Proving nature of roots of polynomial.

Question

Suppose $P(x)$ is an $n$th degree polynomial with real coefficients. Prove that all the roots of $P'(x)$ are real if all the roots of $P(x)$ are real.

Here's my attempt

Questions:

I'm not sure how to prove that for a given polynomial (see link for context), its turning points are real (with or without conditions -be it all or some of its coefficients/ roots are real/ complex)

Is it possible for a turning point to be complex? If so give as example/ proof (or what its conditions are for it to happen or why it cannot happen).

I'm very new to proofs, is the "proof" I've given in any way acceptable?

Answer 1

What you are asking for is a direct consequence of Rolle's Theorem. Basically Rolle's Theorem states that if a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a) = f(b)$, then $\exists c \in (a,b)$ such that $f'(c) = 0$. If you apply Rolle's Theorem to the desired polynomial between any two real roots of $f$, (a polynomial is continuous and differentiable), you will get what you need directly.

For more information on Rolle's Theorem

As @DonAntonio and @Omnomnomnom have pointed out, for repeated roots, the zero will be a root of the polynomial and it's derivatives as well so this can be considered separately.

Answer 2

If you have a polynomial $p(x)$ of degree $n$, and you are given that $p(x)$ has $n$ roots, all of which are real then between any pair of consecutive roots (according to your graph, roots which are one after the other) must have a local extremum in between the roots. And each of these extrema trace back to a particular value of $x$ which will be roots of $p'(x)$. Since there are $n$ roots of $p(x)$, there must be $(n-1)$ extrema and hence $(n-1)$ real roots of $p'(x)$. Since $p'(x)$ will be of the degree $(n-1)$, all of its roots are real.