# How do you factor 2x²+15x-108?

May 11, 2015

$2 {x}^{2} + 15 x - 108 = \left(2 x - 9\right) \left(x + 12\right)$

Noticing the $2 {x}^{2}$ term, I looked for a factorisation of the form:

$2 {x}^{2} + 15 x - 108 = \left(2 x + a\right) \left(x + b\right)$

$= 2 {x}^{2} + \left(2 b + a\right) x + a b$

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

$2 b + a = 15$

$a b = - 108$

I know that $108 = 9 \cdot 12$ (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)

$f \left(x\right) = 2 {x}^{2} + 15 x - 108. \left(1\right) = \left(x - p\right) \left(x - q\right)$

Converted trinomial: $f ' \left(x\right) = {x}^{2} + 15 x - 216 = \left(x - p '\right) \left(x - q '\right)$. 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):
$p = \frac{p '}{a} = - \frac{9}{2} \mathmr{and} q = \frac{q '}{a} = \frac{24}{2} = 12.$

Factored form: $f \left(x\right) = \left(x - \frac{9}{2}\right) \left(x + 12\right) = \left(2 x - 9\right) \left(x + 12\right)$

Check: Develop: $\left(2 x - 9\right) \left(x + 12\right) = {x}^{2} + 15 x - 108.$ Correct.