$f'(x)-xf(x)=0$ has more roots than $f(x)=0$ 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?

$f(x)e^{-x^2/2}$ is zero at $\alpha_1$, and tends to zero at $-\infty$. So it must have a zero derivative somewhere in $(-\infty,\alpha_1)$.
Edited to reply to Gobi's comment  
You can use Rolle's theorem after a little work. Let us write $g(x)$ for $f(x)e^{-x^2/2}$. Take any point $t \in (-\infty,\alpha_1)$. Since $g(x)$ tends to zero at $-\infty$, there is a point $c < t$ such that $g(c) < g(t)/2$. Then by the Intermediate Value Theorem, there exist points $a \in (c,t)$ and $b \in (t,\alpha_1)$ such that $g(a) = g(b) = g(t)/2$. Now you can use Rolle's therem on $(a,b)$.
