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Let's break down and solve this inequality problem step by step.

### Given Inequality:
\[
\frac{x(x+2)}{5-x} \geq 0
\]

The table you need to fill helps determine the signs of different parts of the inequality. You will examine critical points where the numerator and denominator are zero, and analyze intervals around those points.

### Steps:
1. **Identify the critical points** by setting the numerator and denominator equal to zero.

- **Numerator**: \( x(x+2) = 0 \)  
  This gives two critical points:  
  \[
  x = 0 \quad \text{or} \quad x = -2
  \]

- **Denominator**: \( 5 - x = 0 \)  
  This gives another critical point:  
  \[
  x = 5
  \]

2. **Sign Chart Intervals**:
   These critical points divide the real number line into four intervals:
   - \( (-\infty, -2) \)
   - \( (-2, 0) \)
   - \( (0, 5) \)
   - \( (5, \infty) \)

### Completing the Sign Chart:

We evaluate the signs of \( x + 2 \), \( 5 - x \), and \( x(x+2) / (5 - x) \) in these intervals.

| Interval      | Test Point | \( x + 2 \) | \( 5 - x \) | \( \frac{x(x+2)}{5-x} \) |
|---------------|------------|-------------|-------------|-------------------------|
| \( (-\infty, -2) \) | -3         | -          | +           | -                       |
| \( (-2, 0) \)     | -1         | +          | +           | +                       |
| \( (0, 5) \)      | 1          | +          | +           | +                       |
| \( (5, \infty) \) | 6          | +          | -           | -                       |

3. **Critical Points Behavior**:
   - At \( x = -2 \), the numerator is zero, so the expression is **zero**.
   - At \( x = 0 \), the numerator is also zero, so the expression is **zero**.
   - At \( x = 5 \), the denominator is zero, so the expression is **undefined** (vertical asymptote).

### Interval Notation Solution:

You are looking for where the expression is **non-negative**, which means \( \geq 0 \). The expression is positive in the intervals \( (-2, 0) \) and \( (0, 5) \), and zero at \( x = -2 \) and \( x = 0 \).

Thus, the solution in interval notation is:
\[
[-2, 0] \cup (0, 5)
\]

Would you like more details or have any further questions?

### Here are 5 related questions for you to practice:
1. How would the inequality change if the numerator was \( x(x-3) \) instead?
2. What would happen if the inequality was \( \frac{x(x+2)}{x-5} \leq 0 \)?
3. Can you explain why \( x = 5 \) is excluded from the solution set?
4. How do you determine whether to include endpoints in the interval notation?
5. How would you solve \( \frac{x^2 - 1}{x - 2} \geq 0 \)?

### Tip:
Always check the signs of both the numerator and denominator in each interval when working with rational inequalities!

Follow the steps to find all solutions for the following inequality: x(x + 2) / (5 - x) ≥ 0.

Learn how to solve the inequality x(x + 2) / (5 - x) ≥ 0 step by step, using a sign chart to determine intervals and analyze the behavior of the rational expression. This problem explores critical points, interval testing, and interval notation. Suitable for Grades 10-12.

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