Suppose a polynomial $P$ of degree $n$ has $n$ distinct real roots then $P'$ (the derivative of $P$) has $n-1$ distinct real roots.
Proof by Induction:
Base case: For $n=1$, $P_1 (x)=a_0+a_1x, a_1\neq 0$ has $1$ real (distinct) root, $x=\frac{-a_0}{a_1}$. Then $P_1 ' (x)=a_1$ has $1-1=0 $ real distinct root.
Induction step: Assume that the claim holds for some $n\in\mathbb{N}, n>1$. That is, $$P_n (x) = a_0+a_1x+a_2x^2+…+a_nx^n$$ has $n$ distinct real roots implies that $$P_n'(x)=a_1+2a_2x+3a_3x^2+…+na_nx^{n-1}$$ has $n-1$ real distinct roots.
Now, suppose that for $n+1$, $$P_{n+1} (x) = a_0+a_1x+a_2x^2+…+a_nx^n+a_{n+1}x^{n+1}$$ has $n+1$ real distinct roots. Claim is that $P_{n+1}'(x)$ has $n$ real distinct roots.
We know: $$P_{n+1}'(x)=a_1+2a_2x+3a_3x^2+…+na_nx^{n-1} +(n+1)a_{n+1} x^n$$. I know by Induction Hypothesis that $$a_1+2a_2x+3a_3x^2+…+na_nx^{n-1}$$ has $n-1$ real distinct roots.
But I cannot think how to use this fact to argue the case for $P_{n+1}'(x)$.
Best Answer
Just draw a picture of $P(x).$ You will see that between any two zeros there is a critical point (UNLESS the zero is degenerate, in which case it is a critical point itself). The way to justify the picture is Rolle's theorem as Andres says.