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

Apr 24, 2016

$9 {t}^{3} + 18 t - {t}^{2} - 2$

$= \left(9 t - 1\right) \left({t}^{2} + 2\right)$

$= \left(9 t - 1\right) \left(t - \sqrt{2} i\right) \left(t + \sqrt{2} i\right)$

Explanation:

Normally I would rearrange in standard form first, but we can factor by grouping as it is, so...

$9 {t}^{3} + 18 t - {t}^{2} - 2$

$= \left(9 {t}^{3} + 18 t\right) - \left({t}^{2} + 2\right)$

$= 9 t \left({t}^{2} + 2\right) - 1 \left({t}^{2} + 2\right)$

$= \left(9 t - 1\right) \left({t}^{2} + 2\right)$

The remaining quadratic factor can be factored using Complex coefficients...

$= \left(9 t - 1\right) \left(t - \sqrt{2} i\right) \left(t + \sqrt{2} i\right)$