Actually let $g(x)=0$ when $f(x)=0$ otherwise $g(x)=f(x)/f'(x)$. Seems clear to me that if $x_0$ is an $n$-order root of $f(x)$ where $n$ is a positive integer, and $f(x)$ can be expressed as a Taylor series about $x_0$, then $x_0$ is a first-order root of $g(x)$. Does this also hold under les strict conditions? Perhaps when the first $n+1$ derivatives are continuous at $x_0$.
f(x) need not be a polynomial. I defined g(x) to ensure it is zero when f(x)=f'(x)=0. Notice we could have f(x)=sin(x) in which case g(x)=tan(x).