Factors of polynomial not passing the Bezout's identity test

Question

When factoring $x^3 - 2x^2 - 4x - 8$ the result you get is $(x-2)(x^2 - 4)$ or $(x-2)^2 (x+2)$ , meaning that the mentioned polynomial is divisible by each of these factors. When using the Bezout's identity regarding factoring polynomials backwards and plugging values ±2 into the polynomial, the result is never zero. Then how is it possible for it to be divisible with both $(x-2)$ and $(x+2)$?

Thanks in advance.

Answer 1

I made a mistake while pulling out factors. The answer is: Bezout's identity works only when a polynomial has real number roots.

Answer 2

There is actually a remainder if you divide your polynomial by $X-2$: $$X^3-2X^2-4X-8=(X-2)(X^2-4)-16$$

Answer 3

The problem is that the polynomial does not actually factor in that way. We have

$$(x-2)(x^2-4)=x^3-4x-2x^2+8=x^3-2x^2-4x\color{red}{+}8$$

Note that this is different because in this one we have $+8$ instead of $-8$, so it was not an issue with Bezout's identity, but rather with factoring.