= Solution
<Eisenstein's criterion> at two makes $X^5-4X+2$ irreducible: all nonleading coefficients are divisible by two and the constant is not divisible by four. Its <Galois group> is therefore transitive on five roots, and its order is divisible by five, so <Cauchy's theorem for finite groups> supplies a five-cycle.
The <derivative> is $5X^4-4$, with real critical points $\pm c$, $c=(4/5)^{1/4}$. The values are $f(-c)=2+16c/5>0$ and $f(c)=2-16c/5<0$. The <derivative> is positive outside these two points and negative between them, so there are exactly three real roots and one nonreal conjugate pair. <Complex conjugation> therefore acts as a single <transposition>.
A five-cycle together with any <transposition> generates $S_5$: conjugating that <transposition> by powers of the cycle gives edges joining labels with a fixed nonzero difference modulo five. These edges form a connected graph, and <transpositions on a connected graph generate the symmetric group>. Hence
$$
\boxed{\operatorname{Gal}(L/\mathbb Q)=S_5,\qquad[L:\mathbb Q]=120.}
$$
The equality of degree and <group> order uses separability in <characteristic> zero.
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