Decompose $X^8 + X^7 + X^6 + X^4 + 1$ over $\mathbb{Z}_2$ Is there a clever way to decompose $f(X) = X^8 + X^7 + X^6 + X^4 + 1$ into irreducible factors over $\mathbb{Z}_2$? Or to see that it is irreducible itself?
This is what I have thought about so far: $f(0) \ne 0$ and $f(1) \ne 0$, so $f$ does not have linear factors. It also does not have quadratic factors, because the only irreducible quadratic polynomial over $\mathbb{Z}_2$, $X^2 + X + 1$, does not divide $f$. I could continue finding all irreducible polynomials of degree 3 and 4 and carrying out the corresponding polynomial divisions.
Is there a more clever solution?

Note that $(x^8+x^7+x^6+x^4+1)(x+1)=x^9 + x^6 + x^5 + x^4 + x + 1$.
Thus, if $\alpha^5=1$, then $\alpha$ is a root of the latter polynomial. That is, the latter polynomial is divisible by $x^5+1=(x^4 + x^3 + x^2 + x + 1)(x+1)$.  But this means that 
$x^4 + x^3 + x^2 + x + 1$ is a factor of your original polynomial. Now polynomial division leads us to
$$x^8+x^7+x^6+x^4+1=(x^4 + x + 1)(x^4 + x^3 + x^2 + x + 1).$$
It is easy to see that both fators on the RHS are irreducible over $\mathbb{Z}_2$. In fact, they have no roots in $\mathbb{Z}_2$ and are not divisible by $x^2+x+1$.
