- #1

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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.

- Thread starter morbello
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- #1

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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

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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

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im learning so all help is ok.thank you for your help.

- #5

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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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