calculus - 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.

Answer

This follows by Rolle's theorem by essentially the same argument. Let $a_1