How to factor this polynomial over $\mathbb{F}_2$ On page 587 in Dummit and Foote, the authors say the polynomial $\frac{x^{16}-x}{x(x-1)(x^2+x+1)}$ can be factored into quartics over $\mathbb{F}_2$ as $(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$.
I am having trouble seeing this. When I divide the polynomial $x^{16}-x$ by $x(x-1)(x^2+x+1)$ using long division, I get $x^{12}+x^9+x^6+x^3+1$. However, I am not sure how to factor this polynomial into quartics. How do I do this? 

Over $\mathbb Z$ we have
$$
\begin{align}
x^{16}-x &= x(x^{15}-1)
\\&= x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x)
\\&= x (x - 1) (x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
\end{align}
$$
and so your polynomial is $(x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$.
You only need to factor $(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$ mod $2$.
