How do I factor the polynomial x^5-4x^3-8x^2+32 completely?

I'll suggest to group the 1st term with the last
one.

We'll use the
formula:

a^n + b^n = (a+b)(a^n-1 - a^n-2*b + ... +
b^n)

x^5 + 32 = x^5 + 2^5 = (x+2)(x^4 - 2x^3 + 4x^2 - 8x +
16)

We'll group the middle
terms:

-4x^3-8x^2 = -4x^2(x +
2)

We'll re-write the
polynomial:

(x+2)(x^4 + 2x^3 + 4x^2 + 8x + 16) - 4x^2(x +
2)

We'll factorize by
x+2:

(x+2)(x^4 - 2x^3 + 4x^2 - 8x + 16 -
4x^2)

(x+2)(x^4 - 2x^3 - 8x +
16)

We'll regroup x^4 - 2x^3 =
x^3(x-2)

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

(x+2)(x^3(x-2) - 8(x-2)) = (x+2)*(x-2)*(x^3 -
8)

But x^3 - 8 = (x-2)(x^2 + 2x +
4)

The polynomial will
become:

(x+2)*(x-2)*(x^3 - 8) = (x+2)*(x-2)^2*[(x^2 + 2x +
4)]

(x+2)*(x-2)*(x^3 - 8) = (x-2)*(x^2 - 4)*(x^2 + 2x +
4)

The factorized polynomial is (x-2)*(x^2 -
4)*(x^2 + 2x + 4).