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Math Help - Descartes Rule and Fundamental Theorem

  1. #1
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    Descartes Rule and Fundamental Theorem

    Hello Hello!

    I've been doing some homework and I really don't understand a few things.

    Can anyone explain to me how the Descartes' Rule and Fundamental theorem can be used to determine the number of complex roots (in a polynomial)?
    Examples would be absolutely stellar!

    Also, does anyone know what setting a "good window" (on a graphing calculator) is?
    and what features of a polynomial should be given?

    Thank you! :]
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Deleted
    Last edited by Mathstud28; December 22nd 2008 at 08:37 PM.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Descartes rule of signs: Let p(x) be defined as before. The amount of sign changes in p(x) is the number of real, positive roots of p(x). The number of sign changes in p(-x) is the number of real, negative roots of p(x).

    So let's combine these two rules. As before let the polynomial p(x) be of degree n. Let there be m sign changes in p(x) and e sign changes in p(-x). By the FTA we must have n roots exactly, so the amount of roots left is n-m-e=i. This number i is the number of imaginary roots of p(x)
    a little comment here..

    the m stated there is the maximum number ONLY of positive real roots and not necessarily the number of positive real roots. (and e respectively..)

    why maximum and not the exact? if it is \geq 2, the polynomial may possibly have complex roots.. (complex roots always come in pairs.)

    Quote Originally Posted by Mathstud28 View Post
    ... so the amount of roots left is n-m-e=i. This number i is the number of imaginary roots of p(x)
    this statement is not necessarily true..

    counter-example: p(x)=x^4-2x^3+2x^2-2x+1 here, we have m=4 and e=0.. our n=4.. thus i=0?

    however, p(x)=(x^2+1)(x-1)^2 for which there are 2 positive real roots, no negative real roots and 2 imaginary number roots.. thus, the equation is not necessarily true..
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by kalagota View Post
    a little comment here..
    Thank you...I have deleted my post. I definitely made an error as you pointed out it should be there are AT LEAST i imaginary roots.
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