Factoring Trinomials by Grouping
Factoring a Trinomial of the Form ax^{2} + bx + c by Grouping
In this method, we rewrite the middle term of the trinomial using two
terms. Then we factor by grouping.
For example, weâ€™ll use this method
to factor 15x^{2}  16x + 4.
Write the middle term, 16x, as
10x  6x.
Group the first two terms and group
the last two terms.
Factor 5x out of the first group;
factor 2 out the second group.
Finally, factor out (3x  2). 
15x^{2}  16x + 4 = 15x^{2}  10x  6x + 4
= (15x^{2}  10x) + (6x + 4)
= 5x(3x  2) + (2)(3x  2)
= (3x  2)(5x  2) 
The key to this method is knowing how to rewrite the middle term of the
trinomial. The following procedure describes a way to do this.
Procedure â€” To Factor ax^{2} + bx + c by Grouping
Step 1 Factor out common factors (other than 1 or 1).
Step 2 List the values of a, b, and c.
Then find two integers whose product is ac and whose sum
is b. If no two such integers exist then the trinomial is not
factorable over the integers.
Step 3 Replace the middle term, bx, with a sum or difference using
the two integers found in Step 2.
Step 4 Factor by grouping.
To check the factorization, multiply the binomial factors.
Note:
This method also works when a = 1. However, in those cases The ProductSum
method requires fewer steps.
Example 1
Factor: 6x^{2 }+ 7x + 2.
Solution
Step 1 Factor out common factors (other than 1 or 1).
There are no common factors other than 1 and 1.
Step 2 List the values of a, b, and c. Then find two integers whose
product is ac and whose sum is b.
6x^{2 }+ 7x + 2 has the form ax^{2} + bx + c where a = 6, b =
7, and c = 2.
The product ac is 6 Â· 2 = 12.
Thus, find two integers whose product, ac, is 12 and whose sum, b, is 7.
â€¢ Since their product is positive, the integers must have the same sign.
â€¢ Since their sum is also positive, the integers must both be positive.
Here are the possibilities:
Product
1 Â· 12
2 Â· 6
3 Â· 4 
Sum
13
8
7 
The integers 3 and 4 satisfy the requirements that their product is 12 and
their sum is 7.
Step 3 Replace the middle term, bx,
with a sum or difference using the
two integers found in Step 2.

6x^{2 }+ 7x + 2 
Replace 7x with 3x + 4x. 
= 6x^{2 }+ 3x + 4x + 2 
Step 4 Factor by grouping.
Group the first pair of terms and group
the second pair of terms.
Factor 3x out of the first group; factor 2
out of the second group.
Factor out the common factor (2x + 1). 
= (6x^{2 }+ 3x) + (4x + 2) =
3x(2x^{ }+ 1) + 2(x + 1)
= (2x^{ }+ 1)(3x + 2) 
The result is: 6x^{2 }+ 7x + 2 = (2x^{ }+ 1)(3x + 2). You can multiply to check the factorization. We leave the check to you.
Note:We replaced 7x with 3x + 4x.
If we switch 3x and 4x, we can still group
and factor:
= 6x^{2} + 4x + 3x + 2
= (6x^{2} + 4x) + (3x + 2)
= 2x(3x + 2) + 1(3x + 2)
= (3x + 2)(2x + 1)
