# How do you factor by grouping t^3 + 6t^2 - 2t - 12?

Apr 28, 2015

Group ${t}^{3}$ and 6t^2 # like this,

${t}^{2} \left(t + 6\right)$

then group $- 2 t$ and $- 12$ like this,

$- 2 \left(t + 6\right)$

${t}^{2} \left(t + 6\right) - 2 \left(t + 6\right) \implies \left({t}^{2} - 2\right) \left(t + 6\right)$
Now, further ${t}^{2} - 2$ is a difference of two squares so you factorize thus,
${t}^{2} - 2 = \left(t - \sqrt{2}\right) \left(t + \sqrt{2}\right)$
So, the final result is $\left(t - \sqrt{2}\right) \left(t + \sqrt{2}\right) \left(t + 6\right)$