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The inequality to solve is:

\[
(x - 2)^2 (x + 8) < 0
\]

### Step 1: Analyze the structure of the inequality

- \((x - 2)^2\) is a squared term, which means it is always non-negative (i.e., \((x - 2)^2 \geq 0\) for all \(x\)).
- \((x + 8)\) is a linear term.

We are looking for where the product of these two terms is less than zero.

### Step 2: Determine when the expression can be negative

For the inequality \((x - 2)^2 (x + 8) < 0\) to hold, the sign of the expression depends on \( (x + 8) \) because \((x - 2)^2 \geq 0\) for all \(x\). So, the only way for the product to be less than 0 is for \(x + 8\) to be negative while \((x - 2)^2\) is positive.

Thus, \((x - 2)^2 \neq 0\) and \(x + 8 < 0\).

### Step 3: Solve for \(x\)

1. \((x - 2)^2 \neq 0\) implies \(x \neq 2\).
2. \(x + 8 < 0\) implies \(x < -8\).

### Step 4: Write the solution in interval notation

The inequality holds when \(x < -8\) and \(x \neq 2\), but since \(x = 2\) is outside this range (it's greater than \(-8\)), it doesn't affect the solution.

Thus, the solution in interval notation is:

\[
(-\infty, -8)
\]

Would you like any further clarification or more details?

Here are 5 related questions for deeper understanding:
1. How do we determine the sign of quadratic terms in inequalities?
2. Why can't \((x - 2)^2\) be negative?
3. How would the solution change if the inequality were non-strict (i.e., \(\leq 0\))?
4. What role does the \(x = 2\) play in this problem, and why does it not affect the interval?
5. Can you apply this same reasoning to inequalities with cubic or higher-degree terms?

**Tip:** Always pay attention to the role of squared terms in inequalities; they are non-negative and greatly affect the overall sign of the expression.

Learn how to solve the quadratic inequality (x - 2)^2 (x + 8) < 0 step-by-step. This problem involves key concepts of algebra and inequalities, suitable for students in Grades 9-11.

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