= 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
$$
\boxed{\operatorname{Gal}(f/\mathbb F_2)\simeq C_4.}
$$
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