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Roots of a polynomial

by ehrenfest
Tags: polynomial, roots, solved
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ehrenfest
#1
May16-08, 05:42 PM
P: 1,996
1. The problem statement, all variables and given/known data
Let P(x) be a polynomial of odd degree with real coefficients. Show that the equation P(P(x))=0 has at least as many real roots as the equation P(x) = 0, counted without multiplicities.


2. Relevant equations
By the FTC, P(x) and P(P(x)) factor into complex linear factors.


3. The attempt at a solution
Please just give me hint.

By the odd degree, we know that both P(x) and P(P(x)) have at least one real root.

By the FTC, P(x) and P(P(x)) factor into complex linear factors.

Oh wait, let \alpha_1,...,\alpha_m be the roots of P(x)=0. Because P(x) has odd degree, we know that p(R) = R. So, we can find distinct \beta_1,...,\beta_n such that P(\beta_i) = \alpha_i. That was easy. I guess I will post it anyway.
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cristo
#2
May16-08, 05:45 PM
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Quote Quote by ehrenfest View Post
I guess I will post it anyway.
ehrenfest
#3
May16-08, 06:06 PM
P: 1,996
Quote Quote by cristo View Post
OK fine delete it.


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