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.