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 $P_{n+1}'(x)$.

  • 4
    $\begingroup$ Use Rolle's theorem. $\endgroup$ – Andrés E. Caicedo Dec 14 '13 at 17:09
  • $\begingroup$ Thanks Andreas. Here is my idea: Write $P_{n+1}$ as $$P_{n+1}(x) = (x-r_1)(x-r_2)...(x-r_n)(x-r_{n+1})$$ where $r_i$ are distinct real numbers. Then by Rolle's theorem, I will get $p_i \in (r_i,r_{i+1})$ s.t. $P_{n+1}'(p_i)=0$. In total I will get $n$ such $p_i$, all distinct. Thus, $P_{n+1}'(x)$ has n distinct real roots. Is this fine? $\endgroup$ – ugstudent1243 Dec 14 '13 at 17:31
  • $\begingroup$ @ugstudent1243 just a question. How do you apply the induction hypothesis? To deduce that $P_{n+1}''$ has $n-1$ distinct real roots you have to know that $P_{n+1}'$ has $n$ distinct real roots. $\endgroup$ – P.. Dec 14 '13 at 17:43
  • $\begingroup$ I guess, I don't need induction-proof after I used Rolle's theorem. $\endgroup$ – ugstudent1243 Dec 14 '13 at 17:52
  • $\begingroup$ Is my idea correct? I need your confirmation before I can be happy. :D $\endgroup$ – ugstudent1243 Dec 14 '13 at 18:47

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.


Between any two roots of $p(x)$ there'll be a root of $p'(x)$ by Rolle's Theorem , so $p'(x)$ has at least $n-1$ roots and they are distinct , but since $p'(x)$ has degree $n-1$ thus these are only $n-1$ distinct roots of $p'(x)$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.