# Why x^p-1 Factors over GF(2) the Way It Does

Let $p$ be a prime number. We know that $F_2=GF(2)=\{0,1\}$. Let $F_2[x]$ be the ring  the polynomials with indeterminate $x$ and coefficients in $F_2$. The polynomial $(x^p-1) \in F_2[x]$ factors over $F_2$ as  follows for various $p$:

$x^3-1=(1+x) (1+x+x^2).$
$x^5-1=(1+x) (1+x+x^2+x^3+x^4).$
$x^7-1=(1+x) (1+x+x^3) (1+x^2+x^3).$
$x^{11}-1=(1+x) (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}).$
$x^{13}-1=(1+x) (1+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}).$
$x^{17}-1=(1+x) (1+x^3+x^4+x^5+x^8) (1+x+x^2+x^4+x^6+x^7+x^8).$

None of the factors above can be factored anymore, hence they are irreducible over $F_2$. Let us call $x+1$ the trivial factor since the root $x= 1$ already belongs to $F_2$. But why do we have two nontrivial irreducible factors of $x^7-1$, each of degree 3, whereas $x^{13}-1$ has only one non-trivial irreducible factor of degree 12? It appears that either there is only one nontrivial factor, or the degree of all nontrivial factors is the same. Why?