Find the solution set of `((x-1)(x-2)^(2)(x+4))/((x+2)(x-3)) ge 0`

Answer

To solve the inequality \(\frac{(x-1)(x-2)^{2}(x+4)}{(x+2)(x-3)} \geq 0\), we will follow these steps:

### Step 1: Identify the critical points
The critical points are the values of \(x\) where the expression is either zero or undefined. We find these by setting the numerator and denominator to zero.

- **Numerator**: \((x-1)(x-2)^{2}(x+4) = 0\)
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x - 2 = 0 \Rightarrow x = 2\) (with multiplicity 2)
- \(x + 4 = 0 \Rightarrow x = -4\)

- **Denominator**: \((x+2)(x-3) = 0\)
- \(x + 2 = 0 \Rightarrow x = -2\)
- \(x - 3 = 0 \Rightarrow x = 3\)

Thus, the critical points are \(x = -4\), \(x = -2\), \(x = 1\), \(x = 2\), and \(x = 3\).

### Step 2: Create a number line
We will plot the critical points on a number line:

```
---|---|---|---|---|---|---|---|---|---|---
-4 -2 1 2 3
```

### Step 3: Test intervals
We will test the sign of the expression in each interval created by the critical points:

1. **Interval \((-∞, -4)\)**: Choose \(x = -5\)
\[ \frac{(-5-1)(-5-2)^{2}(-5+4)}{(-5+2)(-5-3)} = \frac{(-6)(49)(-1)}{(-3)(-8)} > 0 \]

2. **Interval \((-4, -2)\)**: Choose \(x = -3\)
\[ \frac{(-3-1)(-3-2)^{2}(-3+4)}{(-3+2)(-3-3)} = \frac{(-4)(25)(1)}{(-1)(-6)} < 0 \]

3. **Interval \((-2, 1)\)**: Choose \(x = 0\)
\[ \frac{(0-1)(0-2)^{2}(0+4)}{(0+2)(0-3)} = \frac{(-1)(4)(4)}{(2)(-3)} > 0 \]

4. **Interval \((1, 2)\)**: Choose \(x = 1.5\)
\[ \frac{(1.5-1)(1.5-2)^{2}(1.5+4)}{(1.5+2)(1.5-3)} = \frac{(0.5)(0.25)(5.5)}{(3.5)(-1.5)} < 0 \]

5. **Interval \((2, 3)\)**: Choose \(x = 2.5\)
\[ \frac{(2.5-1)(2.5-2)^{2}(2.5+4)}{(2.5+2)(2.5-3)} = \frac{(1.5)(0.25)(6.5)}{(4.5)(-0.5)} < 0 \]

6. **Interval \((3, ∞)\)**: Choose \(x = 4\)
\[ \frac{(4-1)(4-2)^{2}(4+4)}{(4+2)(4-3)} = \frac{(3)(4)(8)}{(6)(1)} > 0 \]

### Step 4: Determine the solution set
From the tests, we find:
- The expression is non-negative in the intervals \((-∞, -4)\), \((-2, 1)\), and \((3, ∞)\).
- At the critical points:
- \(x = -4\): \(0\) (included)
- \(x = -2\): undefined (not included)
- \(x = 1\): \(0\) (included)
- \(x = 2\): \(0\) (included)
- \(x = 3\): undefined (not included)

### Final Solution Set
Combining these results, the solution set is:
\[ (-\infty, -4] \cup (-2, 1] \cup [2, 3) \cup (3, \infty) \]