I was asked this question on my exam but I didn't know how to solve it. Eisenstein criterion can't be applied, rational roots theorem shows that it doesn't have a linear factor because possible roots are $\lbrace 1,-1,2,-2,4,-4 \rbrace$ and it doesn't seem to be irreducible over $\mathbf Z/2\mathbf Z$ or $\mathbf Z/3\mathbf Z$
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$\begingroup$Hint: $f(x)=x^5-2x+4$ has only one real root. By rational root test, $f(x)$ has no linear factor. If $f(x)$ is reducible, it must have a quadratic factor of the form $x^2+Ax+B,$ where $B$ divides $4$. But one has $A^2-4B<0$ and hence $B>0$, otherwise it will yield at least two real roots. There are just a few cases to consider.
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