factor completely

One change to Tenille's answer: the signs in the result are not correct. If you FOIL out her result, you get

2(-2x - 1)(x + 3) = (2)[(-2x)(x) + (-2x)(3) + (-1)(x) + (-1)(3)] = (2)[-2x^{2} -6x - x - 3] = (2)(-2x^{2} - 7x - 3)

= -4x^{2}-14x- 6 <----NOT what we started with!

I find it easier to factor if my leading coefficient is positive. That is, if we factor out a -2 instead of a 2, we get

-4x^{2} + 14x - 6 = (-2)(2x^{2} - 7x + 3)

Then factoring (-2)(2x^{2} - 7x + 3) gives us (-2)(2x - 1)(x - 3)

Check by FOIL:

(-2)(2x-1)(x-3) = (-2)[(2x)(x) + (2x)(-3) + (-1)(x) + (-1)(-3)] = (-2)(2x^{2} - 6x - x + 3) =

(-2)(2x^{2} - 7x + 3) = -4x^{2} + 14x - 6 <---- What we started with.

## Comments

my mistake. Thank you Kevin for catching that. This is why one should always double their work.

-4x^2 + 14x - 6

2(-2x^2 + 7x - 3)

2(2x - 1)(-x + 3)