$\gcd(f,f')=1$ Does this imply that f has not multiply irreducible factors in $\mathbb{C}[x]$?

Question

I want to find out if this affermation is true: let $f\in \mathbb{Q}[x]$ such that $\gcd(f,f')=1$ Does this imply that f has not multiply irreducible factors in $\mathbb{C}[x]$? (We know that it has not multiply roots in $\mathbb{Q}[x]$) Can anyone help me? Thank you in advance.

Answer 1

Hints: The only monic irreducible polynomials in $\mathbb{C}[x]$ are of the form $x-\lambda.$ The proof given for $\mathbb{Q}[x]$ should work for $\mathbb{C}[x],$ but if you are unsure, consider $f^{\prime}(x)$ when $f(x) = (x-\lambda)^{2}g(x)$ for some polynomial $g(x).$