Question 47148
One way to do this is to factorise this into linear factors and count the times -1 appear. However, while that is the ultimate aim, the question tempts you to take out the factors one at a time.


First, to show that -1 is a root, you plug -1 into the equation.
So if {{{f(x) = x^3 + 2x^2 - x - 2 }}}
Then f(-1) = -1 + 2 + 1 - 2 = 0


This means that -1 is a root of f(x) at least once. This is an application of the remainder theorem. Anyway, knowing this, we can divide the polynomial f(x) by (x -(-1))=(x+1), knowing that there is no remainder. 

{{{f(x) = x^3 + 2x^2 - x - 2 = (x+1)(x^2 + x - 2)}}}

At this point, one can directly factorise the quadratic, or apply remainder theorem again. If you apply the remainder theorem to the quadratic with x = -1, you get (-1)^2 -1 - 2 = -2. So, no, -1 is not a root again. In the end, to find the other roots, we have to factorise the quadratic ourselves to get

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

So the roots are 1, -1, -2