Finding the correct value of k so a root is a factor of P(x)

[Potatochip911]

Homework Statement

P(x)=x^24-2kx^6+k^2

Find all values of k so that (x-(sqrt(3)/2+i/2)) is a factor of P(x)

Homework Equations

The Attempt at a Solution

Letting z=sqrt(3)/2+i/2
mod(z)=1
arg(z)=π/6

z^(24) gives e^(24π/6)i=e^(2πi)=1

z^(6) gives e^(6π/6)i=e^(π)i=-1

If (x-(sqrt(3)/2+i/2)) is to be a factor then P((sqrt(3)/2+i/2))=0
Substituting z^(24) and z^(6) into P(x) gives
0=1-2*-1*k+k^2
0=k^2+2k+1
0=(k+1)^2
k=-1

Unfortunately when I plug the new equation with the value of k=-1 into wolfram alpha it doesn't give (x-(sqrt(3)/2+i/2)) as a factor.

 

[haruspex]

Unfortunately when I plug the new equation with the value of k=-1 into wolfram alpha it doesn't give (x-(sqrt(3)/2+i/2)) as a factor.

Strange... it clearly is a factor.