 Solution

The splitting field over ${\mathbb{F}}_2$ is ${\mathbb{F}}_{16}$, and it contains the unique subfield ${\mathbb{F}}_4$. Over ${\mathbb{F}}_4$, Frobenius is the square of the order-four ${\mathbb{F}}_2$ Frobenius, so it has two orbits of length two on the four roots. Thus $f$ factors into two irreducible quadratics, andgenerated by $x\mapsto x^4$ on ${\mathbb{F}}_{16}$.