



 
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) = 4x^{3}  6x^{2} + 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 4x^{3}  6x^{2} + 8x  6 : [gif] So f(x) = (x  1) (4x^{2}  2x + 6). (If you need to review how synthetic division works, see Melanie's explanation) Hope this helps,  


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