I found the following question here.
Let $K $ be an extension field of $k $. Suppose $f(x) \in k[x]$ is irreducible with degree $n$ and let $[K:k]=m$, where $(n,m)=1$. Show that $f$ is irreducible in $K[x]$.
For the sake of argument, assume that $f$ is reducible in $K[x]$. Then there exists an $\alpha\in K$ such that $f(\alpha)=0$. Note that $$k\subseteq k (\alpha)\subseteq K .$$ By Tower rule, we have $$[K:k]=[K:k(\alpha)][k (\alpha):k]. $$
In other words, $m=[K:k(\alpha)]\cdot n$. But this contradicts $(n,m)=1$. Hence, $f$ is irreducible over $k $.
But, if $f$ is reducible in $K[x]$, it only means that there exists some irreducible polynomial $p(x)\in K[x]$ such that $p(x)\mid f(x)$. So I don't think my claim
Then there exists an $\alpha\in K$ such that $f(\alpha)=0$
is correct. Also note that the claim will work if $K$ is algebraically closed. As I mentioned above, I don't have full information regarding the problem.
So do I need any further information to complete the solution? If not, is there any possibility of improving this solution?
Hints and alternate solutions are appreciated.