I am currently working with $GF(2)$, in particular with $GF(2^4)$. One task is to find all irreducible polynomials. I have found ways of reducing the list of all candidates drastically. In my current list of irreducibles there is still one polynomial left which shouldn't be there:
$x^4+x^2+1$
I don't know why this one is reducible. How can I factorize this polynomial?
On
for any prime $p$ and $\{a_j \in \mathbb{Z}\}$ you have:
$$ \sum_{j=0}^n a_j x^{b_j p} \equiv_p \left(\sum _{j=0}^n a_j x^{b_j}\right)^p $$
using $p=2$ the given polynomial $x^4+x^2+1 \equiv_2 (x^2+x^1+1)^2$
btw, if you are looking at irreducible polynomials in finite fields it is useful to know how many irreducible polynomials of degree $n$ there are over the field with $q$ elements.
this is $$ \rho(n) = \frac1{n} \sum_{d|n} \mu(\frac{n}{d}) q^d $$ in the case $q=2, n=4$ this evaluates to $\frac14(2^4-2^2)=3$. these are $x^4+x^3+1, x^4+x+1$ and $x^4+x^3+x^2+x+1$.
$(x^2+x+1)^2 = x^4+x^2+1$ by Freshmen's Dream.
On the other hand, let $f$ be a polynomial of degree $4$ over $\mathbb F_2$, which has no roots. Then $f$ is either irreducible or decomposes into two irreducible factors of degree $2$. But $x^2+x+1$ is the only irreducible of degree $2$, hence we get $f=(x^2+x+1)^2$. So your polynomial is the only reducible polynomial of degree $4$ without roots.