Section 6-10
Factoring Quadratic Trinomials
Given a trinomial x2 + bx + c, you may
or may not be able to factor it. Some polynomials are "irreducible",
others can be broken up into linear factors.
To find linear factors, you need to look for two numbers p and q so
that p + q = b and pq = c. Then,
using the FOIL, we have:
(x + p)(x + q) = x2 + px + qx + pq
= x2 + (p + q)x + pq
= x2 + bx + c
Example 1:
Factor, if possible.
x2 – 4x – 21
Here, we have b = –4 and c = –21.
So, we need two numbers whose sum is –4 and whose product is –21.
Listing all the integer factor pairs of –21, we get:
| Factors |
Sum |
| 1, –21 |
–20 |
| –1, 21 |
20 |
| 3, –7 |
–4 |
| –3, 7 |
4 |
So, the correct pair is 3, –7.
(x + 3)(x – 7) = x2 – 4x– 21
You can double check this using FOIL.
Example 2:
Factor, if possible.
x2 + 7x + 13
Here, we need two numbers whose sum is 7 and whose product is 13.
Listing all the integer factor pairs of 13, we get:
| Factors |
Sum |
| 1, 13 |
14 |
| –1, –13 |
–14 |
Since none of the factor pairs add up to 7, this polynomial cannot be
factored.
