Question

3n^3-4n^2+9n-12 factor by grouping

Answer 1

3n^3+9n = 3n(n^2+3)
-4n^2-12 = -4(n^2-3)
So, you get (n^2-3)(3n-4)

Answer 2

n^2(3n-4) 3(3n-4) so our factor are n^2 3 and 3n

Answer 3

\[rewrite \to 3n ^{3}+9n-4n ^{2}-12\]
\[3n \left( n ^{2}+3 \right)-4\left( n ^{2}+3 \right)\]

Answer 4

I mistyped n^2-3 instead of n^2+3. Can't copy paste and fix with this tool easily.

Answer 5

It should be as follows:
3n^3+9n = 3n(n^2+3)
-4n^2-12 = -4(n^2+3)
So, you get (n^2+3)(3n-4)