# $x^8-1$ - Irreductible polynomials over $\mathbb{Z}/3\mathbb{Z}$. [duplicate]

How could it possible to factorise $x^8-1$ in product of irreductibles in the rings $\mathbb{Z}/3\mathbb{Z}$?

Theorem : Let $A$ an integral domain and $I$ a proper ideal of $A$. If $f(x) \not \equiv a(x)b(x) (\mod I)$ for any polynomials $a(x)$, $b(x)$ $\in A[x]$ of degree $\in [1, \deg(f))$, then $f(x)$ is irreductible in $A[x]$

I think I have to use this theorem, but I am not certain. Is anyone could help me at this point?

• Do. Not. Reask. The same question. EVER!!!!!! – Jyrki Lahtonen Mar 9 '16 at 8:20