This question originates from Pinter's Abstract Algebra, Chapter 29, Exercise F5.
Let $F$ be a field, and $K$ a finite extension of $F$. Let $p(x)$ be irreducible in $F[x]$. If $[K : F]$ and $\operatorname{deg} p(x)$ are relatively prime, then $p(x)$ is irreducible in $K[x]$.
Let $b$ be a root of $p(x)$. Then $[K(b):F(b)][F(b):F] = [K(b):K][K:F]$, where $[F(b):F] = \operatorname{deg}p(x)$. Let $r=[F(b):F]$ and $s=[K:F]$, so $[K(b):F(b)]r = [K(b):K]s$.
Given $r$ and $s$ are relatively prime, $r$ must divide $[K(b):K]$ and therefore $r \le [K(b):K]$. Suppose the opposite that $p(x)$ is reducible in $K[x]$. This implies there exists an irreducible polynomial $q(x)\in K[x]$ with $\operatorname{deg}q(x)< r$ such that $b$ is a root. But $\operatorname{deg}q(x)=[K(b):K]<r$, which is impossible.
Is this a reasonable argument?