TITLE: Factorize polynomial over $GF(3)$
QUESTION [5 upvotes]: I want to factorize $x^{11}-1$ over $GF(3)$ but I'm stuck at 
$(x-1)(x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1).$
I have tried to do it trial and error but failed. Is
$$
x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1
$$
already irreducible over $GF(3)$?

REPLY [2 votes]: I don't want at all to depreciate the cleverness and the mastery of the previous answer, however, you should be aware that this kind of questions – it is hard to define what is this kind, it grows every day – can be answered by computer algebra.
For example, using sage :
sage: R.<x> = PolynomialRing(GF(3))
sage: factor(x^11 - 1)
(x + 2) * (x^5 + 2*x^3 + x^2 + 2*x + 2) * (x^5 + x^4 + 2*x^3 + x^2 + 2)