factor completely x to the eighth power - 1?

Answer 1

x^8-1?

Well... It should already been factor out... U cant factor furthermore.

Answer 2

x^8 - 1 =( x^4+1)(x^2+1)(x+1)(x- 1)

Here's how it works: (Remember that 1*1=1; also 1/1=1). So factor x^8 -l and you get (x^4+1)(x^4-1).
You cannot factor x^4+1 (try it!), but you can factor x^4-1; you get (x^2+1)(x^2-1). You cannot factor x^2+1, but you can factor x^2-1 to get (x+1)(x-1).

Looking at it from the end backwards:
(x+1)(x-1) = x^2-1
(x^2+1)(x^2-1) = x^4-1
(x^4+1)(x^4-1) = x^8-1

(it is impossible to factor x^2+1) If you do, you get a nonsense answer: (x+1)(x+1). Try to recombine the terms and you get x^2 + 2x + 1. But you wanted x^2+1....

on the other hand when you combine (x+1)(x-1), you get x^2 + 1x - 1x - 1. Combine like terms to get your final answer: x^2 - 1.

good luck

Answer 3

x^8 - 1 = (x^4 - 1)(x^4 + 1)

= (x^2 - 1)(x^2 + 1)(x^4 + 1)

= (x -1)(x+1)(x^2 + 1)(x^4 + 1)

Answer 4

(x -1)(x+1)(x^2 + 1)(x^4 + 1)

Answer 5

x^8 - 1
(x^4 - 1)(x^4 + 1)
(x^2 + 1)(x^2 - 1)(x^4 + 1)
(x - 1)(x + 1)(x^2 + 1)(x^4 + 1)

Answer 6

(x^4+1)(x^2+1)(x+1)(x-1)

Answer 7

Over the integers, rational numbers or real numbers:

(x^4 + 1) (x^2 + 1) (x+1) (x-1)

Over the algebraic numbers or the complex numbers:

(x-1)(x+1)(x-i)(x+i) (x-(1+i)/sqrt(2)) (x-(1-i)/sqrt(2)) (x-(-1+i)/sqrt(2)) (x-(-1-i)/sqrt(2))

It completely factors over these two fields.

Over Z2, the field with two elements 0 and 1:

(x-1)^8

So it depends in which field you are doing your factorization in.

Answer 8

x^8 - 1


This is the difference of two squares, which you can right as:

(x^4 - 1) (x^4 + 1), [foil it out and see, it comes out to (x^8 - 1) ]

But, (x^4 - 1) is also the diff. of squares, so:

(x^4 - 1) = (x^2 - 1)(x^2 + 1)

And,

(x^2 - 1) = (x - 1)(x + 1)

So, altogether,
(x^8 - 1) = (x^4 + 1)(x^2 + 1)(x + 1)(x - 1)

Answer 9

Just ignore the first two answers because they must have forgotten how to factor the difference of perfect squares, then everyone else is steering you totally in the right direction

Answer 10

x^(even number) - 1 is the difference of squares, and x^(2n) -1 = (x^n + 1)(x^n - 1)