A trinomial is a polynomial that has three terms. A polynomial has two or more algebraic terms. The first term has an exponent to the second degree (x2), the second term is x, and the last term is a constant (just any number). Factoring trinomials is an extremely important algebra skill. Factoring polynomials can be tricky however with the detailed explanation below any problem can be solved.
Trinomial
ax2 + bx + c
a = the x2 term
b = the x term
c = the constant value
Example 1:
a2 + 4a – 12
First identify the values for a, b, and c.
a = 1
b = 4
c = -12
Remember when no number is in front of a variable, we assume the coefficient is 1.
Step 2: Find two numbers that when added together produce b and when multiplied together produce c. In this case we should list the factors of -12 first.
Factors of -12:
Possible Combinations | Does c = -12 when multiplied? | Does b = 4 when added? | Correct Combination |
4 x 3 | 12 | 7 | No |
4 x -3 | -12 | 1 | No |
-4 x -3 | 12 | -7 | No |
-4 x 3 | -12 | -1 | No |
6 x 2 | 12 | 8 | No |
-6 x 2 | -12 | -4 | No |
6 x -2 | -12 | 4 | Yes |
-6 x -2 | 12 | -8 | No |
12 x 1 | 12 | 13 | No |
-12 x 1 | -12 | -11 | No |
12 x -1 | -12 | 11 | No |
-12 x -1 | 12 | -13 | No |
The table above shows all the possible combinations for example 1. However only one combination equals both b and c (highlighted in bold). The correct combination of the two numbers should look like this:
(a – 2) (a + 6)
We can foil the problem to see if it matches the original problem.

Example 2:
Factor the following polynomial:
x2 + 2x -8
First identify the values of a, b, and c.
a = 1
b = 2
c =-8
Now consider the factors of c and when added together equal b.
Possible Combinations | Does c = -8 when multiplied? | Does b = 2 when added? | Correct Combination |
8 x 1 | 8 | 9 | No |
-8 x -1 | 8 | -9 | No |
8 x -1 | -8 | 7 | No |
-8 x 1 | 8 | -7 | No |
2 x 4 | 8 | 6 | No |
-2 x 4 | -8 | 2 | Yes |
2 x -4 | -8 | -2 | No |
-2 x -4 | 8 | -8 | No |
The correct combination should look like this:
(x – 2) (x + 4)
Use the foil method to check it against the original problem.
