How many integers $n$ are there those satisfy the following inequality $n^4 - n^3 - 3n^2 - 3n - 17 < 0$?

A. $4$ B. $6$ C. $8$ D. $10$ E. $12$
1. We start with the inequality \( n^4 - n^3 - 3n^2 - 3n - 17 < 0 \).
2. We rewrite the inequality as \( n^4 - n^3 - 3n^2 - 3n - 18 + 1 < 0 \), which simplifies to \( n^4 - n^3 - 3n^2 - 3n - 18 < -1 \).
3. We factorize the polynomial \( n^4 - n^3 - 3n^2 - 3n - 18 \). We find that \( n = -2 \) and \( n = 3 \) are roots of the polynomial:
   \[
   P(n) = n^4 - n^3 - 3n^2 - 3n - 18
   \]
   Substituting \( n = -2 \):
   \[
   P(-2) = (-2)^4 - (-2)^3 - 3(-2)^2 - 3(-2) - 18 = 16 + 8 - 12 + 6 - 18 = 0
   \]
   Substituting \( n = 3 \):
   \[
   P(3) = 3^4 - 3^3 - 3 \cdot 3^2 - 3 \cdot 3 - 18 = 81 - 27 - 27 - 9 - 18 = 0
   \]
4. We factorize \( P(n) \) as:
   \[
   P(n) = (n - 3)(n + 2)(n^2 + 3)
   \]
   Therefore, the inequality becomes:
   \[
   (n - 3)(n + 2)(n^2 + 3) + 1 < 0
   \]
5. We analyze the sign of \( (n - 3)(n + 2)(n^2 + 3) + 1 \) for different values of \( n \):
   - For \( n > 3 \):
     \[
     (n - 3)(n + 2) > 0 \implies P(n) + 1 > 0
     \]
   - For \( n = 3 \):
     \[
     P(3) + 1 = 0 + 1 > 0
     \]
   - For \( n = 2 \):
     \[
     P(2) + 1 = (-1)(4)(7) + 1 < 0
     \]
   - For \( n = 1 \):
     \[
     P(1) + 1 = (-2)(3)(4) + 1 < 0
     \]
   - For \( n = 0 \):
     \[
     P(0) + 1 = (-3)(2)(3) + 1 < 0
     \]
   - For \( n = -1 \):
     \[
     P(-1) + 1 = (-4)(1)(4) + 1 < 0
     \]
   - For \( n = -2 \):
     \[
     P(-2) + 1 = 0 + 1 > 0
     \]
   - For \( n < -2 \):
     \[
     (n - 3)(n + 2) > 0 \implies P(n) + 1 > 0
     \]
6. From the analysis, the values of \( n \) that satisfy the inequality are \( n = 2, 1, 0, -1 \).

The final answer is \(\boxed{4}\)