The complete set of values of `x` for which
`(x^(3)(x-1)^(2)(x+4))/((x+1)(x-3)) ge 0` is

Options

A. `(-oo,-4] uu (-1,0] uu (3,oo) uu {1}`
B. `(-oo,-4] uu (-1,0] uu (3,oo) uu {-1}`
C. `[-4,-1) uu [0,1) uu (3,oo) `
D. `[-4,-1) uu [0,1] uu (3,oo) `

Explanation

To solve the inequality

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

we will follow these steps:

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

- **Numerator**: \(x^3 (x-1)^2 (x+4) = 0\) gives:
- \(x^3 = 0 \Rightarrow x = 0\)
- \((x-1)^2 = 0 \Rightarrow x = 1\)
- \(x+4 = 0 \Rightarrow x = -4\)

- **Denominator**: \((x+1)(x-3) = 0\) gives:
- \(x+1 = 0 \Rightarrow x = -1\)
- \(x-3 = 0 \Rightarrow x = 3\)

So, the critical points are \(x = -4, -1, 0, 1, 3\).

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

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

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

1. **Interval \((- \infty, -4)\)**: Choose \(x = -5\)
\[ \frac{(-5)^3 (-5-1)^2 (-5+4)}{(-5+1)(-5-3)} = \frac{-125 \cdot 36 \cdot (-1)}{(-4)(-8)} > 0 \]

2. **Interval \((-4, -1)\)**: Choose \(x = -2\)
\[ \frac{(-2)^3 (-2-1)^2 (-2+4)}{(-2+1)(-2-3)} = \frac{-8 \cdot 9 \cdot 2}{(-1)(-5)} < 0 \]

3. **Interval \((-1, 0)\)**: Choose \(x = -0.5\)
\[ \frac{(-0.5)^3 (-0.5-1)^2 (-0.5+4)}{(-0.5+1)(-0.5-3)} = \frac{-0.125 \cdot 2.25 \cdot 3.5}{0.5 \cdot (-3.5)} > 0 \]

4. **Interval \((0, 1)\)**: Choose \(x = 0.5\)
\[ \frac{(0.5)^3 (0.5-1)^2 (0.5+4)}{(0.5+1)(0.5-3)} = \frac{0.125 \cdot 0.25 \cdot 4.5}{1.5 \cdot (-2.5)} < 0 \]

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

6. **Interval \((3, +\infty)\)**: Choose \(x = 4\)
\[ \frac{(4)^3 (4-1)^2 (4+4)}{(4+1)(4-3)} = \frac{64 \cdot 9 \cdot 8}{5 \cdot 1} > 0 \]

### Step 4: Compile the results
From the tests, we have:

- Positive in \((- \infty, -4)\)
- Negative in \((-4, -1)\)
- Positive in \((-1, 0)\)
- Negative in \((0, 1)\)
- Negative in \((1, 3)\)
- Positive in \((3, +\infty)\)

### Step 5: Include critical points
Now we need to check the critical points:
- At \(x = -4\): The expression is \(0\) (included).
- At \(x = -1\): The expression is undefined (not included).
- At \(x = 0\): The expression is \(0\) (included).
- At \(x = 1\): The expression is \(0\) (included).
- At \(x = 3\): The expression is undefined (not included).

### Final Solution
The complete set of values of \(x\) for which the inequality holds is:

\[ (-\infty, -4] \cup (-1, 0] \cup [1, 3) \cup (3, +\infty) \]