Hi Sean.

Factoring cubic generally is not simple. I would suggest using the factor theorem. This tells us that given a cubic f(x), if we can guess a value "n" for which f(n) = 0, then (x - n) is a factor of f(x). Then we can use synthetic division or long division to give us a quadratic factor to go with the (x - n). Factoring a quadratic should be straightforward for you if you are working on factoring cubics.

For example:

f(x) = 4x³ - 6x² + 8x - 6

I try guessing x = 1. Does f(1) = 0?

f(1) = 4(1)³ - 6(1)² + 8(1) - 6 = 4 - 6 + 8 - 6 = 0. Getting it equal to zero means (x - 1) is a factor. (I got it on my first guess)

Now I use synthetic division to divide (x - 1) into 4x³ - 6x² + 8x - 6 :

[gif]

So f(x) = (x - 1) (4x² - 2x + 6).

(If you need to review how synthetic division works, see Melanie's explanation)

Hope this helps,
Stephen La Rocque