This is an exercise from Garling's A Course in Galois Theory Ch. 5 on irreducible polynomials.
The other part of the question was to show that $x^5 - 4x + 2$ was irreducible over $\mathbb Q(i)$. I got that one. Clearly the polynomial is irreducible over $\mathbb Q$ by Eisenstein and Gauss' lemma, and, since 5 and 2 are relatively prime, it follows from a previous exercise that a root of the polynomial must have degree 5 over $\mathbb Q(i)$. So the given polynomial must be the root's minimal polynomial over $\mathbb Q(i)$ (and not just over $\mathbb Q$). So in particular the polynomial must be irreducible over $\mathbb Q(i)$.
But that trick doesn't work for $x^4 - 4x + 2$ anymore, and I'm stumped. I convinced myself that it can't have a linear factor over $\mathbb Z[i]$ because that would mean a root that was a factor of $2$, and none of the factors of $2$ (namely, $\pm 2, \pm 1, \pm 2i, \pm i, \pm (1 - i), \pm (1 + i)$) solve the polynomial, but I don't see why it can't have a quadratic factor. Help!