Let's walk through how to solve this inequality and fill out the sign chart.

Problem:

We are given the inequality: $\frac{(x - 2)(x + 3)}{-x + 4} < 0$

The goal is to determine where this rational expression is negative and positive by analyzing the sign changes across different intervals, based on critical points.

Step 1: Find the critical points

The critical points occur where:

• The numerator $(x - 2)(x + 3) = 0$

  • $x = 2$
  • $x = -3$

• The denominator $-x + 4 = 0$

  • $x = 4$

These points split the real number line into the following intervals:

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

Note: At $x = 4$, the expression is undefined since the denominator becomes 0.

Step 2: Choose test points within each interval

We’ll test the signs of the factors and the overall expression at points within each interval:

• In $(-\infty, -3)$, use $x = -4$
• In $(-3, 2)$, use $x = 0$
• In $(2, 4)$, use $x = 3$
• In $(4, \infty)$, use $x = 5$

Step 3: Analyze the signs

Let's check the sign of each factor in the expression for the chosen points.

Step 4: Identify where the expression is negative

We are interested in where the expression is less than 0.

From the table:

• The expression is negative in the intervals $(-3, 2)$ and $(4, \infty)$.

Step 5: Write the solution

Since the expression is undefined at $x = 4$, we exclude it from the solution. The final solution is:

$(-3, 2) \cup (4, \infty)$

Final Answer:

Now you can fill the sign chart accordingly:

1. For $x + 3$:

   • Negative in $(- \infty, -3)$
   • Positive in $(-3, 2)$, $(2, 4)$, and $(4, \infty)$

2. For $x - 2$:

   • Negative in $(- \infty, 2)$
   • Positive in $(2, \infty)$

3. For $-x + 4$:

   • Positive in $(- \infty, 4)$
   • Negative in $(4, \infty)$

4. For the overall expression $\frac{(x - 2)(x + 3)}{-x + 4}$:

   • Positive in $(- \infty, -3)$ and $(2, 4)$
   • Negative in $(-3, 2)$ and $(4, \infty)$

Would you like further details on how to fill the chart?
Here are five related questions to explore:

1. How do we handle rational inequalities involving multiple critical points?
2. What is the role of test points in sign charts?
3. Why is the expression undefined at $x = 4$?
4. How would the solution change if the inequality were $\geq 0$?
5. What happens if the denominator has multiple roots?

Tip: Always be mindful of where the expression is undefined when dealing with rational inequalities!