# How do you factor 9t^3+18t-t^2-2?

$9 {t}^{3} + 18 t - {t}^{2} - 2 = \left(9 {t}^{3} + 18 t\right) - \left({t}^{2} + 2\right)$
$= 9 t \left({t}^{2} + 2\right) - \left({t}^{2} + 2\right)$
$= \left(9 t - 1\right) \left({t}^{2} + 2\right)$
Since ${t}^{2} + 2 > 0$ for all $t \in \mathbb{R}$, it has no linear factors with real coefficients. So our factorization is complete.