Factoring Trinomials as a Product of Binomials

Binomials are simplified polynomials with two terms like x² + 2 or 3ab² – 5b³. Trinomials are simplified polynomials with three terms like 3x² + 14x + 7 or ab + b – c. It’s possible to sometimes multiply two binomials and get a trinomial, like the following:

(2x – 3)(x + 5)
= (2x-3)(x) + (2x-3)(5)
= 2x² – 3x + 10x – 15
= 2x² + 7x – 15

Our goal is to be able to factor such a trinomial. To do this, we’re going to need to somehow reverse the process of multiplication, and effectively go from the “= 2x² + 7x – 15″ step to the “= 2x² – 3x + 10x – 15″ step for any given trinomial. The problem is that there is no clear, logical way to do this. In light of this problem, we need a small trick to get us started.

Our trick is called the “magic number”. Recall that the standard form for a trinomial is something like ax² + bx + c. We are going to call the product of a and c the “magic number”. The trick is based on finding two integers that multiply together to give us the “magic number”, but that add together to give us b. Here’s an example of finding the “magic number” and the two integers that multiply to give us the “magic number” while adding together to give us the coefficient b:

6x² – x – 15

Note that a = 6, b = -1 and c = -15. The “magic number” is ac = 6(-15) = -90. So what are two numbers that multiply together to give us -90 while adding together to give us -1. The answer is -10 and 9.

The point of this process is that the two numbers we get as a result are the integers we use as coefficients to break up the middle term of the trinomial. In action, this means:

6x² – x – 15 becomes 6x² – 10x + 9x – 15 or 6x² + 9x – 10x – 15

Note that the order of the two numbers doesn’t matter; we’ll get the same answer either way. So now that we have broken up the middle term, we want to use the GCF method to factor the first two terms and “pull out” the GCF like this:

6x² – 10x + 9x – 15
2x(3x – 5) + 9x – 15

Now you want to do the same thing to the last two terms:

2x(3x – 5) + 9x – 15
2x(3x – 5) + 3(3x – 5)

What you’ll notice is that what’s left from the grouping of the first two terms and the grouping of the second two terms should be the same binomial. Now we can apply the distributive property to “pull out” the binomial (3x-5) like so:

2x(3x – 5) + 3(3x – 5)
(3x – 5)(2x + 3)

And we’ve successfully factored the trinomial. We can check our answer by multiplying these two binomial factors together to see if we get the original polynomial.

(3x – 5)(2x + 3)
2x(3x – 5) + 3(3x – 5)
6x² – 10x + 9x – 15
6x² – x – 15

Not only do we get the same answer, but the steps end up being in the exact reverse order of how we did the factoring.