# No complex roots of a polynomial implies its derivative has no complex roots

It's a well known result that if a degree $n \geq 2$ polynomial $P(x)$ with real coefficients has $n$ distinct real roots, then its derivative $P'(x)$ will have $n-1$ distinct real roots. This is a consequence of Rolle's Theorem.

I would like to know if the following statement true:

If a degree $n \geq 2$ polynomial $P(x)$ with real coefficients has $n$ real roots (not necessarily distinct), may consist of multiplicities $>1$), then $P'(x)$ will have $n-1$ real roots (not necessarily distinct).

I've spent a while looking for a proof or a counterexample to this online but had no luck in doing so. I've also tried working out a proof for myself but ran into trouble when considering multiplicities of roots greater than one.

• Hint: a root of $P(x)$ of multiplicity $k \gt 1$ must be a root of $P'(x)$ of multiplicity $k-1$. – dxiv Nov 22 '16 at 2:02

## 1 Answer

This follows by Rolle's theorem by essentially the same argument. Let $a_1<a_2<\dots<a_k$ be the distinct roots of $P(x)$, with multiplicities $d_1,d_2,\dots,d_k$. By Rolle's theorem, $P'(x)$ has a root between $d_i$ and $d_{i+1}$ for each $i$, giving $k-1$ real roots. In addition, each $a_i$ is a root of $P'(x)$ of multiplicity $d_i-1$ (if $d_i=1$ this means that $a_i$ is not actually a root of $P'(x)$ at all). So counting with multiplicity, this gives $$k-1+\sum_{i=1}^k(d_i-1)=-1+\sum_{i=1}^kd_i$$ real roots of $P'(x)$. But $\sum_{i=1}^kd_i=n$ since this is just the number of roots of $P(x)$ (counted with multiplicity). Thus $P'(x)$ has $n-1$ real roots.