How many real roots does the polynomial $x^5 + x^4 - x^3 - x^2 - 2x - 2$ have?

$ 
\textbf{a)}\ 1
\qquad\textbf{b)}\ 2
\qquad\textbf{c)}\ 3
\qquad\textbf{d)}\ 4
\qquad\textbf{e)}\ \text{None of above}
$
1. We start with the polynomial \( P(x) = x^5 + x^4 - x^3 - x^2 - 2x - 2 \).

2. We test for possible rational roots using the Rational Root Theorem, which states that any rational root of the polynomial \( P(x) \) must be a factor of the constant term (-2) divided by a factor of the leading coefficient (1). The possible rational roots are \( \pm 1, \pm 2 \).

3. We test \( x = -1 \):
   \[
   P(-1) = (-1)^5 + (-1)^4 - (-1)^3 - (-1)^2 - 2(-1) - 2 = -1 + 1 + 1 - 1 + 2 - 2 = 0
   \]
   Since \( P(-1) = 0 \), \( x = -1 \) is a root.

4. We perform polynomial division of \( P(x) \) by \( x + 1 \):
   \[
   P(x) = (x + 1)Q(x)
   \]
   where \( Q(x) \) is the quotient. Using synthetic division:
   \[
   \begin{array}{r|rrrrrr}
     -1 & 1 & 1 & -1 & -1 & -2 & -2 \\
        &   & -1 & 2 & -1 & 2 & 0 \\
     \hline
        & 1 & 0 & 1 & -2 & 1 & 0 \\
   \end{array}
   \]
   The quotient is \( Q(x) = x^4 + x^2 - 2 \).

5. We factor \( Q(x) = x^4 + x^2 - 2 \):
   \[
   Q(x) = x^4 + x^2 - 2 = (x^2 + 2)(x^2 - 1)
   \]
   Note that \( x^2 - 1 \) can be further factored:
   \[
   x^2 - 1 = (x - 1)(x + 1)
   \]

6. Thus, the polynomial \( P(x) \) can be factored as:
   \[
   P(x) = (x + 1)(x^2 + 1)(x^2 - 2)
   \]

7. We identify the real roots of the factors:
   - \( x + 1 = 0 \) gives \( x = -1 \)
   - \( x^2 + 1 = 0 \) gives no real roots (since \( x^2 = -1 \) has no real solutions)
   - \( x^2 - 2 = 0 \) gives \( x = \sqrt{2} \) and \( x = -\sqrt{2} \)

8. Therefore, the real roots of the polynomial \( P(x) \) are \( -1, \sqrt{2}, \) and \( -\sqrt{2} \).

The final answer is \( \boxed{ 3 } \).