Solving Z^7+128

  • Thread starter morbello
  • Start date
  • #1
73
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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.



Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
11
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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.
 
  • #3
Hurkyl
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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?
 
  • #4
73
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im learning so all help is ok.thank you for your help.
 
  • #5
10
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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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