irreducibility in $F_2[X]/X^5+X^2+1$ Let $K=F_2[X]/(X^5+X^2+1)$ and $Q=T^2+T+1$.
I am trying to figure out whether $Q $ is irreducible in $K[X]$.
There is a hint in my exercise (show that a root of $Q$ verifies $X^3=1$)
What I did:
I showed that indeed, a root T of Q verifies $T^3=1$:
Let T be a root of Q. We have $T^2+T+1=0$ so $T^3+T^2+T=0$.
Substracting the first equation from the second gives $T^3=1$.
Now, I don't know how to use that to prove that Q is reducible (what I suspect) in $K[T]$
Many thanks for any help or hint.
 A: Hint: Since $X^2 + X + 1$ has degree $2$, it is reducible over $K$ iff it has a root $r$ in $K$. As you showed, then $r^3 = 1$, so $r$ has order $3$ in the group of units $K^\times$. But what is the order of $K^\times$? What does Lagrange's Theorem say?
A: To avoid confusion, I'll write $\alpha$ for the residue class of $X$ in $K = {\mathbb F}_2[X]/(X^5 + X^2 + 1)$, so $K = {\mathbb F}_2(\alpha)$ with $\alpha^5 + \alpha^2 + 1 = 0$. Additionally, I'll look at the polynomial $Q = T^2 + T + 1$ in $K[T]$.
The question is to figure out whether or not $Q$ is reducible.
The first point to realize is that because $Q$ is of degree $2$, it is reducible if and only if it has a root in $K$. 
Secondly, as the hint suggests and as the OP already figured out, every root $r \in K$ of $Q$ satisfies $r^3 = 1$. This implies that the order of $r$ in the group $K^*$ of invertible elements of $K$ is $1$ or $3$. Now $1$ cannot really happen: that corresponds to $r = 1$, but that is not a root of $Q$. However, $K^*$ is a cyclic group of $2^5 - 1 = 31$ elements and because $3 \not\mid 31$ it does not have any elements of order $3$. 
