If `((x^(2)-1)(x+2)(x+1)^(2))/((x-2))lt0` , then complete solution set of inequation is

Options

A. `(-2,-1)uu(1,2)`
B. `(-oo, -2)uu(2,oo)`
C. `(-2,-1)uu(2,oo)`
D. `(-2,-1)uu(1,oo)`

Explanation

To solve the inequality

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

we will follow these steps:

### Step 1: Factor the numerator and denominator
The numerator can be factored as follows:
- \(x^2 - 1 = (x - 1)(x + 1)\)
- The term \((x + 1)^2\) is already factored.
- The term \((x + 2)\) is already factored.

Thus, we can rewrite the expression as:

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

### Step 2: Identify the critical points
The critical points occur where the numerator or denominator is zero. Setting each factor to zero gives us:
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x + 1 = 0 \Rightarrow x = -1\) (with multiplicity 3)
- \(x + 2 = 0 \Rightarrow x = -2\)
- \(x - 2 = 0 \Rightarrow x = 2\)

The critical points are \(x = -2, -1, 1, 2\).

### Step 3: Determine the intervals
The critical points divide the real number line into the following intervals:
1. \((-∞, -2)\)
2. \((-2, -1)\)
3. \((-1, 1)\)
4. \((1, 2)\)
5. \((2, ∞)\)

### Step 4: Test each interval
We will test a point from each interval to determine the sign of the expression in that interval.

1. **Interval \((-∞, -2)\)**: Test \(x = -3\)
\[ \frac{(-3 - 1)(-3 + 1)^3(-3 + 2)}{(-3 - 2)} = \frac{(-4)(-2)^3(-1)}{-5} = \frac{-32}{-5} > 0 \]

2. **Interval \((-2, -1)\)**: Test \(x = -1.5\)
\[ \frac{(-1.5 - 1)(-1.5 + 1)^3(-1.5 + 2)}{(-1.5 - 2)} = \frac{(-2.5)(-0.5)^3(0.5)}{-3.5} = \frac{(-2.5)(-0.125)(0.5)}{-3.5} < 0 \]

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

4. **Interval \((1, 2)\)**: Test \(x = 1.5\)
\[ \frac{(1.5 - 1)(1.5 + 1)^3(1.5 + 2)}{(1.5 - 2)} = \frac{(0.5)(2.5)^3(3.5)}{-0.5} < 0 \]

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

### Step 5: Compile the solution set
The expression is negative in the intervals \((-2, -1)\) and \((1, 2)\).

### Final Solution Set
The complete solution set of the inequality is:

\[ (-2, -1) \cup (1, 2). \]