Factoring Trinomials of the Form
x
^{2} + bx + c
and
x
^{2}  bx + c
Just as the product of two binomials can often be rewritten as a
trinomial, trinomials of the form
ax
^{2} + bx + c
can often be
rewritten as the product of two binomials. For example,
x
^{2} + 3x + 2 = (x + 1)(x + 2)
.
We now know that the product of two binomials of the form
(x + d )
and
(x + e)
is given by:
(x + d )(x + e) = x
^{2} + xe + dx + de = x
^{2} + (d + e)x + de


Thus, in order to rewrite a binomial
x
^{2} + bx + c
as the product of
two binomials (
b
positive or negative), we must find numbers
d
and
e
such that
d + e = b
and
de = c
. Since
c
is positive,
d
and
e
must have the same sign.
Here are the steps to factoring a trinomial of the form
x
^{2} + bx + c
, with
c > 0
. We assume that the coefficients are integers, and
that we want to factor into binomials with integer coefficients.
 Write out all the pairs of numbers which can be multiplied to produce
c
.
 Add each pair of numbers to find a pair that produce
b
when added. Call the
numbers in this pair
d
and
e
.
 If
b > 0
, then the factored form of the trinomial is
(x + d )(x + e)
. If
b < 0
,
then the factored form of the trinomial is
(x  d )(x  e)
.
 Check: The binomials, when multiplied, should equal the original trinomial.
Note: Some trinomials cannot be factored. If none of the pairs total
b
, then
the trinomial cannot be factored.
Example 1: Factor
x
^{2} + 5x + 6
.
 Pairs of numbers which make 6 when multiplied:
(1, 6)
and
(2, 3)
.

1 + 6≠5
. 2 + 3 = 5. Thus,
d = 2
and
e = 3
(or vice versa).

(x + 2)(x + 3)
 Check:
(x + 2)(x + 3) = x
^{2} +3x + 2x + 6 = x
^{2} + 5x + 6
Thus,
x
^{2} + 5x + 6 = (x + 2)(x + 3)
.
Example 2: Factor
x
^{2}  7x + 12
.
 Pairs of numbers which make 12 when multiplied:
(1, 12)
,
(2, 6)
, and
(3, 4)
.

1 + 12≠7
.
2 + 6≠7
. 3 + 4 = 7. Thus,
d = 3
and
e = 4
.

(x  3)(x  4)
 Check:
(x  3)(x  4) = x
^{2} 4x  3x + 12 = x
^{2}  7x + 12
Thus,
x
^{2}  7x + 12 = (x  3)(x  4)
.
Example 3: Factor
2x
^{3} +4x
^{2} + 2x
.
First, remove common factors:
2x
^{3} +4x
^{2} +2x = 2x(x
^{2} + 2x + 1)
 Pairs of numbers which make
1
when multiplied:
(1, 1)
.
 1 + 1 = 2. Thus,
d = 1
and
e = 1
.

2x(x + 1)(x + 1)
(don't forget the common factor!)
 Check:
2x(x + 1)(x + 1) = 2x(x
^{2} +2x + 1) = 2x
^{3} +4x
^{2} + 2x
Thus,
2x
^{3} +4x
^{2} +2x = 2x(x + 1)(x + 1) = 2x(x + 1)^{2}
.
x
^{2} + 2x + 1
is a perfect square trinomial.
Factoring Trinomials of the Form
x
^{2} + bx  c
and
x
^{2}  bx  c
In the equation
(x + d )(x + e) = x
^{2} + (d + e)x + de = x
^{2} + bx + c
(mentioned in Heading ),
c
is negative if
and only if
de
is negative; that is, if and only if
d
is negative
or
e
is negative. Thus, the equation becomes
(x + d )(x  e) = x
^{2} + (d  e)x  de = x
^{2} + bx  c
, which
c > 0
.
In order to rewrite the trinomial
x
^{2} + bx  c
as the product of two
binomials, we must find numbers
d
and
e
such that
d  e = b
and
 de = c
. If
b
is positive, then
d  e > 0
, so
d > e
. If
b
is negative, then
d  e < 0
, so
d < e
.
Here are the steps to factoring a trinomial of the form
x
^{2} + bx  c
, with
c > 0
:
 Write out all the pairs of numbers which can be multiplied to
produce
c
.
 Subtract each pair of numbers to find a pair that produce
b
when
one is subtracted from the other. Call the numbers in this pair
d
and
e
. If
b > 0
, let
d
be the larger number, and if
b < 0
,
let
d
be the smaller number.
 The factored form of the trinomial is
(x + d )(x  e)
.
 Check: The binomials, when multiplied, should equal the original
trinomial. If the middle term has the wrong sign, you most likely
switched
d
and
e
. Switch the "+" and "" sign in your binomials
and check again.
Note: Again, not every trinomial can be factored.
Example 1: Factor
x
^{2} + 6x  16
.
 Pairs of numbers which make 16 when multiplied:
(1, 16)
,
(2, 8)
, and
(4, 4)
.

16  1≠6
. 8  2 = 6. Since
b = 6 > 0
,
d = 8
and
e = 2
.

(x + 8)(x  2)
 Check:
(x + 8)(x  2) = x
^{2} 2x + 8x  16 = x
^{2} + 6x  16
Thus,
x
^{2} + 6x  16 = (x + 8)(x  2)
.
Example 2: Factor
x
^{2}  x  20
.
 Pairs of numbers which make 20 when multiplied:
(1, 20)
,
(2, 10)
, and
(4, 5)
.

20  1≠1
.
10  2≠1
. 5  4 = 1. Since
b =  1 < 0
,
d = 4
and
e = 5
.

(x + 4)(x  5)
 Check:
(x + 4)(x  5) = x
^{2} 5x + 4x  20 = x
^{2}  x  20
Thus,
x
^{2}  x  20 = (x + 4)(x  5)
.
Example 3: Factor
x
^{2}  16
(Here,
x
has an implied coefficient of
b = 0
).
 Pairs of numbers which make 16 when multiplied:
(1, 16)
,
(2, 8)
, and
(4, 4)
.

16  1≠ 0
.
8  2≠ 0
. 4  4 = 0. Thus,
d = 4
and
e = 4
.

(x + 4)(x  4)
 Check:
(x + 4)(x  4) = x
^{2} 4x + 4x  16 = x
^{2}  16
Thus,
x
^{2}  16 = (x + 4)(x  4)
.