Factoring ax + bx + c

Introduction
Factoring ax 2 + bx + c
You have learned how to factor a standard trinomial of the form ax 2 + bx + c, where a = 1. If
a = 2 or more, what is the methodology used to solve then?
One method is to continue using guess and check, however using this method can get more
complicated as there are more possibilities to work with. Here is a review of Guess &
Check.
Guess & Check
Example #1: Factor x 2 + 11x + 24
Step 1: Create two binomials from the trinomial, placing x in each: (x + _)(x + _) (Note that
signs may differ in each binomial.
Step 2: Figure what two numbers multiply to the constant (24) and add up to the co-efficient of
the second term (11): 8 x 3 = 24 and 8 + 3 = 11
Step 3: Solve: (x + 8)(x + 3) are the factors for x 2 + 11x + 24
Check: Mutliply the two binomials out using FOIL to check your work.
What if the co-effient for the first term is greater than one?
Example #2: Factor 6x 2 – 9x - 15
Step 1: Can a greatest common factor (GCF) be removed? If no, skip to Step 2. If yes, proceed
… 3(2x 2 – 3x - 5)
Step 2: Now, you must factor the remainder trinomial 2x 2 – 3x - 5: So, referring to Step 1 in the
previous example, create two binomials, this time … (2x )(x ) (Note that there may be
differences of signs so I have left the signs out at this point (+ or -).
Step 3: Now, you must figure what two numbers will multiply to the constant (5) and the coefficient
of the second term (-3), but also must use the factors that will multiply to the coefficient
of the first term: 5 x 1 is the only possibility for the constant 5 as it is prime, and the
co-efficient of the first term, 2, is also prime, so 2 x 1 is the only possibility there: You must use
2x and x and 5 and 1 as the four terms in the two binomials.
Step 4: Solve (arrange the signs): (2x – 5)(x + 1) are the factors for 2x 2 – 3x – 5. The final answer
is 3(2x – 5)(x + 1). Don’t forget about the GCF, if there was one.

Check: Mutliply the two binomials out (and, in this case, the GCF 3) using FOIL to check your
work.
What if Guess & Check really doesn’t work because there are just too many possibilities when
the co-efficient has a term greater than 1? Now, we come to the Grouping Method.
Grouping Method
Example: Factor 8a 2 + 18a – 5
Step 1: Is there a GCF? No, so skip to Step 2.
Step 2: Take the co-efficient of the first term and multiply it to the constant: 8 x -5 = -40.
Step 3: List all of the factors for -40: -40 x 1, -20 x 2, -10 x 4, -8 x 5 (and there opposites)
Step 4: Select the two factors that will add up to the co-efficient of the second term (18) and rewrite
the polynomial with those factors:
a. 20 and – 2 = 18
b. 8a 2 + 20a – 2a – 5
Step 5: Separate the polynomial into two groups: 8a 2 + 20a and -2a – 5
Step 6: Take the greatest common factor out for each group: 4a(2a + 5) – 1(2a + 5) … 4a is the
GCF for the first group and -1 is the GCF for the second group.
Step 7: Solve. Take the two sets of factors and re-write them. The first group is 4a – 1. The
second group is the pair 2a + 5. So, the final answer uses the factors for 8a 2 + 18a – 5 as
(4a – 1)(2a + 5). Note that the pair of factors in the grouping is just re-written as one binomial
in the final answer.
Check: If you have a pair of factors that are the same in one group (i.e. 2a + 5 in this example),
you likely have the correct answer. Mutliply the two binomials out using FOIL to check your
work.