Is $t^4 + 7$ irreducible over $\mathbb{Z}_{17}$?

Question

There aren't any linear factors, $-7$ doesn't have a square root in $\mathbb{Z}_{17}$ so I've ruled out those 'easy' factorisations. I'm not sure how to continue. Looking for a hint, not a solution.

Answer 1

Extended hints:

1. Calculate the order of $-7$ in $\Bbb{Z}_{17}^*$. In what follows I call it $n$.
2. Because $n$ is a power of two (how did I know that in advance?) you can deduce that any zero $\alpha$ of $x^4+7$ has order $4n$ ($\alpha$ is in some finite extension field $\Bbb{F}_{17^k}$).
3. Recall that the multiplicative group $\Bbb{F}_{17^k}^*$ is cyclic of order $17^k-1$. Use this to determine $k$.
4. What's the connection between $k$ and the minimal polynomial of $\alpha$?