Solve `((x+1)(x-3))/((x-2)) ge 0`.

Answer

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

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

1. Set the numerator equal to zero:
\[ (x+1)(x-3) = 0 \]
This gives us:
- \(x + 1 = 0 \Rightarrow x = -1\)
- \(x - 3 = 0 \Rightarrow x = 3\)

2. Set the denominator equal to zero:
\[ x - 2 = 0 \Rightarrow x = 2 \]

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

### Step 2: Create a number line
We will plot the critical points on a number line:
\[ -\infty \quad -1 \quad 2 \quad 3 \quad +\infty \]

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

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

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

3. **Interval \((2, 3)\)**: Choose \(x = 2.5\)
\[ \frac{(2.5 + 1)(2.5 - 3)}{2.5 - 2} = \frac{(3.5)(-0.5)}{0.5} = \frac{-1.75}{0.5} < 0 \]

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

### Step 4: Determine the sign of the expression
From our tests, we have:
- For \(x < -1\): Negative
- For \(-1 < x < 2\): Positive
- For \(2 < x < 3\): Negative
- For \(x > 3\): Positive

### Step 5: Include critical points
Now we check the critical points:
- At \(x = -1\):
\[ \frac{(0)(-4)}{-3} = 0 \quad (\text{included since } \geq 0) \]
- At \(x = 2\):
The expression is undefined (denominator is zero).
- At \(x = 3\):
\[ \frac{(4)(0)}{1} = 0 \quad (\text{included since } \geq 0) \]

### Step 6: Write the solution
The solution set where the expression is non-negative is:
\[ x \in [-1, 2) \cup [3, +\infty) \]

### Final Answer
\[ \text{The solution is } x \in [-1, 2) \cup [3, +\infty) \]