# How do you factor x^2+ 11x = 180?

May 24, 2015

First subtract $180$ from both sides to get:

${x}^{2} + 11 x - 180 = 0$

Noticing that the coefficient of ${x}^{2}$ is $1$ and the signs of the other two coefficients, this may factorize as:

$\left(x + a\right) \left(x - b\right) = {x}^{2} + \left(a - b\right) x - a b$ with $a > 0$, $b > 0$, $a - b = 11$ and $a b = 180$.

$180 = 20 \times 9$ and $11 = 20 - 9$

So we can let $a = 20$ and $b = 9$ to get:

${x}^{2} + 11 x - 180 = \left(x + 20\right) \left(x - 9\right)$

Hence ${x}^{2} + 11 x = 180$ has solutions $x = - 20$ and $x = 9$