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

1 Answer
Apr 24, 2016

Answer:

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

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

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

Explanation:

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)#