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)[-2x² -6x - x - 3] = (2)(-2x² - 7x - 3)

= -4x²-14x- 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

-4x² + 14x - 6 = (-2)(2x² - 7x + 3)

Then factoring (-2)(2x² - 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)(2x² - 6x - x + 3) =

(-2)(2x² - 7x + 3) = -4x² + 14x - 6   <---- What we started with.