factoring 256x^8 - 1 completely

[jennal]

Hi I am having difficulty with a math problem I would appreciate any help.

Factor completely:

256x^8 - 1

 

[wantedcriminal03]

Don't know if this is right or not, but I randomly get:

$\displaystyle 256x^8 - 1$

$\displaystyle =16^2(x^8)-1$

$\displaystyle =(16x^4)^2-1^2$

$\displaystyle y^2-x^2=(y-x)(y+x)$

$\displaystyle =(16x^4+1)(16x^4-1)$

 

[Denis]

The (16x^4-1) factors to (4x^2+1)(4x^2-1)

Then the (4x^2-1) factors to (2x+1)(2x-1)

 

[wantedcriminal03]

Oh yeh! Thanks for the finish Denis :wink:

 

[jennal]

Thank you so much for the help. We were going in the right direction just got stumped.