a
(
b
+
c
)
a
b
+
a
c
==
a
b
+
a
c
Multiplyinga
(
b
+
c
)
Factoring
Consider factoring the result of the opening example:
12
x
2
y
3
+ 6
xy
2
===
6
xy
2
⋅
2
xy
+
6
xy
2
⋅
1
Factoring
6
xy
2
(
?
)
6
xy
2
(2
xy
+1)
We see that the distributive property allows us to write the polynomial
12
x
2
y
3
+ 6
xy
2 as a
product of the two factors
6
xy
2 and
(2
xy
+1).
Note that in this case,
6
x
2
y
is the GCF of
the terms of the polynomial.
GCF(12
x
2
y
3
, 6
xy
2
)=6
xy
2

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Factoring out the greatest common factor (GCF)The process of rewriting a polynomial as a
product using the GCF of all of its terms. of a polynomial involves rewriting it as a product where
a factor is the GCF of all of its terms.
8
x
3
+4
x
2
−16
x
9
ab
2
−18
a
2
b
−3
ab
==
4
x
(2
x
2
+
x
−4)
3
ab
(3
b
−6
a
−1)
⎫⎭⎬
Fac
toring out the GCF
To factor out the GCF of a polynomial, we first determine the GCF of all of its terms. Then we
can divide each term of the polynomial by this factor as a means to determine the remaining
factor after applying the distributive property in reverse.
Example 3
Factor out the GCF:
18
x
7
−30
x
5
+6
x
3
.
Solution:
In this case, the GCF(18, 30, 6) = 6, and the common variable factor with the smallest exponent
is
x
3
.
The GCF of the polynomial is
6
x
3
.
18
x
7
−30
x
5
+6
x
3
=
6
x
3
(?)
The missing factor can be found by dividing each term of the original expression by the GCF.
18
x
7
6
x
3
=3
x
4
−30
x
5
6
x
3
=−5
x
2
+6
x
3
6
x
3
=+1
Apply the distributive property (in reverse) using the terms found in the previous step.
18
x
7
−30
x
5
+6
x
3
=
6
x
3
(3
x
4
−5
x
2
+1)
If the GCF is the same as one of the terms, then, after the GCF is factored out, a constant term
1 will remain. The importance of remembering the constant term becomes clear when
performing the check using the distributive property.
6
x
3
(3
x
4
−5
x
2
+1)==
6
x
3
⋅
3
x
4
−
6
x
3
⋅
5
x
2
+
6
x
3
⋅
118
x
7
−30
x
5
+6
x
3
✓
Answer:
6
x
3
(3
x
4
−5
x
2
+1)
Example 4
Factor out the GCF:
27
x
5
y
5
z
+54
x
5
yz
−63
x
3
y
4
.
Solution:
The GCF of the terms is
9
x
3
y
.
The last term does not have a variable factor of
z
, and thus
z
cannot be a part of the greatest common factor. If we divide each term by
9
x
3
y
, we obtain
27
x
5
y
5
z
9
x
3
y
=3
x
2
y
4
z
54
x
5
yz
9
x
3
y
=6
x
2
z
−63
x
3
y
4
9
x
3
y
=−7
y
3
and can write
27
x
5
y
5
z
+54
x
5
yz
−63
x
3
y
4
==
9
x
3
y
(?)9
x
3
y
(3
x
2
y
4
z
+6
x
2
z
−7
y
3
)
Answer:
9
x
3
y
(3
x
2
y
4
z
+6
x
2
z
−7
y
3
)

Try this!
Factor out the GCF:
12
x
3
y
4
−6
x
2
y
3
−3
xy
2
Answer:
3
xy
2
(4
x
2
y
2
−2
xy
−1)
(click to see video)
Of course, not every polynomial with integer coefficients can be factored as a product of
polynomials with integer coefficients other than 1 and itself. If this is the case, then we say that it
is a prime polynomialA polynomial with integer coefficients that cannot be factored as a product
of polynomials with integer coefficients other than 1 and itself.. For example, a linear factor such
as
10
x
−9
is prime. However, it can be factored as follows:
10
x
−9=
x
(10−9
x
)or10
x
−9=5(2
x
−95)
If an
x
is factored out, the resulting factor is not a polynomial. If any constant is factored out, the
resulting polynomial factor will not have integer coefficients. Furthermore, some linear factors