 Solution

Over a finite field ${\mathbb{F}}_q$, the splitting field is a finite extension ${\mathbb{F}}_{q^m}$, where $m$ is the least common multiple of the degrees of the distinct irreducible factors. Its Galois group is cyclic of order $m$, generated by $x\mapsto x^q$.

Over ${\mathbb{F}}_2$, the polynomial has no linear factor: its values at zero and one are both one. The only irreducible quadratic is $x^2+x+1$; reduction modulo it gives $x$, not zero. A reducible quartic without linear factors would split into quadratics, so $f$ is irreducible. Thus its splitting field is ${\mathbb{F}}_{16}$ and