Question

factor the GCF and then the resulting trinomial: 3x^3-3x^2-6x

Answer 1

GCF = 3x

Answer 2

3x^3-3x^2-6x = 3x(x^2 - x - 2)
factoring x^2 - x -2 = x^2 -2x + x -2 = x(x - 2) +1(x - 2) = (x + 1)(x - 2)
so
3x^3-3x^2-6x = 3x(x^2 - x - 2) = (3x)(x + 1)(x - 2)

Answer 3

We have the trinomial
\[3x^3-3x^2-6x\]
We need to factor it, Let's take out x from all the terms
\[x(3x^2-3x-6)\]
We'll now find the factors of
\[3x^2-3x-6)\]
Let's take out 3 common
\[3(x^2-x-2)\]
we have now
\[x^2-x-2]\
We need to find factors of -2 such that their sum is -1
these are -2 and 1
so
\[x^2-2x+x-2\]
We have
\[x(x-2)+1(x-2)\]
So we get
\[(x-2)(x+1)
So
the factors of \(3x^3-3x^2-6x\) are
\[3x(x-2)(x+1)\]