How to factor $x^5-1$ for binary codes

Question

Currently, I have that I can do:

$$ x^5-1 = (x-1)(x^4+x^3+x^2+x+1) $$

Then I thought I could do:

$$ x^5-1 = (x-1)(x(x+1)(x^2+1)+1) $$

However I want to use the factorization to find all cyclic codes in $[5, k]$. For that, I need to find all irreducible factors of $ x^5-1 $. I am not seeing if I can actually reduce $x^4+x^3+x^2+x+1$ any further

Answer 1

Over GF(2)$=\Bbb F_2$ we have $1=-1$ and $2=0$. The factorization (using prime factors in $\Bbb F_2[x]$) of the given polynomial is already $$ x^5+1 =(x+1)(x^4+x^3+x^2+x+1)\ . $$ To see that $f=(x^4+x^3+x^2+x+1)$ is irreducible (= prime in $\Bbb F_2[x]$), it is enough to check there is no factor of degree one or two. The irreducible factors in degrees one and two are $x$, $(x+1)$, and $(x^2+x+1)$. The factors in degree one are easily excluded. (Since $f(0)=f(1)=1$.) And the rest by division with rest of $f$ by $(x^2+x+1)$ is $x+1$, so the factor in degree two is also excluded.

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A computer algebra system is useful in such cases, for instance sage delivers the factorization:

sage: R.<x> = PolynomialRing(GF(2))
sage: factor(x^5+1)
(x + 1) * (x^4 + x^3 + x^2 + x + 1)

sage: # note that x^2 + 1 is reducible... 
sage: factor(x^2+1)
(x + 1)^2