If $F$ is a field with $\operatorname{char} F = 0$ and $f(x) \in F[x]$ irreducible, all zeros of $f(x)$ in any extension has multiplicity $s = 1$. What I have trying is:
Suppose that $f(x)$ has at least one zero $\alpha$ such that $f(x) = (x - \alpha)^sq(x)$, $s > 1$ in some extension. Then I guess that $(x-\alpha)^{s-1} \mid f(x)$.
So, $f(x)$ is not irreducible, where $f(x) = (x-a)^{s-1}h(x)$.
But is seems wrong once I neither used the hypothesis $\operatorname{char} F = 0$.
What I am loosing? Could someone help me?
Thanks a lot.
 A: Outline: We can assume that $f$ has degree $n\gt 1$. Then $f'(x)$ has degree $n-1$. (This is where we use characteristic $0$. In characteristic $p$, this part can fail. For example, the derivative of $x^p+1$ is the $0$ polynomial.) 
Since $f$ is irreducible, $f(x)$ and $f'(x)$ are relatively prime over $F$. By the Bezout Identity, they are relatively prime over any extension field $K$ of $f$. But then $f(x)$ cannot have a root of multiplicity $\gt 1$ over $K$, since if $(x-a)^2$ divides $f(x)$ over $K$, then $f(x)$ and $f'(x)$ are not relatively prime over $K$: each is divisible by $x-a$.
A: This isn't quite right, first of all $(x-\alpha)^{s-1}$ might not be a polynomial with coefficients in $F$ when $\alpha\not\in F$. However, you do know in characteristic $0$ that the derivative of a non-constant polynomial is not $0$. If
$$f(x)=(x-\alpha)^s\prod_{i=1}^n (x-\alpha_i)^{e_j}, s>1, e_j\ge 1$$
then we have that $\gcd(f'(x),f(x))$--both of which have coefficients in $F$--cannot be $1$, since this same relationship would hold in an extension where clearly $(x-\alpha)\big|\gcd(f(x),f'(x))$, i.e. $\alpha$ is a root of this gcd, which is a contradiction.
