If $f$ has more than one root in $K$, then $f$ splits and $K/k$ is Galois? Let $f \in k[x]$ be an irreducible polynomial of prime degree $p$ such that $K \cong k[x]/f(x)$ is a separable extension. How do I see that if $f$ has more than
one root in $K$, then $f$ splits and $K/k$ is Galois?

Let $G=\mathrm{Aut}(K/k)$. Since $f$ has more than one root in $K$, it follows that $|G|>1$. 
Suppose that $|G|<p$, and choose some $1\neq\sigma\in G$ of order $q$. Let $F$ be the subfield of $K$ fixed by $\sigma$. Then by Artin's theorem, it follows that $K/F$ is a Galois extension, with $[K:F]=|\sigma|=q$.
Therefore
$$p=[K:k]=[K:F][F:k]=q[F:k] $$
which is a contradiction because $q<p$ and $p$ is prime.
Hence $|G|=p$, so $K$ is a Galois extension of $k$ and $f$ splits in $K$.
