What are the real zeros of the function g(x) = x3 + 2x2 − x − 2?

Upon a slight rearrangement this problem gets a lot simpler to see.

x^3-x+2x^2-2=0  now factor 1st and 2nd pair of terms...

x(x^2-1)+2(x^2-1)=0

(x+2)(x^2-1)=0  now the second factor is a "difference of square" of the form:

(a^2-b^2) which always factors to (a+b)(a-b), in this case:

(x+2)(x+1)(x-1)=0

So g(x) has three real zero when x={-2, -1, 1}