Question

How do you factor `2x²+15x-108`?

Answer 1

`2x^2 + 15x - 108 = (2x - 9)(x + 12)`

Noticing the `2x^2` term, I looked for a factorisation of the form:

`2x^2 + 15x - 108 = (2x + a)(x + b)`

`= 2x^2 + (2b+a)x + ab`

Comparing coefficients, we are looking for `a` and `b` such that:

`2b + a = 15`

`ab = -108`

I know that `108 = 9 * 12` (amongst other possibilities) and found that putting `a = -9` and `b = 12` worked.

Ultimately, it is always possible to use the general formula when solving quadratics like this, but spotting the factors of the constant term and trying a few possibilities will often get you the answer quicker.

Answer 2

There is another way that doesn't require guessing: the new AC Method to factor trinomials (Google, Yahoo Search)

`f(x) = 2x^2 + 15x - 108. (1) = (x - p)(x - q)`

Converted trinomial: `f'(x) = x^2 + 15x - 216 = (x -p')(x - q')`. Find p' and q'.

Compose factor pairs of (a.c = -216). Proceed:....(-8, 27)(-9, 24). This last sum is 24 - 9 = 15 = b. Then, p' = -9 and q' = 24. Back to original
trinomial f(x) (1):
`p= (p')/a = -9/2 and q = (q')/a = 24/2 = 12.`

Factored form: `f(x) = (x - 9/2)(x + 12) = (2x - 9)(x + 12)`

Check: Develop: `(2x - 9)(x + 12) = x^2 + 15x - 108.` Correct.