# Roots of a polynomial

by ehrenfest
Tags: polynomial, roots, solved
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 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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P: 8,325
 Quote by ehrenfest I guess I will post it anyway.
P: 1,996
 Quote by cristo
OK fine delete it.

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