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The inequality provided is:

\[
\frac{x^2 + x - 12}{x - 1} \leq 0
\]

Let's solve this inequality step-by-step:

### Step 1: Factor the Numerator

First, let's factor the quadratic expression in the numerator:

\[
x^2 + x - 12
\]

To factor this expression, we look for two numbers that multiply to \(-12\) (constant term) and add up to \(1\) (coefficient of \(x\)). These numbers are \(4\) and \(-3\). So, we can factor the quadratic as:

\[
x^2 + x - 12 = (x + 4)(x - 3)
\]

### Step 2: Rewrite the Inequality

Now, rewrite the inequality with the factored numerator:

\[
\frac{(x + 4)(x - 3)}{x - 1} \leq 0
\]

### Step 3: Determine Critical Points

The critical points occur where the numerator or denominator is zero:

1. \(x + 4 = 0 \implies x = -4\)
2. \(x - 3 = 0 \implies x = 3\)
3. \(x - 1 = 0 \implies x = 1\) (Note: \(x = 1\) is a vertical asymptote as it makes the denominator zero)

### Step 4: Analyze the Intervals

We now have four critical points dividing the number line into intervals: \((- \infty, -4)\), \((-4, 1)\), \((1, 3)\), and \((3, \infty)\).

We will test the sign of the expression \(\frac{(x + 4)(x - 3)}{x - 1}\) in each of these intervals.

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

2. **Interval \((-4, 1)\):**  
   Choose \(x = 0\):  
   \[
   \frac{(0 + 4)(0 - 3)}{0 - 1} = \frac{(4)(-3)}{-1} = \frac{-12}{-1} = 12 > 0
   \]

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

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

### Step 5: Determine Solution

The expression is less than or equal to zero in the intervals where it is negative or zero. We include \(x = -4\) and \(x = 3\) because the inequality is non-strict (≤), but we exclude \(x = 1\) because it results in a division by zero. Thus, the solution is:

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

### Final Answer

\[
(-\infty, -4] \cup (1, 3]
\]

Would you like more details or have any questions? 

### Related Questions

1. How do you solve quadratic inequalities?
2. How do you find the critical points of a rational expression?
3. What is the significance of a vertical asymptote in a rational inequality?
4. How do you determine which intervals satisfy a rational inequality?
5. What are the steps to graph a rational function and its inequalities?

### Tip

When solving rational inequalities, always carefully analyze the critical points, especially where the denominator is zero, to avoid including non-permissible values in the solution set.

Solve the inequality: (x^2 + x - 12) / (x - 1) ≤ 0

Learn how to solve the rational inequality (x^2 + x - 12) / (x - 1) ≤ 0 with a step-by-step solution. This problem involves factorization of quadratic expressions, critical point identification, and interval sign analysis. Suitable for students in Grades 10-12.

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