To solve the cubic equation \( x^3 + 2x^2 - x - 2 = 0 \), we can begin by using the Rational Root Theorem to look for possible rational roots, which are factors of the constant term (-2) divided by factors of the leading coefficient (1). This gives us potential rational roots of ±1, ±2.

Let's test these possible roots:

1. **Testing \( x = 1 \):**
\[
f(1) = 1^3 + 2(1^2) - 1 - 2 = 1 + 2 - 1 - 2 = 0
\]
Thus, \( x = 1 \) is a root.

Now that we have found a root, we can use synthetic division to factor the polynomial by \( x - 1 \).

2. **Synthetic Division of \( x^3 + 2x^2 - x - 2 \) by \( x - 1 \):**
\[
\begin{array}{r|rrrr}
1 & 1 & 2 & -1 & -2 \\
& & 1 & 3 & 2 \\
\hline
& 1 & 3 & 2 & 0 \\
\end{array}
\]
This gives us a quotient of \( x^2 + 3x + 2 \).

3. **Factoring \( x^2 + 3x + 2 \):**
We can factor this quadratic polynomial:
\[
x^2 + 3x + 2 = (x + 1)(x + 2)
\]

4. **Putting it all together:**
The original polynomial can now be expressed as:
\[
x^3 + 2x^2 - x - 2 = (x - 1)(x + 1)(x + 2) = 0
\]

5. **Finding the roots:**
Setting each factor equal to zero gives us the roots:
\[
x - 1 = 0 \Rightarrow x = 1
\]
\[
x + 1 = 0 \Rightarrow x = -1
\]
\[
x + 2 = 0 \Rightarrow x = -2
\]

Thus, the solutions to the equation \( x^3 + 2x^2 - x - 2 = 0 \) are:
\[
\boxed{1, -1, -2}
\]