# Solving Z^7+128

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.

## The Attempt at a Solution

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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.

Hurkyl
Staff Emeritus
Gold Member

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?

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

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.