Expert: David Hemmer - 3/28/2007

Question
I am having trouble remembering the steps in factorising numbers/pronumerals raised to powers 3 and beyond. If I could pose the specific question: is it possible to factorise [x^3 - 8]?????

Answer
Factoring polynomials is not easy! One helpful fact to know is the following theorem:

Theorem: Suppose a polynomial has a root r, i.e. p(r)=0, then it has a factor x-r.

Notice in your polynomial x^3-8 it has 2 as a root. Therefore it has x-2 as a factor. If you factor this out you get (by long division):

x^3-8=(x-2)(x^2+2x+4)

If the remaining quadratic factored it would have two root but the quadratic formula tells you the roots are both imaginary, so it won't factor any further.