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math>philosophy

  • 3 years ago

factorize x^5 - x + 1 = 0 i'm not sure how to factor beyond the 3rd/4th degree. how do i do this one?

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  1. anonymous
    • 3 years ago
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    fairly sure this will not factor over the rationals, since a quick graph shows it has only one real zero

  2. math>philosophy
    • 3 years ago
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    but how does a "quick graph [that] shows it has one zero" show anything? y=x^5 also has one zero as well. not sure what the point is

  3. anonymous
    • 3 years ago
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    \(y=x^5\) is factored

  4. anonymous
    • 3 years ago
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    if you can find the zeros then you can factor. i am assuming you mean factor over the rational numbers, although maybe i am mistaken about that

  5. amistre64
    • 3 years ago
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    i dont spose theres a simple easy way to "complete a quintic"?

  6. anonymous
    • 3 years ago
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    not according to galois

  7. amistre64
    • 3 years ago
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    you can try newtons method of trial and error to get closer and closer to a real root

  8. asnaseer
    • 3 years ago
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    hmmm - do you really want to factorise this or are you looking for the solutions? if its the solutions you want then use the method suggested by @amistre64

  9. math>philosophy
    • 3 years ago
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    factor

  10. asnaseer
    • 3 years ago
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    the only factorisation I can think of here is as follows:\[\begin{align} x^5-x+1&=0\\x(x^4-1)+1&=0\\ x(x^2-1)(x^2+1)+1&=0\\ x(x-1)(x+1)(x^2+1)+1&=0\\ \end{align}\]but I don't think that helps towards getting to the solution. :(

  11. cinar
    • 3 years ago
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    since f(-1)>0 and f(-2)<0 one root is between -1 and -2 by the Bolzano theorem..

  12. cinar
    • 3 years ago
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    since it is odd degree poly. we can say that it has at least one real root.. but I do not know how to say it has only one real root without looking its graph and I also wanna learn it..

  13. cinar
    • 3 years ago
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    wow man look at the roots!! http://www.wolframalpha.com/input/?i=factor+x^5-x%2B1%3D0 http://www.wolframalpha.com/input/?i=factor+x^5-x%2B1

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