Solve `((x-5)^(2)(x+2)^(3)(x-4))/((x-3)^(4))le0`

Answer

To solve the inequality

\[ \frac{(x-5)^{2}(x+2)^{3}(x-4)}{(x-3)^{4}} \leq 0, \]

we will follow these steps:

### Step 1: Identify the critical points
The critical points occur when the numerator is zero or the denominator is zero.

- The numerator \((x-5)^{2}(x+2)^{3}(x-4) = 0\) gives us the points:
- \(x = 5\) (double root, even power)
- \(x = -2\) (triple root, odd power)
- \(x = 4\) (single root, odd power)

- The denominator \((x-3)^{4} = 0\) gives us the point:
- \(x = 3\) (even power, not defined)

Thus, the critical points are \(x = -2, 3, 4, 5\).

### Step 2: Plot the critical points on a number line
We will mark the critical points on a number line:

```
---|---|---|---|---|---|---|---|---|---
-2 3 4 5
```

### Step 3: Determine the sign of the expression in each interval
We will analyze the sign of the expression in the intervals formed by these critical points:

1. **Interval \((-∞, -2)\)**:
- Choose \(x = -3\):
\[ \frac{(-3-5)^{2}(-3+2)^{3}(-3-4)}{(-3-3)^{4}} = \frac{(8)^{2}(-1)^{3}(-7)}{(6)^{4}} > 0 \]

2. **Interval \((-2, 3)\)**:
- Choose \(x = 0\):
\[ \frac{(0-5)^{2}(0+2)^{3}(0-4)}{(0-3)^{4}} = \frac{(5)^{2}(2)^{3}(-4)}{(3)^{4}} < 0 \]

3. **Interval \((3, 4)\)**:
- Choose \(x = 3.5\):
\[ \frac{(3.5-5)^{2}(3.5+2)^{3}(3.5-4)}{(3.5-3)^{4}} = \frac{(1.5)^{2}(5.5)^{3}(-0.5)}{(0.5)^{4}} < 0 \]

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

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

### Step 4: Compile the results
The signs in the intervals are as follows:
- \((-∞, -2)\): Positive
- \((-2, 3)\): Negative
- \((3, 4)\): Negative
- \((4, 5)\): Positive
- \((5, ∞)\): Positive

### Step 5: Include critical points
- At \(x = -2\), the expression is \(0\) (included).
- At \(x = 3\), the expression is undefined (not included).
- At \(x = 4\), the expression is \(0\) (included).
- At \(x = 5\), the expression is \(0\) (included).

### Final Solution
The solution to the inequality is:
\[ x \in [-2, 3) \cup [4, 5]. \]