Question involving polynomials and fields I'm studying Galois Theory and I'm stuck in a problem, let's see:

Let $P(x) = X^5 - p \in \mathbb{Q}[X]$, where $p$ is prime. Find a Splitting field $L$ of $P$, five subfields of $L$ and the factoration of $P$ in that fields.

Now, let $\epsilon = e^{2i\pi/5}$, of course $L=\mathbb{Q}(\sqrt[5]{p},\epsilon)$ and the subfields may be $\mathbb{Q}_i(\epsilon^i \cdot \sqrt[5]{p})$, for $i=0,1,2,3,4$.
I'm stuck on the factorization, for instance, if $i=0$ then in $\mathbb{Q}_{0}(\sqrt[5]{p})$ we can write $$P(x) = (x - \sqrt[5]{p})(x^4 + x^3(\sqrt[5]{p}) + x^2 (\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4)$$
My question is:
Is $x^4 + x^3(\sqrt[5]{p}) + x^2 (\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4$ irreductible over $\mathbb{Q}_0(\sqrt[5]{p})$??
I've tried to solve a system or thinking in Binomial Theorem, but well... I'm here. Can you give me a hint? Also, I need to find that in another four fields...
 A: As we have $x^4 + x^3(\sqrt[5]{p}) + x^2(\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4 = (x-\sqrt[5]{p}\epsilon)(x-\sqrt[5]{p}\epsilon^2)(x-\sqrt[5]{p}\epsilon^3)(x-\sqrt[5]{p}\epsilon^4)$ implies if one of the linear reducible factor lies in $\mathbb{Q}(\sqrt[5]{p})$ then $5th$ primitive root of unity also lies in $\mathbb{Q}(\sqrt[5]{p})$ which is not true. Then we have only possibility left is degree 2 irreducible factors. As none of the $5th$ primitive roots of unity lies in $\mathbb{Q}(\sqrt[5]{p})$, so we can club the roots $\sqrt[5]{p}\epsilon$ and $\sqrt[5]{p}\epsilon^4$ together and see whether $x^2 - \sqrt[5]{p}(\epsilon + \epsilon^4)x + (\sqrt[5]{p})^2$  but $(\epsilon + \epsilon^4) \notin \mathbb{Q}(\sqrt[5]{p}) $ as if you adjoin $\epsilon + \epsilon^4$ in $\mathbb{Q}$ the resulting field have even dimension over $\mathbb{Q}$. So the polynomial $x^4 + x^3(\sqrt[5]{p}) + x^2(\sqrt[5]{p})^2 + x(\sqrt[5]{p})^3 + (\sqrt[5]{p})^4$ is irreducible over $\mathbb{Q}[\sqrt[5]{p}]$.
