$X^2-5$ irreducible in $\mathbb{Q}(\sqrt[5]{2})[X]$

Question

Is there a smart way to show that $X^2-5$ is irreducible in $\mathbb{Q}(\sqrt[5]{2})[X]$?

If it wasn't irreducible we'd have $\sqrt{5}=a\cdot 1+b\cdot\sqrt[5]{2}+c\cdot\sqrt[5]{8}+d\cdot\sqrt[5]{16}$ which I'm sure somehow leads to a contradiction but on paper that would be time-consuming.

Answer 1

Any (finite) algebraic field extension of $\Bbb Q$ that contains $\sqrt 5$ has even degree over $\Bbb Q$. What is the degree of $\Bbb Q(\sqrt[5]2)$ over $\Bbb Q$?