Finding an irreducible quartic in $\mathbb{Z}_2[x]$ I'm trying to solve the following question, can you please help me?

Find an irreducible polynomial $p(x) \in \mathbb{Z}_2[x]$  with degree 4.

Thank you!
 A: An irreducible biquadratic cannot have a root, so the constant term must be $1$, and the number of nonzero coefficients must be odd. So we would have
$$x^4 + ax^3 + bx^2 + cx + 1$$
with $a+b+c=1$ (addition is modulo $2$, of course, since $a,b,c\in\mathbb{Z}_2$).
Now, we also must avoid the case in which it is a product of two irreducible quadratics. What are the irreducible quadratics in $\mathbb{Z}_2[x]$? We have $x^2+x+1$... and that's it, because the other three quadratics, $x^2$, $x^2+1$, and $x^2+x$, all have roots.
So you need to make sure that $x^4+ax^3+bx^2+cx+1\neq (x^2+x+1)^2$. 
So... can you find a polynomial that satisfies those two conditions ($a+b+c=1$ and $x^4+ax^3+bx^2+cx+1\neq (x^2+x+1)^2$) ?
A: Hint $\rm\:f(x) = (x^5\!-\!1)/(x\!-\!1) = x^4\! +\! x^3\! +\! x^2\! +\! x\! +\! 1\:$ has no linear factors by $\rm\:f(0) = 1 = f(1).\:$ And $\rm\:f\:$ has no quadratic factors: the only irreducible quadratic is $\rm\:g(x) = x^2\!+\!x\!+\!1,\:$ but $\rm\:g\:|\:f\:$ $\Rightarrow$ $\rm\:(x\!-\!1)g\:|\:(x\!-\!1)f,\:$ i.e. $\rm\:x^3\!-\!1\:|\:x^5\!-\!1\:$ $\Rightarrow$ $\rm\:x^3\!-\!1\:|\:x^2\!-\!1 = (x^5\!-\!1)-x^2(x^3\!-\!1),\,$ contra degree. 
