We are tasked with solving the inequality:

\[
x^3 - 5x^2 - 6x < 0
\]

### Step 1: Factor the expression
First, factor the left-hand side:

\[
x(x^2 - 5x - 6) < 0
\]

Next, factor the quadratic expression \( x^2 - 5x - 6 \):

\[
x(x - 6)(x + 1) < 0
\]

### Step 2: Determine critical points
The critical points occur where the expression equals zero, which are:

\[
x = 0, \quad x = 6, \quad x = -1
\]

These divide the real number line into four intervals:
\[
(-\infty, -1), \quad (-1, 0), \quad (0, 6), \quad (6, \infty)
\]

### Step 3: Test the sign in each interval
We now determine the sign of \( x(x - 6)(x + 1) \) in each of these intervals by picking a test point from each.

1. For \( x \in (-\infty, -1) \), take \( x = -2 \):
   \[
   (-2)((-2) - 6)((-2) + 1) = (-2)(-8)(-1) = -16 < 0
   \]
2. For \( x \in (-1, 0) \), take \( x = -\frac{1}{2} \):
   \[
   \left(-\frac{1}{2}\right)\left(-\frac{1}{2} - 6\right)\left(-\frac{1}{2} + 1\right) = \left(-\frac{1}{2}\right)(-6.5)(0.5) > 0
   \]
3. For \( x \in (0, 6) \), take \( x = 1 \):
   \[
   (1)(1 - 6)(1 + 1) = 1(-5)(2) = -10 < 0
   \]
4. For \( x \in (6, \infty) \), take \( x = 7 \):
   \[
   (7)(7 - 6)(7 + 1) = (7)(1)(8) = 56 > 0
   \]

### Step 4: Solution
The inequality is satisfied when the expression is negative, which occurs in the intervals:

\[
(-\infty, -1) \cup (0, 6)
\]

Since the inequality is strict (i.e., \( < 0 \) and not \( \leq 0 \)), the critical points \( x = -1 \), \( x = 0 \), and \( x = 6 \) are not included.

Thus, the solution is:

\[
x \in (-\infty, -1) \cup (0, 6)
\]

### Would you like more details or have any questions?

Here are some related questions:
1. How would the solution change if the inequality were \( \leq 0 \)?
2. How do you factor quadratic expressions systematically?
3. What is the graphical interpretation of this inequality?
4. How do you solve inequalities involving higher powers of \( x \)?
5. What if the inequality were \( x^3 - 5x^2 - 6x > 0 \)?

**Tip:** When solving inequalities, always check the sign in each interval created by the critical points.

Learn how to solve the polynomial inequality x^3 - 5x^2 - 6x < 0 step by step. This problem covers factoring polynomials, finding critical points, and determining solution intervals. Ideal for students in Grades 9-12 studying Algebra and polynomial inequalities.

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