Factor completely?

Question

http://i36.photobucket.com/albums/e16/Screch/3.png

Answer 1

9t⁶+57t⁵+18t⁴

Every term is divisible by 3 and t⁴, so the first set of factors is 3t⁴.  That leaves:

3t²+19t+6

This can be factored to:

(3t+a)(t+b) = 3t²+19t+6

3t²+(3b+a)t+ab=3t²+19t+6

So,

3a+b=19

ab=6

So,

a=6; b=1

So, the additional factors are (3t+6)(t+1).

The full set are 3t⁴(3t+6)(t+1)

Answer 2

9t^6 + 57t^5 + 18t^4

= (3t^4) (3t^2 + 19t + 6)

= (3t^4) (t + 6) (3t + 1)

Answer 3

3t^4(3t^2+19t+6)