= 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 $af+bf'=1$. This identity remains valid over any extension $E/F$. A repeated root $\gamma\in E$ would satisfy $f(\gamma)=f'(\gamma)=0$, 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.
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