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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.

| Interval      | \(x + 3\) | \(x - 2\) | \(-x + 4\) | \(\frac{(x - 2)(x + 3)}{-x + 4}\) |
|---------------|-----------|-----------|-----------|-----------------------------------|
| \((- \infty, -3)\) | \(-\)     | \(-\)     | \(+\)     | \(+\)                             |
| \((-3, 2)\)        | \(+\)     | \(-\)     | \(+\)     | \(-\)                             |
| \((2, 4)\)         | \(+\)     | \(+\)     | \(+\)     | \(+\)                             |
| \((4, \infty)\)    | \(+\)     | \(+\)     | \(-\)     | \(-\)                             |

---

### 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!

Solve (x - 2)(x + 3) / (-x + 4) < 0 by filling in the sign chart with a - where you want to indicate negative and a + where you want to indicate positive. Use -inf for −∞ and inf for ∞.

Learn how to solve the rational inequality (x - 2)(x + 3) / (-x + 4) < 0 using sign analysis. This involves determining critical points, dividing the number line into intervals, and testing signs of the function in each interval. Suitable for students in Grades 10-12.

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