We are given the inequality:

$\frac{x^2 + 3x + 2}{x^2 - 2x - 3} > 0$

Let's go through the steps systematically.

Step 1: Identify the critical values

First, we need to factor both the numerator and denominator to identify the critical points (where the expression is equal to 0 or undefined).

Factor the numerator:

$x^2 + 3x + 2 = (x + 1)(x + 2)$

Factor the denominator:

$x^2 - 2x - 3 = (x - 3)(x + 1)$

Now the inequality becomes: $\frac{(x + 1)(x + 2)}{(x - 3)(x + 1)} > 0$

Step 2: Simplify the expression

Notice that $(x + 1)$ is common in both the numerator and the denominator, so we cancel it out. However, we must remember that $x = -1$ is a critical value because the expression would be undefined at this point.

This simplifies the inequality to: $\frac{x + 2}{x - 3} > 0$

Step 3: Critical points and test intervals

The critical points are $x = -2$ (where the numerator is zero), $x = 3$ (where the denominator is zero), and $x = -1$ (where the original expression is undefined).

Thus, the critical points are: $x = -2, -1, 3$

These divide the number line into the following intervals:

1. $(-\infty, -2)$
2. $(-2, -1)$
3. $(-1, 3)$
4. $(3, \infty)$

Step 4: Test points in each interval

Now, we choose a test point from each interval to determine the sign of the expression.

Interval $(-\infty, -2)$:

Choose $x = -3$: $\frac{-3 + 2}{-3 - 3} = \frac{-1}{-6} = \frac{1}{6} > 0$ The expression is positive in this interval.

Interval $(-2, -1)$:

Choose $x = -1.5$: $\frac{-1.5 + 2}{-1.5 - 3} = \frac{0.5}{-4.5} = \frac{-1}{9} < 0$ The expression is negative in this interval.

Interval $(-1, 3)$:

Choose $x = 0$: $\frac{0 + 2}{0 - 3} = \frac{2}{-3} = -\frac{2}{3} < 0$ The expression is negative in this interval.

Interval $(3, \infty)$:

Choose $x = 4$: $\frac{4 + 2}{4 - 3} = \frac{6}{1} = 6 > 0$ The expression is positive in this interval.

Step 5: Analyze the inequality

We need the expression to be positive, so we look for intervals where the expression is greater than 0. Based on our test points, the inequality is satisfied in the intervals: $(-\infty, -2) \cup (3, \infty)$

Conclusion:

The solution to the inequality is: $x \in (-\infty, -2) \cup (3, \infty)$

Do you have any questions, or would you like more details?

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Here are 5 related questions you might consider:

1. How would the solution change if the inequality were $\geq 0$?
2. How does the behavior of rational functions change around vertical asymptotes?
3. Can you solve a similar inequality where the numerator has a higher degree than the denominator?
4. What if both the numerator and denominator were cubic polynomials?
5. How would you solve this inequality graphically?

Tip: Always check for undefined points in rational inequalities, as they can affect the solution set.