Solving Z^7+128: Finding Factors & Easier Ways

[morbello]

ive been working on the seventh root off -128 still have not got it.

but now I am 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

 

[DLPhysics]

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]

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?

 

[morbello]

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

 

[mrkuo]

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.