that appear in common.
Therefore, the greatest common divisor of 360 and 756 is:
GCD = 22 · 32
=4·9
= 36
Therefore, the greatest common divisor (factor) is gcd(360, 756) = 36. Note
what happens when we write each of the given numbers as a product of the
greatest common factor and a second factor:
360 = 36 · 10
756 = 36 · 21
In each case, note how the second second factors (10 and 21) contain no additional common factors.
!
Finding the Greatest Common Factor of Monomials
Example 4 reveals the technique used to find the greatest common factor of
two or more monomials.
Finding the GCF of two or more monomials. To find the greatest common factor of two or more monomials, proceed as follows:
1. Find the greatest common factor (divisor) of the coefficients of the given
monomials. Use prime factorization if necessary.
2. List each variable that appears in common in the given monomials.
3. Raise each variable that appears in common to the highest power that
appears in common among the given monomials.
5
5A. COMMON FACTORS
You Try It!
EXAMPLE 5. Find the greatest common factor of 6x3 y 3 and 9x2 y 5 .
Solution: To find the GCF of 6x3 y 3 and 9x2 y 5 , we note that:
1. The greatest common factor (divisor) of 6 and 9 is 3.
2. The monomials 6x3 y 3 and 9x2 y 5 have the variables x and y in common.
3. The highest power of x in common is x2 . The highest power of y in
common is y 3 .
Thus, the greatest common factor is gcf(6x3 y 3 , 9x2 y 5 ) = 3x2 y 3 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
6x3 y 3 = 3x2 y 3 · 2x
9x2 y 5 = 3x2 y 3 · 3y
Note how the set of second monomial factors (2x and 3y) contain no additional
common factors.
!
You Try It!
EXAMPLE 6. Find the greatest common factor of 12x4 , 18x3 , and 30x2 .
Solution: To find the GCF of 12x4 , 18x3 , and 30x2 , we note that:
1. The greatest common factor (divisor) of 12, 18, and 30 is 6.
2. The monomials 12x4 , 18x3 , and 30x2 have the variable x in common.
3. The highest power of x in common is x2 .
Thus, the greatest common factor is gcf(12x4 , 18x3 , 30x2 ) = 6x2 . Note what
happens when we write each of the given monomials as a product of the greatest
common factor and a second monomial:
12x4 = 6x2 · 2x2
18x3 = 6x2 · 3x
30x2 = 6x2 · 5
Note how the set of second monomial factors (2x2 , 3x, and 5) contain no
additional common factors.
!
6
MODULE 5. FACTORING
Factor Out the GCF
Previously we multiplied a monomial and polynomial by distributing the monomial times each term in the polynomial.
2x(3x2 + 4x − 7) = 2x · 3x2 + 2x · 4x − 2x · 7
= 6x3 + 8x2 − 14x
In this section we reverse that multiplication process. THat is, you are given
the final product and are asked to find original multiplication problem. In the
case 6x3 + 8x2 − 14x, the greatest common factor of 6x3 , 8x2 , and 14x is 2x.
We then use the distributive property to factor out 2x from each term of the
polynomial.
6x3 + 8x2 − 14x = 2x · 3x2 + 2x · 4x − 2x · 7
= 2x(3x2 + 4x − 7)
Factoring. Factoring is “unmultiplying.” You are given the product, then
asked to find the original multiplication problem.
First rule of factoring. If the terms of the given polynomial have a greatest
common factor (GCF), then factor out the GCF.
You Try It!
EXAMPLE 7. Factor: 6x2 + 10x + 14
Solution: The greatest common factor (GCF) of 6x2 , 10x and 14 is 2. Factor
out the GCF.
6x2 + 10x + 14 = 2 · 3x2 + 2 · 5x + 2 · 7
= 2(3x2 + 5x + 7)
Checking your work. Every time you factor a polynomial, remultiply to
check your work.
Check: Multiply. Distribute the 2.
2(3x2 + 5x + 7) = 2 · 3x2 + 2 · 5x + 2 · 7
= 6x2 + 10x + 14
That’s the original polynomial, so we factored correctly.
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5A. COMMON FACTORS
!
You Try It!
EXAMPLE 8. Factor: 12y 5 − 32y 4 + 8y 2
Solution: The greatest common factor (GCF) of 12y 5 , 32y 4 and 8y 2 is 4y 2 .
Factor out the GCF.
12y 5 − 32y 4 + 8y 2 = 4y 2 · 3y 3 − 4y 2 · 8y 2 + 4y 2 · 2
= 4y 2 (3y 3 − 8y 2 + 2)
Check: Multiply. Distribute the monomial 4y 2 .
4y 2 (3y 3 − 8y 2 + 2) = 4y 2 · 3y 3 − 4y 2 · 8y 2 + 4y 2 · 2
= 12y 5 − 32y 4 + 8y 2
That’s the original polynomial. We have factored correctly.
!
You Try It!
EXAMPLE 9. Factor: 12a3 b + 24a2 b2 + 12ab3
Solution: The greatest common factor (GCF) of 12a3 b, 24a2 b2 and 12ab3 is
12ab. Factor out the GCF.
12a3 b + 24a2 b2 + 12ab3 = 12ab · a2 − 12ab · 2ab + 12ab · b2
= 12ab(a2 + 2ab + b2 )
Check: Multiply. Distribute the monomial 12ab.
12ab(a2 + 2ab + b2 ) = 12ab · a2 − 12ab · 2ab + 12ab · b2
= 12a3 b + 24a2 b2 + 12ab3
That’s the original polynomial. We have factored correctly.
!
Speeding Things Up a Bit
After showing your work on a number of examples such as those in Examples 7,
8, and 9, you’ll need to learn how to perform the process mentally.
You Try It!
EXAMPLE 10. Factor each of the following polynomials: (a) 24x + 32,
(b) 5x3 − 10x2 − 10x, and (c) 2x4 y + 2x3 y 2 − 6x2 y 3 .
Solution: In each case, factor out the greatest common factor (GCF):
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MODULE 5. FACTORING
a) The GCF of 24x and 32 is 8. Thus, 24x + 32 = 8(3x + 4)
b) The GCF of 5x3 , 10x2 , and 10x is 5x. Thus:
5x3 − 10x2 − 10x = 5x(x2 − 2x − 2)
c) The GCF of 2x4 y, 2x3 y 2 , and 6x2 y 3 is 2x2 y. Thus:
2x4 y + 2x3 y 2 − 6x2 y 3 = 2x2 y(x2 + xy − 3y 2 )
When mentally factoring out the GCF, it is even more important that you check
your results. The check can also be done mentally. For example, in checking
the third result, mentally distribute 2x2 y times each term of x2 + xy − 3y 2 .
For example, multiplying 2x2 y times the first term x2 produces 2x4 y, the first
term in the original polynomial.
2x2 y (x2 + xy − 3y 2 ) = 2x4 y + 2x3 y 2 − 6x2 y 3
Continue in this manner by checking the product of 2x2 y with each term of
x2 + xy − 3y 2 . Make sure that each result agrees with the corresponding term
of the original polynomial.
!
The distributive property allows us to pull the GCF out in front or to pull
it out in back. In symbols:
ab + ac = a(b + c)
or
ba + ca = (b + c)a
You Try It!
EXAMPLE 11. Factor: 2x(3x + 2) + 5(3x + 2)
Solution: In this case, the greatest common factor (GCF) is 3x + 2.
2x(3x + 2) + 5(3x + 2) = 2x · (3x + 2) + 5 · (3x + 2)
= (2x + 5)(3x + 2)
!
You Try It!
EXAMPLE 12. Factor: 15a(a + b) − 12(a + b)
Solution: In this case, the greatest common factor (GCF) is 3(a + b).
15a(a + b) − 12(a + b) = 3(a + b) · 5a − 3(a + b) · 4
= 3(a + b)(5a − 4)
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5A. COMMON FACTORS
Alternate solution: It is possible that you might fail to notice that 15 and
12 are divisible by 3, factoring out only a common factor a + b.
15a(a + b) − 12(a + b) = 15a · (a + b) − 12 · (a + b)
= (15a − 12)(a + b)
However, you now need to notice that you can continue, factoring out a 3 from
both 15a and 12.
= 3(5a − 4)(a + b)
Note that the order of factors differs from the first solution, but because of
the commutative property of multiplication, the order does not matter. The
answers are the same.
!
Factoring by Grouping
The final factoring skill in this section involves four-term expressions. The
technique for factoring a four-term expression is called factoring by grouping.
You Try It!
EXAMPLE 13. Factor: 2x2 + 3x + 2xy + 3y
Solution: We “group” the first and second terms, noting that we can factor
an x out of both of these terms. Then we “group” the third and fourth terms,
noting that we can factor a y out of both of these terms.
2x2 + 3x + 2xy + 3y = x (2x + 3) + y (2x + 3)
Now we note that we can factor a 2x + 3 out of both of these terms.
= (x + y)(2x + 3)
!
You Try It!
EXAMPLE 14. Factor: ac − bd + ad − bc
Solution: In the current order, there really isn’t any common factor in the
first and second terms that we can factor out, nor is there a common factor in
the third and fourth terms that we can factor out.
ac − bd + ad − bc
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MODULE 5. FACTORING
Upon rearranging the terms, we note that we can factor out an a from the first
and second terms and a b from the third and fourth terms.
ac + ad − bd − bc = a (c + d) + b (−d − c)
However, this is not helpful, because it does not produce a common factor.
Let’s try factoring a −b from the third and fourth terms instead.
ac + ad − bd − bc = a (c + d) − b (d + c)
At first glance, this still doesn’t seem helpful until we realize that d + c is the
same as c + d.
= a (c + d) − b (c + d)
Now we can factor a c + d out of both of these terms.
= (a − b)(c + d)
!
You Try It!
EXAMPLE 15. Factor: 2xy − 2xz + yz − y 2
Solution: Note that we can factor a 2x out of the first two terms and a y out
of the second two terms.
2xy − 2xz + yz − y 2 = 2x (y − z) + y (z − y)
However, this is not helpful, because it does not produce a common factor.
Let’s try factoring a −y from the third and fourth terms instead.
2xy − 2xz + yz − y 2 = 2x (y − z) − y (y − z)
That produced a common factor of y − z, which we can now factor out.
= (2x + y)(y − z)
!