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}