College algebra
posted by amy on .
What is the fundamental theorem of algebra for x^664?

I would understand it to say that
x^6  64 = 0 has 6 roots
Proof:
first treat x^6 as a difference of squares
x^6  64
= (x^3  8)(x^3 + 8)
now as a difference of cubes for the first factor, and a sum of cubes for the second part
= (x2)(x^2 + 2x + 4)(x+2)(x^2  2x + 4)
so x^6  64 = 0
(x2)(x^2 + 2x + 4)(x+2)(x^2  2x + 4) = 0
x2=0 > x=2
x+2=0 > x=2
x^2+2x+4)=0 > x = (2 ± √12)/2 = 1 ± √3 i
x^2  2x + 4=0 > x = (2 ± √12)/2 = 1 ± √3 i
6 roots!