# How do you factor by grouping x^3+3x^2-118x-120?

May 10, 2015

${x}^{3} + 3 {x}^{2} - 118 x - 120$
$= \left({x}^{3} + 2 {x}^{2} - 120 x\right) + \left({x}^{2} + 2 x - 120\right)$
$= \left(x + 1\right) \left({x}^{2} + 2 x - 120\right)$
$= \left(x + 1\right) \left(\left({x}^{2} + 2 x + 1\right) - 121\right)$
$= \left(x + 1\right) \left({\left(x + 1\right)}^{2} - {11}^{2}\right)$
$= \left(x + 1\right) \left(x - 10\right) \left(x + 12\right)$

I found the first grouping by noticing that $x = - 1$ is a solution of ${x}^{3} + 3 {x}^{2} - 118 x - 120 = 0$, so $\left(x + 1\right)$ is a factor.

The second grouping to $\left({x}^{2} + 2 x + 1\right) - 121$ is completing the square.