Factoring polynomials
The following two theorems are of use in factoring polynomials.
The Remainder Theorem
When a linear expression
=
( )
d
x

x
a
is divided into a polynomial
( )
P
x
to give a quotient
( )
Q
x
and remainder
R
, so that
=
( )
P
x
+
(
)

x
a
( )
Q
x
R
 (i),
then
=
R
( )
P
a
, that is, the
remainder
when a polynomial
( )
P
x
is divided by
=
( )
d
x

x
a
, is the
value
of
( )
P
x
at
=
x
a
, namely
( )
P
a
.
The Factor Theorem
A linear expression of the form
=
( )
d
x

x
a
is a
factor
of a polynomial
( )
P
x
exactly when
=
( )
P
a
0.
Example 1
: Factor the polynomial
=
( )
P
x
+


x
3
5
x
2
2
x
24, and find the zeros of
( )
P
x
.
Solution
:
Suppose that
=
( )
P
x
(
)
+
x
p
(
)
+
x
q
(
)
+
x
r
, where
p
,
q
and
r
are integers. Then
=
p q r

24
and
p
,
q
and
r
are integer factors of the
constant term

24
in
( )
P
x
. The associated zeros

p
,

q
and

r
are also integer factors of

24.
If
( )
P
x
has only one linear factor
=
( )
A
x
+
x
p
, with
p
an integer and a companion quadratic factor
=
( )
B
x
+
+
x
2
m x
n
, with no integer zeros, such that
=
( )
P
x
( )
A
x
( )
B
x
=
(
)
+
x
p
(
)
+
+
x
2
m x
n
,
we must have
=
p n

24, so that
p
, or equivalently

p
,
is still an integer factor of

24.
It is therefore reasonable to look for a zero of
(
)
P
x
which is an
integer factor of the constant term

24.
The possiblities to consider for such an integer zero of
( )
P
x
are:
+
1,
+
2,
+
3,
+
4,
+
6,
+
8,
+
12,
+
24.
It is probably easier to check whether
+
1
and
+
2 are zeros by direct substitution rather than by using synthetic division.
=
( )
P 1
+


1
5
2
24
=
≠

20
0,
=
(
)
P

1
 +
+

1
5
2
24 =
≠

18
0 ,
=
( )
P 2
+


8
20
4
24
=
0.
Since
=
x
2
is a zero of
( )
P
x
, it follows by the Factor Theorem that

x
2
is a factor of
( )
P
x
. We can check this by synthetic division,
and also find the quotient
( )
Q
x
when
( )
P
x
is divided by

x
2.
2

1
5

2

24
2
14
24
________________________
1
7
12

0
=
+


x
3
5
x
2
2
x
24
(
)

x
2 (
)
+
+
x
2
7
x
12 .
Hence, by factoring the
residual quadratic
+
+
x
2
7
x
12, we see that
=
( )
P
x
(
)

x
2 (
)
+
x
3 (
)
+
x
4 .
____________
The corresponding
zeros
of
( )
P
x
are
2,

3
and

4.
Example 2
: Factor the polynomial
=
( )
P
x
+

+
x
3
2
x
2
13
x
10, and find the zeros of
( )
P
x
.
Solution
:
The
integer factors
of 10 provide candidates for zeros of
( )
P
x
, that is,
+
1,
+
2,
+
5,
+
10.
Straight away
=
( )
P 1
+

+
1
2
13
10
=
0, so it follows by the Factor Theorem that

x
1
is a factor of
( )
P
x
.
We can find the quotient when
( )
P
x
is divided by

x
1 by using synthetic division.
1

1
2

13
10
1
3

10
________________________
1
3

10

0
=
+

+
x
3
2
x
2
13
x
10
(
)

x
1 (
)
+

x
2
3
x
10
=
(
)

x
1 (
)

x
2 (
)
+
x
5
__________
The
zeros
of
( )
P
x
are 1, 2
and

5.