The problem is stated as follows: Let $K$ be a finite field extension of $F$ and consider $P\in K[x]$ monic irreducible. Then, there exists $Q\in K[x]$ such that $PQ\in F[x]$.
This question is essentially asking us to prove that $(P)\cap F[x]\neq\emptyset$, and I'm not sure where to begin if I take a more algebraic route. Another approach would maybe be to take a basis of $K$, i.e. $\{\alpha_1,\dots,\alpha_n\}$. If $P(x)=a_0+a_1x+\dots+a_mx^m$, then there is some formula for the coefficients $a_i=\sum_{k=1}^{n} b_{i_k}\alpha_{i_k}$ where $b_{i_k}\in F$. This will give us that $P(x)=\sum_{i=1}^m\sum_{k=1}^mb_{i_k}\alpha_{i_k}x^i$. I still don't see the proof - any suggestions would be appreciated.
There is a very simple argument (as it doesn’t require separability or Galois ideas) – consider the reduction mod $P$ map from $F[X]$ to $K_{d-1}[X]$ (where $d$ is the degree of $P$). It’s $F$-linear and for dimensionality reasons it can’t be injective. Thus it has a nontrivial kernel and you are done.