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Math Help - Random Polynomial Question

  1. #1
    Senior Member
    Joined
    Jul 2008
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    Exclamation Random Polynomial Question

    Hi

    If P(x) is monic of degree 7, if P(x) is odd, prove P(x) = ax + bx^3 + Cx^5 + x^7.
    I think we need a reasoning where the eqn. has infintely many roots so it's the zero polynomial or something?

    Could someone please help?

    Thanx a lot!
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  2. #2
    Super Member
    Joined
    May 2009
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    No, it's not a zero-polynomial.

    A monic 7th-degree polynomial would look like this:
    P(x) = x^7 + a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0.
    (You should write polynomials in standard form, ie. with exponents in descending order. I'm changing the coefficients to a_0, a_1,... a_6.)

    If P(x) is odd, then P(-x) = -P(x):
    P(-x)
    = ({\color{red}-x})^7 + a_6 ({\color{red}-x})^6 + a_5 ({\color{red}-x})^5 + a_4 ({\color{red}-x})^4 + a_3 ({\color{red}-x})^3 + a_2 ({\color{red}-x})^2 + a_1 ({\color{red}-x}) +a_0
    = {\color{red}-}x^7 + a_6 x^6 \;{\color{red}-}\; a_5 x^5 + a_4 x^4 \;{\color{red}-}\; a_3 x^3 + a_2 x^2 \;{\color{red}-}\; a_1 x +a_0

    -P(x)
    = {\color{red}-}(x^7 + a_6 x^6 + a_5 x^5 + a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 x +a_0)
    = \;{\color{red}-}\;x^7 \;{\color{red}-}\; a_6 x^6 \;{\color{red}-}\; a_5 x^5 \;{\color{red}-}\; a_4 x^4 \;{\color{red}-}\; a_3 x^3 \;{\color{red}-}\; a_2 x^2 \;{\color{red}-}\; a_1 x \;{\color{red}-}\; a_0)

    In order for P(-x) to equal -P(x),
    a_6 = a_4 = a_2 = a_0 = 0,
    so
    P(x) = x^7 + a_5 x^5 + a_3 x^3 + a_1 x.


    01
    Last edited by yeongil; August 4th 2009 at 05:10 AM.
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