Furthermore some linear factors are not prime for

resulting polynomial factor will not have integer coefficients. Furthermore, some linear factors
are not prime. For example,
5 x −10=5( x −2)
In general, any linear factor of the form
ax + b
, where a and b are relatively prime integers, is
prime.
Factoring by Grouping
In this section, we outline a technique for factoring polynomials with four terms. First, review a
preliminary example where the terms have a common binomial factor.
Example 5
Factor:
7 x (3 x −2)−(3 x −2).
Solution:
Begin by rewriting the second term
−(3 x −2)
as
−1(3 x −2).
Next, consider
(3 x −2)
as a
common binomial factor and factor it out as follows:
7 x (3 x −2)−(3 x −2)===7 x (3 x −2) −1 (3 x −2)(3 x −2) ( ? )(3 x −2)(7 x −1)
Answer:
(3 x −2)(7 x −1)
Factoring by groupingA technique for factoring polynomials with four terms. is a technique that
enables us to factor polynomials with four terms into a product of binomials. This involves an
intermediate step where a common binomial factor will be factored out. For example, we wish to
factor
3 x
3
−12 x
2
+2 x −8
Begin by grouping the first two terms and the last two terms. Then factor out the GCF of each
grouping:

In this form, the polynomial is a binomial with a common binomial factor,
( x −4).
=( x −4)( ? )=( x −4)( 3 x
2
+2 )
Therefore,
3 x
3
−12 x
2
+2 x −8=( x −4)(3 x
2
+2)
We can check by multiplying.
( x −4)(3 x
2
+2)==3 x
3
+2 x −12 x
2
−83 x
3
−12 x
2
+2 x −8 ✓
Example 6
Factor by grouping:
24 a
4
−18 a
3
−20 a +15.
Solution:
The GCF for the first group is
6 a
3
.
We have to choose 5 or −5 to factor out of the second
group.
Factoring out +5 does not result in a common binomial factor. If we choose to factor out −5, then
we obtain a common binomial factor and can proceed. Note that when factoring out a negative
number, we change the signs of the factored terms.
Answer:
(4 a −3)(6 a
3
−5).
Check by multiplying; this is left to the reader as an exercise.
Sometimes we must first rearrange the terms in order to obtain a common factor.
Example 7
Factor:
ab −2 a
2
b + a
3
−2 b
2
.
Solution:
Simply factoring the GCF out of the first group and last group does not yield a common binomial
factor.

We must rearrange the terms, searching for a grouping that produces a common factor. In this
example, we have a workable grouping if we switch the terms
a
3 and
ab .
Answer:
( a −2 b )( a
2
+ b )
Try this! Factor:
x
3
− x
2
y − xy + y
2
.
Answer:
( x − y )( x
2
− y )
(click to see video)
Not all factorable four­term polynomials can be factored with this technique. For example,
3 x
3
+5 x
2
− x +2
This four­term polynomial cannot be grouped in any way to produce a common binomial factor.
Despite this, the polynomial is not prime and can be written as a product of polynomials. It can
be factored as follows:
3 x
3
+5 x
2
− x +2=( x +2)(3 x
2
− x +1)
Factoring such polynomials is something that we will learn to do as we move further along in our
study of algebra. For now, we will limit our attempt to factor four­term polynomials to using the
factor by grouping technique.