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

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Answer 1 · 2016-04-24T21:11:57

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

$=(9t-1)(t^2+2)$

$=(9t-1)(t-sqrt(2)i)(t+sqrt(2)i)$

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

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

$=(9t^3+18t)-(t^2+2)$

$=9t(t^2+2)-1(t^2+2)$

$=(9t-1)(t^2+2)$

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

$=(9t-1)(t-sqrt(2)i)(t+sqrt(2)i)$

How do you factor completely 9t^3+18t-t^2-29t^3+18t-t^2-2?