How do you factor 2x²+15x-108?

2 Answers
May 11, 2015

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.

May 12, 2015

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.