# When polynomials f(x) and f'(x) are relatively prime, f(x) has no repeated roots. Why?

The problem is to show that a polynomial $f(x) \in F[x]$ (F is a field) has no repeated roots if and only if f(x) and f'(x) (the derivative of f(x)) are relatively prime. I've managed to prove one direction of this equivalence (if there are no repeated roots, f(x) and f'(x) are relatively prime), but the other one gives me headaches... can anyone help out?

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You can look at the factorization of $f$ in some algebraic closure $\overline{F}$ of $F$. –  Brandon Carter Apr 19 '11 at 21:42
Use Bezout's theorem. –  Qiaochu Yuan Apr 19 '11 at 22:05

HINT $\rm\ \ g^2\ |\ f\ \Rightarrow\ g\ |\ gcd(f,f{\:'})\$ since $\rm\ (g^2\:h)'\:=\ g\ (g\:h' + 2\:g'\:h)\:.$
Denote the repeated root by r. Then $f(x) = (x-r)g(x)$ and $f'(x) = (x-r)h(x)$, for some $g(x)$ and some $h(x)$.
But here $x-r$ doesn't have to be in $F[x]$. –  user9077 Apr 19 '11 at 22:33