ex. 6^2 – x – 12
6x^2 – 9x + 8x – 12
3x( 2x – 3) + 4(2x – 3)
(3x + 4)(2x – 3)

I can see my students forgetting to put “a” back and divide

Here’s an example. We’ll factor

6x^2 – x – 12

We take the 6 from the first term and multiply the last by it:

x^2 – x – 72

Factoring this, we see that 72 = 8*9 and 8 – 9 = -1, so

x^2 – x – 72 = (x + 8)(x – 9)

Replacing x with 6x and pulling out common factors,

(6x + 8)(6x – 9)

becomes

(3x + 4)(2x – 3)

This is the desired factorization.

Now I’ll prove that it works:

Suppose you are factoring

ax^2 + bx + c

You factor

x^2 + bx + ac = (x-m)(x-n)

(In my example, m = -8, n = 9.)

Now you replace x with ax:

(ax-m)(ax-n) = (ax)^2 + b(ax) + ac

We’ll have to assume that when you remove common factors, the product
of the factors you divide out is equal to a. (In my example I took
out 2 in the first factor and 3 in the second, whose product is 6.)
I think this is a consequence of the assumption that a, b, and c have
no common factors, but I’m not going to bother to prove it.

With this assumption, the resulting factorization is the same as

(ax-m)(ax-n)/a = [a^2x^2 + abx + ac]/a = ax^2 + bx + c

So if you can follow your process, you have indeed factored the
original trinomial.

From Ask Dr. Math