# 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 $x^2+x+1$.
• Neat, thank you! How did you know that multiplying by $X + 1$ would yield a useful polynomial? Is that a common trick in $\mathbb{Z}_2$? – bbrot Jan 7 '17 at 19:35
• I started looking for roots $\alpha$, for which $\alpha^n=1$ (for some $n$). Since $x-1$ is always a factor of $x^n-1$, that led me to consider this multiplication. – math Jan 7 '17 at 20:06