SOLUTION: Factor completely x^6 - 6x - 16 I factored it to (x^3-8)*(x^3+2) and then to(x-2)*(x^2+2x+4)*(x^3+2) but I don't know how to factor it further. Any help would be appreciated.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Factor completely x^6 - 6x - 16 I factored it to (x^3-8)*(x^3+2) and then to(x-2)*(x^2+2x+4)*(x^3+2) but I don't know how to factor it further. Any help would be appreciated.       Log On


   

Question 273924: Factor completely
x^6 - 6x - 16
I factored it to (x^3-8)*(x^3+2) and then to(x-2)*(x^2+2x+4)*(x^3+2) but I don't know how to factor it further. Any help would be appreciated.
Thank you
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
As long as the expression is
x%5E6+-+6x%5E3+-+16
and not the
x%5E6+-+6x+-+16
you posted, then you 100% correct. %28x-2%29%2A%28x%5E2%2B2x%2B4%29%2A%28x%5E3%2B2%29 is as far as your expression will factor...

unless you want to push things a bit and consider 2 to be the perfect cube of root%283%2C+2%29. Then you could use the sum of cubes pattern on the last factor: