The
Factor;
Factoring
FindGreatest
the greatestCommon
common factor
of a list
of terms. by Grouping
6.1
1
2
Factor out the greatest common factor.
3
Factor by grouping.
4
Factor trinomials with a coefficient of 1 for the second-degree
term.
5
Factor trinomials with a coefficient greater than 1 for the
second degree term.
6
Factor such trinomials after factoring out the greatest common
factor.
7
Factor difference of squares
8
Factor perfect square trinomial
The Greatest Common Factor: Factoring by Grouping
To factor means “to write a quantity as a product.” That is, factoring is
the opposite of multiplying. For example,
Multiplying
Factoring
6 · 2 = 12
12 = 6 · 2
Factors
Product
Product
other factored forms of 12 are
− 6(−2), 3 · 4, −3(−4), 12 · 1,
and
Factors
−12(−1).
More than two factors may be used, so another factored form of 12 is
2 · 2 · 3. The positive integer factors of 12 are
1, 2, 3, 4, 6, 12.
Find the greatest common factor of a list of terms.
An integer that is a factor of two or more integers is called a common
factor of those integers. For example, 6 is a common factor of 18 and
24. Other common factors of 18 and 24 are 1, 2, and 3. The greatest
common factor (GCF) of a list of integers is the largest common
factor of those integers. Thus, 6 is the greatest common factor of 18
and 24.
Recall that a prime number has only itself and 1 as factors. Factoring
numbers into prime factors is the first step in finding the greatest
common factor of a list of numbers.
Find the greatest common factor of a list of terms.
(cont’d)
Factors of a number are also divisors of the number. The greatest
common factor is actually the same as the greatest common
divisor. The are many rules for deciding what numbers to divide into
a given number. Here are some especially useful divisibility rules for
small numbers.
Find the greatest common factor of a list of terms.
(cont’d)
Finding the Greatest Common Factor (GCF)
Step 1: Factor. Write each number in prime factored form.
Step 2: List common factors. List each prime number or each
variable that is a factor of every term in the list. (If a prime
does not appear in one of the prime factored forms, it cannot
appear in the greatest common factor.)
Step 3: Choose least exponents. Use as exponents on the common
prime factors the least exponent from the prime factored
forms.
Step 4: Multiply. Multiply the primes from Step 3. If there are no
primes left after Step 3, the greatest common factor is 1.
EXAMPLE 1 Finding the Greatest Common Factor for Numbers
Find the greatest common factor for each list of numbers.
Solution:
50, 75
GCF = 25
12, 18, 26, 32
GCF = 2
50  2  5  5
75  3  5  5
12  2  2  3
26  2 13
18  2  3  3
32  2  2  2  2  2
22  2 11
24  2  2  2  3
22, 23, 24
GCF = 1
23  1 23
Find the greatest common factor of a list of terms.
(cont’d)
The GCF can also be found for a list of variable terms. The exponent
on a variable in the GCF is the least exponent that appears in all the
common factors.
EXAMPLE 2 Finding the Greatest Common Factor for Variable Terms
Find the greatest common factor for each list of terms.
Solution:
16r 9 , 10r15 , 8r12
16r 9  1 2  2  2  2  r 9
GCF = 2r 9
10r15  1 2  5  r15
8r12  2  2  2  r12
s 4t 5 , s3t 6 , s9t 2
s 4t 5  s 4  t 5
GCF = s 3t 2
s 3t 6  s 3  t 6
s 9t 2  s 9  t 2
Factor out the greatest common factor.
Writing a polynomial (a sum) in factored form as a product is called
factoring. For example, the polynomial
3m + 12
has two terms: 3m and 12. The GCF of these terms is 3. We can write
3m + 12 so that each term is a product of 3 as one factor.
3m + 12 = 3 · m + 3 · 4
= 3(m + 4)
GCF = 3
Distributive property
The factored form of 3m + 12 is 3(m + 4). This process is called
factoring out the greatest common factor.
The polynomial 3m + 12 is not in factored form when written as 3 · m + 3 · 4.
The terms are factored, but the polynomial is not. The factored form of 3m +12 is
the product 3(m + 4).
EXAMPLE 3 Factoring Out the Greatest Common Factor
Write in factored form by factoring out the greatest common factor.
Solution:
 6x 2  x 2  2 
6 x 4  12 x 2
30t  25t  10t
6
5
r r
12
 5t 4  6t 2  5t  2 
4
r
10
10
r
2
 1
8 p q  16 p q  12 p q  4 p q  2 p  4 p q  3q
5
2
6 3
4
7
4
2
2
5

Be sure to include the 1 in a problem like r12 + r10. Always check that the
factored form can be multiplied out to give the original polynomial.
EXAMPLE 4 Factoring Out the Greatest Common Factor
Write in factored form by factoring out the greatest common factor.
Solution:
6 p  q  r  p  q
 p  q  6  r 
y 4  y  3  4  y  3
 y  3  y 4  4 
Factor by grouping.
When a polynomial has four terms, common factors can
sometimes be used to factor by grouping.
Factoring a Polynomial with Four Terms by Grouping
Step 1: Group terms. Collect the terms into two groups so that each
group has a common factor.
Step 2: Factor within groups. Factor out the greatest common factor
from each group.
Step 3: Factor the entire polynomial. Factor out a common binomial
factor from the results of Step 2.
Step 4: If necessary, rearrange terms. If Step 2 does not result in a
common binomial factor, try a different grouping.
EXAMPLE 5 Factoring by Grouping
Factor by grouping.
Solution:
pq  5q  2 p  10
q  p  5  2  p  5   p  5 q  2 
2 xy  3 y  2 x  3
y  2 x  3  1 2 x  3   2 x  3 y  1
2a  4a  3ab  6b
2
2a  a  2  3b  a  2   a  2 2a  3b 
x3  3x 2  5 x  15
x2  x  3  5  x  3
  x  3  x 2  5 
EXAMPLE 6 Rearranging Terms before Factoring by Grouping
Factor by grouping.
Solution:
6 y 2  20w  15 y  8 yw
 6 y 2  15 y  20w  8 yw
 3 y  2 y  5   4w  2 y  5 
  2 y  5 3 y  4w
9mn  4  12m  3n
 9mn  12m  3n  4
 3m  3n  4   1 3n  4 
  3m  1 3n  4 
Factoring Trinomials
Using the FOIL method, we see that the product of the binomial k − 3
and k +1 is
(k − 3)(k + 1) = k2 − 2k − 3.
Multiplying
Suppose instead that we are given the polynomial k2 − 2k − 3 and want
to rewrite it as the product (k − 3)(k + 1). That is,
k2 − 2k − 3 = (k − 3)(k + 1).
Factoring
Recall that this process is called factoring the polynomial. Factoring
reverses or “undoes” multiplying.
Factor trinomials with a coefficient of 1 for the
second-degree term.
When factoring polynomials with integer coefficients, we use only
integers in the factors. For example, we can factor x2 + 5x + 6 by
finding integers m and n such that
x2 + 5x + 6 = (x + m)(x + n).
To find these integers m and n, we first use FOIL to multiply the two
binomials on the right side of the equation:
 x  m x  n   x22  nx  mx  mn.
 x   n  m x  mn.
Comparing this result with x2 + 5x + 6 shows that we must find integers
m and n having a sum of 5 and a product of 6.
Product of m and n is 6.
x2  5x  6  x2   n  m x  mn.
Sum of m and n is 5.
Factor trinomials with a coefficient of 1 for the
second-degree term. (cont’d)
Since many pairs of integers have a sum of 5, it is best to begin by
listing those pairs of integers whose product is 6. Both 5 and 6 are
positive, so consider only pairs in which both integers are positive.
Both pairs have a product of 6, but only the pair 2 and 3 has a sum of
5. So 2 and 3 are the required integers, and
x2 + 5x + 6 = (x + 2)(x + 3).
Check by using the FOIL method to multiply the binomials. Make sure
that the sum of the outer and inner products produces the correct
middle term.
EXAMPLE 1 Factoring a Trinomial with All Positive Terms
Factor y2+ 12y + 20.
Solution:
Factors of 20
1, 20
2, 10
4, 5
Sums of
Factors
1 + 20 = 21
2 + 10 = 12
4+5 =9
  y  10 y  2
You can check your factoring by graphing both the unfactored and factored
forms of polynomials on your graphing calculators.
EXAMPLE 2 Factoring a Trinomial with a Negative Middle Term
Factor y2 − 10y + 24.
Solution:
Factors of 24
− 1 , −24
−2 , −12
−3 , −8
−4 , −6
Sums of Factors
−1 + (−24) = −25
−2 + (−12) = −14
−3 + (−8) = −11
−4 + (−6) = −10
  y  6 y  4
EXAMPLE 3 Factoring a Trinomial with a Negative Last (Constant) Term
Factor z2 + z − 30.
Solution:
Factors of − 30 Sums of Factors
− 1 , 30
−1 + (30) = 29
1 , − 30
1 + (−30) = −29
5,−6
5 + (− 6) = −1
−5 , 6
−5 + (6) = 1
  z  6 z  5
EXAMPLE 4 Factoring a Trinomial with Two Negative Terms
Factor a2 − 9a − 22.
Solution:
Factors of −22
Sums of Factors
−1 , 22
1, −22
−2 , 11
2 , −11
−1 + 22 = 21
1 + (−22) = −21
−2 + 11 = 9
2 + (−11) = −9
  a  11 a  2
Factor trinomials with a coefficient of 1 for the
second-degree term. (cont’d)
Some trinomials cannot be factored by using only integers. We call
such trinomials prime polynomials.
Summarize the signs of the binomials when factoring a trinomial
whose leading coefficient is positive.
1. If the last term of the trinomial is positive, both binomials will
have the same “middle” sign as the second term.
2. If the last term of the trinomial is negative, the binomials will have
one plus and one minus “middle” sign.
EXAMPLE 5 Deciding Whether Polynomials Are Prime
Factor if possible.
m  8m  14
2
Prime
Solution:
Factors of 14
Sums of
Factors
−1 , −14
−1 + (−14) =
−15
−2 + (−7) = −9
−2 , −7
y  y2
2
Prime
Factors of 2 Sums of Factors
1, 2
1+2=3
Factor trinomials with a coefficient of 1 for the
second-degree term. (cont’d)
Guidelines for Factoring x2 + bx + c
Find two integers whose product is c and whose sum is b.
1. Both integers must be positive if b and c are positive.
2. Both integers must be negative if c is positive and b is negative.
3. One integer must be positive and one must be negative if c is
negative.
Factor trinomials with a coefficient greater than 1 for
the second-degree term. (Slide and Divide)
Factor : 10x2 – 21x -10
Step 1: Multiply second degree coefficient and constant
term. Remove second degree term and replace constant
term with new constant term.
x2 – 21x – 100
Step 2: Factor as if second degree coefficient is 1.
(x–25)(x+4)
-25 + 4 = -21; -25 x 4 = -100
Step 3: Divide each binomials constant term by original
second degree coefficient. Reduce fractions if possible.
(x-25/10)(x+4/10)  (x-5/2)(x+2/5)
Step 4: Multiply each binomial by the denominator of its constant
term.
(2x-5)(5x+2) Check by multiplying
EXAMPLE 6 Factoring a Trinomial with Two Variables
Factor r2 − 6rs + 8s2.
Solution:
Factors of 8s2
−1s , −8s
−2s , −4s
Sums of Factors
− 1s + (−8s) =
−9s
−2s + (−4s) = −6s
  r  4s  r  2s 
EXAMPLE 6 Factoring a Trinomial with Two Variables
Factor 6m2 + 11mn – 10n2.
Solution:
3m  2n  2m  5n 
4mn
15mn
11mn
EXAMPLE 7 Factoring a Trinomial with a Common Factor
Factor such trinomials after factoring out the greatest
common factor.
If a trinomial has a common factor, first factor it out.
Factor 3x4 − 15x3 + 18x2.
Solution:
 3x 2  x 2  5 x  6 
 3x2  x  3 x  2
When factoring, always look for a common factor first. Remember to
include the common factor as part of the answer. As a check, multiplying out
the factored form should always give the original polynomial.
EXAMPLE 7 Factoring Trinomials with Common Factors
Factor.
28 x 4  58 x3  30 x 2
24 x3  32 x 2 y  6 xy 2
Solution:
 2 x 14 x  29 x  15 
2
2
 2 x2  7 x  3 2 x  5
6x
35x
29x
 2 x 12 x 2  16 xy  3 y 2 
 2 x  6 x  y  2 x  3 y 
2xy
18xy
16xy
Factor a difference of squares.
The formula for the product of the sum and difference of the same two
2
2
terms is
 x  y  x  y   x
y .
Factoring a Difference of Squares
x2  y 2   x  y  x  y 
For example,
m2 16  m2  42   m  4 m  4 .
The following conditions must be true for a binomial to be a difference
of squares:
1. Both terms of the binomial must be squares, such as
x2,
9y2,
25,
1,
m4.
2. The second terms of the binomials must have different signs (one
positive and one negative).
EXAMPLE 1 Factoring Differences of Squares
Factor each binomial if possible.
Solution:
t 2  81
  t  9 t  9
r 2  s2
  r  s  r  s 
y 2  10
prime
q 2  36
prime
After any common factor is removed, a sum of squares cannot
be factored.
EXAMPLE 2 Factoring Differences of Squares
Factor each difference of squares.
Solution:
49 x 2  25
  7 x  5 7 x  5
64a 2  81b 2
 8a  9b 8a  9b 
You should always check a factored form by multiplying.
EXAMPLE 3 Factoring More Complex Differences of Squares
Factor completely.
Solution:
50r  32
 2  25r 2  16 
z 4  100
  z 2  10  z 2  10 
z 4  81
  z  9  z  9 
2
2
2
 2  5r  4 5r  4 
  z 2  9   z  3 z  3
Factor again when any of the factors is a difference of squares as in
the last problem.
Check by multiplying.
Factor a perfect square trinomial.
The expressions 144, 4x2, and 81m6 are called perfect squares
2
because
2
2
4
x

2
x
, and 81m6  9m3
144  12 ,

 

2
.
A perfect square trinomial is a trinomial that is the square of a
binomial. A necessary condition for a trinomial to be a perfect square
is that two of its terms be perfect squares.
Even if two of the terms are perfect squares, the trinomial may not be
a perfect square trinomial.
Factoring Perfect Square Trinomials
2
2
2
x  2 xy  y   x  y 
x 2  2 xy  y 2   x  y 
2
EXAMPLE 4 Factoring a Perfect Square Trinomial
Factor k2 + 20k + 100.
Solution:
 k 2  20k  100
  k  10 
2
Check :
2  k 10  20k
EXAMPLE 5 Factoring Perfect Square Trinomials
Factor each trinomial.
Solution:
x 2  24 x  144
  x  12 
2
25 x 2  30 x  9
  5 x  3
2
36a 2  20a  25
prime
18 x  84 x  98 x
 2 x  9 x 2  42 x  49 
3
2
 2 x  3x  7 
2
Factoring Perfect Square Trinomials
1. The sign of the second term in the squared binomial is always the
same as the sign of the middle term in the trinomial.
2. The first and last terms of a perfect square trinomial must be
positive, because they are squares. For example, the polynomial
x2 – 2x – 1 cannot be a perfect square, because the last term is
negative.
3. Perfect square trinomials can also be factored by using grouping or
the FOIL method, although using the method of this section is often
easier.
Factor a difference of cubes.
Factoring a Difference of Cubes
positive
x3  y 3   x  y   x 2  xy  y 2 
same sign
opposite sign
This pattern for factoring a difference of cubes should be
memorized.
The polynomial x3 − y3 is not equivalent to (x − y )3,
 x  y
whereas
3
  x  y  x  y  x  y 
  x  y   x 2  2 xy  y 2 
x3  y 3   x  y   x 2  xy  y 2 
EXAMPLE 6 Factoring Differences of Cubes
Factor each polynomial.
Solution:
x3  216
  x  6   x 2  6 x  36 
27 x  8
  3x  2   9 x 2  6 x  4 
5x  5
 5  x 3  1
3
3
64 x  125 y
3
6
 5  x  1  x 2  x  1
  4 x  5 y 2 16 x 2  20 xy 2  25 y 4 
A common error in factoring a difference of cubes, such as
x3 − y3 = (x − y)(x2 + xy + y2), is to try to factor x2 + xy + y2. It is easy to
confuse this factor with the perfect square trinomial x2 + 2xy + y2. But because
there is no 2, it is unusual to be able to further factor an expression of the
form x2 + xy +y2.
Factor a sum of cubes.
A sum of squares, such as m2 + 25, cannot be factored by using real
numbers, but a sum of cubes can.
Factoring a Sum of Cubes
positive
x3  y 3   x  y   x 2  xy  y 2 
same sign
opposite sign
Note the similarities in the procedures for factoring a sum of cubes and a
difference of cubes.
1. Both are the product of a binomial and a trinomial.
2. The binomial factor is found by remembering the “cube root, same sign,
cube root.”
3. The trinomial factor is found by considering the binomial factor and
remembering, “square first term, opposite of the product, square last
term.”
EXAMPLE 7 Factoring Sums of Cubes
Factor each polynomial.
Solution:
  p  4   p 2  4 p  16 
p3  64
27 x  64 y
3
512a  b
6
3
3
  3 x  4 y   9 x  12 xy  16 y
2
  8a 2  b  64a 4  8a 2b  b 2 
2

Methods of factoring discussed in this section.