Question 76992
In this case we have a common factor of x, so we can factor out an x

{{{2x^3+3x^2-2x}}}

{{{x(2x^2+3x-2)}}}

Now we can factor the quadratic in the parenthesis

*[invoke quadratic_factoring 2, 3, -2]

So the quadratic in the parenthesis factors to this:
{{{(2x-1)(x+2)}}}

Now our factored polynomial looks like this


{{{x((2x-1)(x+2))}}}

Notice if we foil the terms and distribute the x we get {{{2x^3+3x^2-2x}}} again