Let $f(x)$ be a polynomial with real coefficients. Show that the equation $f'(x)-xf(x)=0$ has more roots than $f(x)=0$.
I saw the hint, nevertheless I can't prove it clearly. The hint is that $f(x)e^{-x^2/2}$ has a derivative $(f'(x)-xf(x))e^{-x^2/2}$, and use the Rolle's theorem.
My outline: I think that $f'-xf$ have zeros between distinct zeros of $f$, and if $f$ has a zero of multiplicity $k$, then $f'-xf$ has the same zero with multiplicity $k-1$. But how can I show that $f'-xf$ have zeros outside of zeros of $f$, i.e. $(-\infty,\alpha_1)$ and $(\alpha_n,\infty)$ where $\alpha_1$, $\alpha_n$ are the first, last zero of $f$ respectively?