Factoring Completely x^8-16?

Question

Want to make sure I know what I am doing I am going for my Finals and this is a sample of what will be on the test, Thank you.

Answer 1

Actually, the other factorizations are not complete over the rationals, let alone the reals:

(x⁸-16)
(x⁴-4)(x⁴+4)
(x²-2) (x²+2) (x²-2x+2) (x²+2x+2)

And if a factorization is required over the reals:

(x-√2) (x+√2) (x²+2) (x²-2x+2) (x²+2x+2)

Answer 2

x^8 - 16
= (x^4 + 4)(x^4 - 4)
= (x^4 + 4)(x^2 + 2) (x^2 - 2)

This is simply difference of squares over and over. Remember, x^4 times x^4 is equal to x^ (4+4) = x^8. Hope that helps!

Answer 3

The key of factoring this is the identity a^2 - b^2 = (a-b)(a+b)

You can write it as
(x^4)^2 - 4^2 = (x^4 - 4)(x^4 + 4)=
((x^2)^2 - 2^2)(x^4+4) = (x^2 - 2)(x^2 + 2)(x^4 + 4)=
(x^2 - (sqrt(2))^2)(x^2 + 2)(x^4 + 4)=
(x - sqrt(2))(x + sqrt(2))(x^2 + 2)(x^4 + 4)

The terms x^2 + 2 and x^4 + 4 can not be further decomposed (at least not in real numbers).

Answer 4

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

Answer 5

x^8-16
= (x^4 -4)(x^4 + 4)
= (x^2 -2)(x^2+2)(x^4+4)
= (x + sqrt2)(x-sqrt2)(x^2+2)(x^4+4)