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Roots of derivative of polynomial.

  1. Dec 23, 2012 #1
    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. jcsd
  3. Dec 23, 2012 #2

    vela

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    Start by writing down P(x) in factored form.
     
  4. Dec 23, 2012 #3

    Dick

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    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.
     
  5. Dec 26, 2012 #4
    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?
     
  6. Dec 26, 2012 #5
    Won't P'(x) have n-1 real roots (number of xn+1-xn's)?
     
  7. Dec 26, 2012 #6
    I have managed, thanks :-)
     
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