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Solving Z^7+128

  1. Sep 27, 2009 #1
    ive been working on the seventh root off -128 still have not got it.

    but now im trying to work out the factors off z^7+128 do i have to work off z-2 and the quadratic z^2-z-c to get the answers or do you think there is an easyer way.



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 27, 2009 #2
    Re: z^7+128

    I believe the seventh root of -128 is -2, so the factor would be (x + 2), so use synthetic division to see what you have left, then work from there.
     
  4. Sep 27, 2009 #3

    Hurkyl

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    Re: z^7+128

    Hrm. I would have suggested factoring 128 rather than just giving him that answer.

    For the O.P. could you give more context? Is there a specific form for the answers you're looking for (e.g. product of linears and quadratics that don't have real roots)? Do you know about complex numbers?
     
  5. Sep 28, 2009 #4
    Re: z^7+128

    im learning so all help is ok.thank you for your help.
     
  6. Sep 29, 2009 #5
    Re: z^7+128

    There should be only one real root to z^7 + 128. This occurs at a z of -2. The other 6 roots are complex roots that can be determined by De Moivre's Theorem, the first of which is approximately: 1.80 + 0.87i.
     
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