Roots of derivative of polynomial.

peripatein
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Hi,

Homework Statement


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.

Homework Equations





The Attempt at a Solution


I have tried proving that using induction, but got stuck.
I'd appreciate some assistance.
 
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Start by writing down P(x) in factored form.
 
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.
 
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?
 
Won't P'(x) have n-1 real roots (number of xn+1-xn's)?
 
I have managed, thanks :-)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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