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Roots of a polynomial

  • Thread starter ehrenfest
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  • #1
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[SOLVED] roots of a polynomial

Homework Statement


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.


Homework Equations


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


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.
 

Answers and Replies

  • #2
cristo
Staff Emeritus
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  • #3
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:uhh:
OK fine delete it.
 

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