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The inequality provided is:

\[
(x + 1)(x - 5)(x^2 - 9) < 0
\]

Let's break it down and solve step-by-step:

### Step 1: Factorize the quadratic term.
The term \(x^2 - 9\) is a difference of squares, and it can be factored as:

\[
x^2 - 9 = (x - 3)(x + 3)
\]

Thus, the inequality becomes:

\[
(x + 1)(x - 5)(x - 3)(x + 3) < 0
\]

### Step 2: Find the critical points.
The critical points occur where the expression equals zero, which happens when any of the factors is zero:

\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\]
\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]

The critical points are \(x = -3, -1, 3, 5\).

### Step 3: Analyze the sign of the expression in each interval.
The critical points divide the real line into the following intervals:
- \( (-\infty, -3) \)
- \( (-3, -1) \)
- \( (-1, 3) \)
- \( (3, 5) \)
- \( (5, \infty) \)

To determine where the product of the factors is negative, we test the sign of the expression in each interval by choosing a test point in each region.

1. **For \(x < -3\)** (e.g., \(x = -4\)):
   \[
   (-4 + 1)(-4 - 5)(-4 - 3)(-4 + 3) = (-3)(-9)(-7)(-1) > 0
   \]
   The product is positive.

2. **For \( -3 < x < -1 \)** (e.g., \(x = -2\)):
   \[
   (-2 + 1)(-2 - 5)(-2 - 3)(-2 + 3) = (-1)(-7)(-5)(1) < 0
   \]
   The product is negative.

3. **For \( -1 < x < 3 \)** (e.g., \(x = 0\)):
   \[
   (0 + 1)(0 - 5)(0 - 3)(0 + 3) = (1)(-5)(-3)(3) > 0
   \]
   The product is positive.

4. **For \( 3 < x < 5 \)** (e.g., \(x = 4\)):
   \[
   (4 + 1)(4 - 5)(4 - 3)(4 + 3) = (5)(-1)(1)(7) < 0
   \]
   The product is negative.

5. **For \(x > 5\)** (e.g., \(x = 6\)):
   \[
   (6 + 1)(6 - 5)(6 - 3)(6 + 3) = (7)(1)(3)(9) > 0
   \]
   The product is positive.

### Step 4: Conclusion.
We want the expression to be **less than zero**. This happens in the intervals where the product is negative, which occurs in:

\[
(-3, -1) \cup (3, 5)
\]

Thus, the solution is:

\[
\boxed{(-3, -1) \cup (3, 5)}
\]

Would you like further details or have any questions? Here are five related questions for further exploration:

1. How would the solution change if the inequality were \( \geq 0 \) instead?
2. How do we handle inequalities with repeated factors?
3. Can you solve a similar inequality with different critical points?
4. How does factoring affect solving polynomial inequalities?
5. How do interval tests ensure the correctness of inequality solutions?

**Tip:** Always analyze the sign of each factor in each interval to determine the overall sign of the expression.

Solve the following inequality: (x+1)(x-5)(x^2 - 9) < 0

Learn how to solve the inequality (x+1)(x-5)(x^2 - 9) < 0 with a detailed step-by-step breakdown. This algebraic problem focuses on polynomial factorization and sign analysis, essential for students in Grades 9-11.

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