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Math Help - polynomial

  1. #1
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    polynomial

    How can I prove that you can not solve the polynomial : x^8-x^7+x^5-x^4+x^3-x+1
    Over Q
    (No use by Eisenstein theorem , without using polynomial Cyaklotomi))
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  2. #2
    Senior Member eumyang's Avatar
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    Hmm, are you the OP in this thread from the Math Is Fun forum?
    http://www.mathisfunforum.com/viewtopic.php?pid=146808
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Rational Root Theorem. This tells you there are no linear factors.
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Maybe by using calculus...
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
    Rational Root Theorem. This tells you there are no linear factors.
    Thanks for the link... The answer follows almost immediately, leaving only one case to check...
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by melese View Post
    Thanks for the link... The answer follows almost immediately, leaving only one case to check...
    One case? Aren't there three? A quadratic, cubic, or quartic factor? I attempted to solve these with long division, but it got messy real quick.
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  7. #7
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    Quote Originally Posted by chiph588@ View Post
    One case? Aren't there three? A quadratic, cubic, or quartic factor? I attempted to solve these with long division, but it got messy real quick.
    Here's what I meant.

    According to the Rational Root Theorem:
    Let P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x^1+a_0=0 be a polynomial equation with integer coefficients. If P(x) has a rational root p/q , with p, q relatively prime integers, then p|a_0 and q|a_n . (Statement found in PlanetMath: rational root theorem)

    In our case, P(x)=x^8-x^7+x^5-x^4+x^3-x+1=0 and a_n=a_0=1 . Assume there is a rational root p/q , then p|a_0=1 and q|a_n=1. Now, p=q=\pm 1, and so p/q=\pm 1. (Two case after all!)

    Checking the case we have P(1)=1 and P(-1)=1 , contradicting P(x)=0 .
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