 Solution

A nonconstant polynomial is separable when it has distinct roots in a splitting field, equivalently when it is coprime to its formal derivative. For such $f$, Bézout's identity for polynomials gives $a,b\in F[T]$ with $af+b{f}^{\prime }=1$. This identity remains valid over any extension $E/F$. A repeated root $\gamma \in E$ would satisfy $f(\gamma )=f^{\prime }(\gamma )=0$, giving the contradiction $1=0$. Thus there are no repeated roots in any field extension. The argument in fact applies to every separable polynomial, whether irreducible or not.