How do you factor completely: #x^8-9#?

2 Answers
Jul 19, 2015

Answer:

#x^8-9=(x-3^(1/4))(x+3^(1/4))(x-i3^(1/4))(x+i3^(1/4))(x-(1/sqrt(2)+i/sqrt(2))3^(1/4))(x+(1/sqrt(2)+i/sqrt(2))3^(1/4))(x-(1/sqrt(2)-i/sqrt(2))3^(1/4))(x+(1/sqrt(2)-i/sqrt(2))3^(1/4))#

Explanation:

Using the difference of squares factorisation (#a^2-b^2=(a-b)(a+b)#) you have:
#x^8-9=(x^4-3)(x^4+3)#

This is probably all they want but you can factor further allowing complex numbers:
#(x^4-3)(x^4+3)=#
#(x^2-3^(1/2))(x^2+3^(1/2))(x^2-i3^(1/2))(x^2+i3^(1/2))=#
#(x-3^(1/4))(x+3^(1/4))(x-i3^(1/4))(x+i3^(1/4))(x-(1/sqrt(2)+i/sqrt(2))3^(1/4))(x+(1/sqrt(2)+i/sqrt(2))3^(1/4))(x-(1/sqrt(2)-i/sqrt(2))3^(1/4))(x+(1/sqrt(2)-i/sqrt(2))3^(1/4))#

The 8 roots are the 8 solutions to: #x^8=9#

Jul 20, 2015

Answer:

Factor #x^8 - 9#

Explanation:

#x^8 - 9 = (x^4 - 3)(x^4 + 3) = #

= #(x^2 - sqrt3)(x^2 + sqrt3)(x^4 + 3)#

= # (x - root(4)(3))(x + root(4)(3))(x^2 + sqrt3)(x^4 + 3)#