How do you factor by grouping $t^{3} + 6 t^{2} - 2 t - 12$ $t^3 + 6t^2 - 2t - 12$ ?

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Answer 1 · 2015-04-28T23:03:42

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

$t^{2} (t + 6)$ $t^2(t+6)$

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

$- 2 (t + 6)$ $-2(t+6)$

Now add the two,

$t^{2} (t + 6) - 2 (t + 6) \Rightarrow (t^{2} - 2) (t + 6)$ $t^2(t+6)-2(t+6) => (t^2-2)(t+6)$

Now, further $t^{2} - 2$ $t^2-2$ is a difference of two squares so you factorize thus,

$t^{2} - 2 = (t - \sqrt{2}) (t + \sqrt{2})$ $t^2-2=(t-sqrt(2))(t+sqrt(2))$

So, the final result is $(t - \sqrt{2}) (t + \sqrt{2}) (t + 6)$ $(t-sqrt(2))(t+sqrt(2))(t+6)$

How do you factor by grouping t3+6t2−2t−12t^3 + 6t^2 - 2t - 12?