Method for factoring trinomials
3 x 2  4 x  4 EXAMPLE PROBLEM
1) Factor a GCF, if possible, to the trinomial
2) Multiply the coefficient of the first term and the last term
3x 2  4 x  4
3  (4)  12
You can only use common factors of -12
The factors not only need to multiply to a -12, but
they also must add up to the coefficient of the
middle term which in this case is a -4
3x 2  4 x  4
The factors of -12 are –(12 & 1), -(6 & 2), -(3 & 4)
The only 2 that can combine to a negative 4 are -6 + 2 = -4
So we will use -6x & 2x to fill in the missing terms
3) Bring down the 1st and the last terms
3x 2  4 x  4
3x ____________________  4
Then place -6x & +2x in the middle terms
3x 2  6 x  2 x  4
2
Example : 6 x 2  x  15
1) __ no _ GCF
2) __ 6 x 2  x  15
product _ 6  15  90
sum  1
the _ two _ numbers _ are _  10 & 9
3) __ 6 x 2  10 x  9 x  15
group
4 & 5) __(6 x 2  10 x)  (9 x  15)
factor
2 x(3 x  5), _ 3(3 x  5)
6) __ Re group
(2 x  3)(3 x  5)
4) Now group the first two terms and the last two terms
by putting them in parentheses.
(3x 2  6 x)  (2 x  4)
5) Now factor each of the 2 quantities in parentheses
(3 x 2  6 x) _ factors _ to
3 x( x  2)
and _(2 x  4) _ factors _ to
2( x  2)
We get
3x( x  2)and 2( x  2)
Both quantities in parentheses must be the same, or you made a mistake and need to
check your earlier steps.
6) The two terms that factored out of the two quantities then need to be grouped as their
own quantity with the quantity already in the parentheses
3 x( x  2)and 2( x  2)
(3 x  2)( x  2)
Example
1) __ 6 x 2  x  15
product _ 6  15  90
sum  1
the _ two _ numbers _ are _  10 & 9
2 & 3) __ 6 x 2  10 x  9 x  15
group
4) __(6 x 2  10 x)(9 x  15)
factor
2 x(3 x  5), _ 3(3 x  5)
5) __ Re group
(2 x  3)(3x  5)