= Solution
The <polynomial> is irreducible by <Eisenstein's criterion> at $2$. Its <Galois group> therefore acts transitively on five roots, so its order is divisible by $5$ by the <orbit-stabilizer theorem>.
To count the real roots, differentiate:
$$
f'(x)=5x(x^3-4).
$$
It is positive on $(-\infty,0)$, negative on $(0,\sqrt[3]4)$, and positive thereafter. The local maximum is $f(0)=2>0$ and the local minimum is $2-6(\sqrt[3]4)^2<0$. The limits at infinity have opposite signs, so there are exactly three real roots and one conjugate pair of nonreal roots. <Complex conjugation> fixes the three real roots and interchanges the other two; its permutation is a <transposition>. The prime-degree <subgroup> result supplied in the paper now gives
$$
\boxed{\operatorname{Gal}(f)\cong S_5.}
$$
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