The sum of distinct real roots of the polynomial $x^5+x^4-4x^3-7x^2-7x-2$ is 

$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ -2 \qquad\textbf{(E)}\ 7$
1. **Identify the polynomial and its roots:**
   We are given the polynomial \( P(x) = x^5 + x^4 - 4x^3 - 7x^2 - 7x - 2 \).

2. **Use the properties of roots of unity:**
   We know that \( \omega \) is a cube root of unity, satisfying \( \omega^3 = 1 \) and \( \omega^2 + \omega + 1 = 0 \). This implies that \( x^2 + x + 1 \) is a factor of \( P(x) \).

3. **Factor the polynomial:**
   Since \( x^2 + x + 1 \) is a factor, we can divide \( P(x) \) by \( x^2 + x + 1 \) to find the other factor:
   \[
   P(x) = (x^2 + x + 1)Q(x)
   \]
   where \( Q(x) \) is a cubic polynomial. Performing polynomial division, we get:
   \[
   Q(x) = x^3 - 5x - 2
   \]
   Thus,
   \[
   P(x) = (x^2 + x + 1)(x^3 - 5x - 2)
   \]

4. **Find the real roots of \( Q(x) = x^3 - 5x - 2 \):**
   We need to find the roots of \( x^3 - 5x - 2 = 0 \). By inspection or using the Rational Root Theorem, we find that \( x = -2 \) is a root. Therefore, \( x + 2 \) is a factor of \( x^3 - 5x - 2 \).

5. **Factor \( Q(x) \) further:**
   Dividing \( x^3 - 5x - 2 \) by \( x + 2 \), we get:
   \[
   x^3 - 5x - 2 = (x + 2)(x^2 - 2x - 1)
   \]
   Thus,
   \[
   P(x) = (x^2 + x + 1)(x + 2)(x^2 - 2x - 1)
   \]

6. **Find the roots of \( x^2 - 2x - 1 = 0 \):**
   Solving the quadratic equation \( x^2 - 2x - 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get:
   \[
   x = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}
   \]

7. **List all distinct real roots:**
   The distinct real roots of \( P(x) \) are \( -2 \), \( 1 + \sqrt{2} \), and \( 1 - \sqrt{2} \).

8. **Sum the distinct real roots:**
   \[
   -2 + (1 + \sqrt{2}) + (1 - \sqrt{2}) = -2 + 1 + 1 = 0
   \]

The final answer is \(\boxed{0}\)