The set of all values of `x` for which
`((x+1)(x-3)^(2)(x-5)(x-4)^(3)(x-2))/(x) lt 0`

Options

A. `(-oo,-1) uu (0,2) uu (4,5)`
B. `(-1,0) uu (2,4) uu (5,oo)`
C. `(-1,0) uu (2,3) uu (4,5)`
D. `(-oo,-1) uu (0,2) uu [3,5)`

Explanation

To solve the inequality

\[ \frac{(x+1)(x-3)^{2}(x-5)(x-4)^{3}(x-2)}{x} < 0, \]

we will follow these steps:

### Step 1: Identify the critical points
The critical points occur when the numerator or denominator is zero. We find these points by setting each factor in the numerator and denominator to zero:

- \(x + 1 = 0 \Rightarrow x = -1\)
- \(x - 3 = 0 \Rightarrow x = 3\) (note that this factor has an even exponent)
- \(x - 5 = 0 \Rightarrow x = 5\)
- \(x - 4 = 0 \Rightarrow x = 4\) (note that this factor has an odd exponent)
- \(x - 2 = 0 \Rightarrow x = 2\)
- \(x = 0\) (denominator)

Thus, the critical points are:
\(-1, 0, 2, 3, 4, 5\).

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

\[ -\infty \quad -1 \quad 0 \quad 2 \quad 3 \quad 4 \quad 5 \quad +\infty \]

### Step 3: Test intervals
We will test the sign of the expression in each interval defined by the critical points. The intervals are:

1. \((- \infty, -1)\)
2. \((-1, 0)\)
3. \((0, 2)\)
4. \((2, 3)\)
5. \((3, 4)\)
6. \((4, 5)\)
7. \((5, +\infty)\)

### Step 4: Determine the sign in each interval
- **Interval \((- \infty, -1)\)**: Choose \(x = -2\)
\[ \frac{(-2+1)(-2-3)^{2}(-2-5)(-2-4)^{3}(-2-2)}{-2} < 0 \quad \text{(negative)} \]

- **Interval \((-1, 0)\)**: Choose \(x = -0.5\)
\[ \frac{(-0.5+1)(-0.5-3)^{2}(-0.5-5)(-0.5-4)^{3}(-0.5-2)}{-0.5} > 0 \quad \text{(positive)} \]

- **Interval \((0, 2)\)**: Choose \(x = 1\)
\[ \frac{(1+1)(1-3)^{2}(1-5)(1-4)^{3}(1-2)}{1} < 0 \quad \text{(negative)} \]

- **Interval \((2, 3)\)**: Choose \(x = 2.5\)
\[ \frac{(2.5+1)(2.5-3)^{2}(2.5-5)(2.5-4)^{3}(2.5-2)}{2.5} > 0 \quad \text{(positive)} \]

- **Interval \((3, 4)\)**: Choose \(x = 3.5\)
\[ \frac{(3.5+1)(3.5-3)^{2}(3.5-5)(3.5-4)^{3}(3.5-2)}{3.5} < 0 \quad \text{(negative)} \]

- **Interval \((4, 5)\)**: Choose \(x = 4.5\)
\[ \frac{(4.5+1)(4.5-3)^{2}(4.5-5)(4.5-4)^{3}(4.5-2)}{4.5} > 0 \quad \text{(positive)} \]

- **Interval \((5, +\infty)\)**: Choose \(x = 6\)
\[ \frac{(6+1)(6-3)^{2}(6-5)(6-4)^{3}(6-2)}{6} > 0 \quad \text{(positive)} \]

### Step 5: Compile the results
The intervals where the expression is negative are:
- \((- \infty, -1)\)
- \((0, 2)\)
- \((3, 4)\)

### Step 6: Write the solution
Thus, the solution to the inequality is:

\[ x \in (-\infty, -1) \cup (0, 2) \cup (3, 4) \]