Factor the Polynomial 16-x^8?

Question

Thanks for the help!

Answer 1

It's the difference of two squares. (4-x^4)(4+x^4). Factor 4-x^4 again. (2-x^2)(2+x^2)

Final answer is (4+x^4)(2+x^2)(2-x^2)

Answer 2

16 - x^8
= (4)^2 - (x^4)^2 >> A^2 - B^2 = (A + B)(A - B)
= (4 + x^4)(4 - x^4)
= [2^2 + 2 . 2. x^2 + (x^2)^2 - 2 . 2. x^2][2^2 - (x^2)^2]
>>(A + B)^2 = A^2 + 2 . A . B + B^2<<
= [(2 + x)^2 - (2x)^2][(2 + x^2)(2 - x^2)]
= [(2 + x^2 + 2x)(2 + x^2 - 2x)](2 + x^2)(2 - x^2)
= (2 + 2x + x^2)(2 - 2x + x^2)(2 + x^2)(2 - x^2)

You may leave it like this if you use only rational numbers
If you want to see x in its first degree, then we can factorize like the following
2 - x^2
= (sqrt2)^2 - (x)^2
= (sqrt2 + x)(sqrt2 - x)

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

Answer 3

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

if you were to take this further, you would get

(2 - x^2)(2 + x^2)(2i + x^2)(-2i + x^2)
which becomes
(2 - x^2)(2 + x^2)(2i + x^2)(-(2i - x^2))

So another way to put this is

-(2 - x^2)(2 + x^2)(2i + x^2)(2i - x^2)

thats if you are allowed to factor using imaginary values.

Answer 4

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

Answer 5

a^2-b^2= (a+b)(a-b)

4^2 - (x^4)^2

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

we can still factor further from here

(2-x²)(2+x)²(2x²-2x+2)(2x²+2x+2)

Answer 6

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