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Factoring Trinomials
Case 1: lead coefficient is one.
In this case we need to find a pair
of numbers m and n whose product is the
constant term (c), and whose sum is the
coefficient of x (b). The m and n then
become the constant terms in our
binomial factors.
Try:
x 2  3 x  28
x 2  8x  7
x 2  2x  8
x 2  15 x  56
x 2  6x  9
Case 2: Lead coefficient not equal to
one, Trial method.
In this case we observe that the
lead term of the trinomial must be the
product of the lead terms in our binomial
factors. Furthermore the constant term
must be the product of the constants in
our binomial factors. If we look at all
possible factors of the lead term and
constant term we can, by trial and error,
determine which combination will form
the middle term when foiling.
Example: Factor x 2  2 x  15
We must find an m and n such that
m  n  15 and m  n  2
m  5 and n  3 will work so…
x 2  2 x  15  ( x  5)( x  3)
Test with foil:
First:
Outside:
Inside:
Last:
x  x  x2
x  (3)  3 x
5  x  5x
5  (3)  15
Then, x 2  3x  5 x  15  x 2  2 x  15
As needed
(Notice how the -3x and 5x add to the 2x and how the 5 and
the -3 multiply to -15 when foiling)
Example: Factor 4 x 2  11x  6
The possible factorings of 4x 2 are:
4x  1x and 2 x  2 x
The possible factorings of 6 are:
1 6 and 2  3
(It is often necessary to allow your factors to be negative)
(Ex : (2)  (3) )
6x 2  7x  3
Then list all possible factoring that
contain these pieces and find the “oi”
part of foil to determine witch one give
you the 11x you need.
(4 x  6)( x  1)
4x  6x  10x
(4 x  1)( x  6)
24x  x  25x
(4 x  2)( x  3)
12x  2x  14x
(4 x  3)( x  2)
8x  3x  11x
(2 x  1)( 2 x  6)
12x  2x  14x
(2 x  2)( 2 x  3)
6x  4x  10x
5x 2  9x  4
Thus, 4 x 2  11x  6  (4 x  3)( x  2)
Try:
3x 2  13x  4
2 x 2  9 x  10
(In practice you can stop once you find it)
Case 3: Lead coefficient not equal to
one, “ac” method.
The trial method works well
when the lead term and constant term
have a small number of factors. Lots of
factors can mean lots of cases to check.
The “ac” method removes the need for
these long trials in exchange for a more
technical methodology.
To factor first find the product of
the lead coefficient (a) and the constant
term (c). The, like in case one, find a
pair of numbers p and q that multiply to
make this product and add to the
coefficient of the x term (b). Then
replace the x term with the sum px  qx
And factor your new four term
polynomial by grouping.
Example: Factor 4 x 2  2 x  12
First find the product (ac)
4  (12)  48
We must find an p and q such that
p  q  48 and p  q  2
p  8 and q  6 will work so replace
the 2x with 8 x  (6) x or just 8x  6x to
get…
4 x 2  8 x  6 x  12
4 x( x  2)  6( x  2)
( x  2)( 4 x  6)
then factor
by grouping
so,
4 x 2  2 x  12  ( x  2)(4 x  6)
Try:
3x 2  19 x  20
6 x 2  31x  35
24 x 2  14 x  3
10 x 2  29 x  10
(Finding p and q is not always easy. Just
list all the ways to factor (ac) and add up
each pair. Be careful of your signs. If
none of the pairs add up to the right
number then the trinomial is simply not
factorable, (or you left off a pair).)