# Roots of derivative of polynomial.

1. Dec 23, 2012

### peripatein

Hi,
1. The problem statement, all variables and given/known data
I am asked to prove that given all roots of a polynomial P of order n>=2 are real, then all the roots of its derivative P' are necessarily real too.
I am permitted to assume that a polynomial of order n cannot have more than n real roots.

2. Relevant equations

3. The attempt at a solution
I have tried proving that using induction, but got stuck.
I'd appreciate some assistance.

2. Dec 23, 2012

### vela

Staff Emeritus
Start by writing down P(x) in factored form.

3. Dec 23, 2012

### Dick

Think about two consecutive roots of P(x), say x1<x2. What can you say about P'(x) on the interval [x1,x2]? Try and find a theorem that applies.

4. Dec 26, 2012

### peripatein

Based on Rolle's Mean Value Theorem, if P(x1)=P(x2)(=0, in this case) then P'(x3) = 0. I've tried using that but was unable to complete my proof by induction of the above mentioned statement.
Would you kindly assist?

5. Dec 26, 2012

### peripatein

Won't P'(x) have n-1 real roots (number of xn+1-xn's)?

6. Dec 26, 2012

### peripatein

I have managed, thanks :-)