Factoring Trinomials
Factoring a Trinomial of the Form ax2+ bx+ c
Example
Factor: 3x2 + 13x - 30
Solution
| Step 1 |
Factor out common factors (other than 1 or
-1). There are no common factors other than 1 or
-1. |
| Step 2 |
Identify the values of a, b, and c. Then find two integers whose
product is ac and whose sum is b.
3x2 + 13x - 30 has the form ax2
+ bx + c where a
= 3, b = 13, and c
= -30.
The product ac is 3 · (-30)
= -90.
Thus, find two integers whose product, ac, is -90 and whose sum,
b, is 13.
• Because their product is negative, the integers must have
different signs.
• Because their sum is positive, the integer with the greater
absolute value must be positive.
Here are some of the possibilities:
| Product
-1
· 90
-2
· 45
-3
· 30
-5
· 18 |
Sum
89
43
27
13 |
We can stop here since -5 and 18 satisfy the requirements that the
product is -90 and the sum is 13.
|
| Step 3 |
Replace the middle term, bx, with a sum or difference using the
two integers found in step 2.
| Original trinomial.
Equivalent 4-term polynomial. |
3x2 + 13x
- 30
= 3x2 - 5x
+ 18x - 30 |
|
| Step 4 |
Factor by grouping.
Factor x out of the first group;
factor 6 out of the second group.
Factor out the common factor
(3x - 5). |
= (3x2
- 5x) + (18x
- 30)
= x(3x - 5)
+ 6(3x - 5)
= (3x - 5)(x
+ 6) |
So, the factorization is (3x - 5)(x
+ 6).
Note:
We can use FOIL to check the
factorization:
| (3x - 5)(x
+ 6) |
= 3x2
+ 18x - 5x -
30
= 3x2 + 13x
- 30 |
The factorization checks.
|