To solve the inequality \((x+1)(x-3) > 12\), follow these steps:

### Step 1: Expand the left side of the inequality.
\[
(x+1)(x-3) = x^2 - 3x + x - 3 = x^2 - 2x - 3
\]
So, the inequality becomes:
\[
x^2 - 2x - 3 > 12
\]

### Step 2: Move 12 to the left side to set the inequality to zero.
\[
x^2 - 2x - 3 - 12 > 0
\]
\[
x^2 - 2x - 15 > 0
\]

### Step 3: Factor the quadratic expression.
\[
x^2 - 2x - 15 = (x - 5)(x + 3)
\]
So, the inequality becomes:
\[
(x - 5)(x + 3) > 0
\]

### Step 4: Solve the inequality using a sign analysis.
We now need to find where \((x - 5)(x + 3)\) is greater than 0. To do this, consider the critical points where the expression equals zero, i.e., when:
\[
x - 5 = 0 \quad \text{or} \quad x + 3 = 0
\]
So, \(x = 5\) and \(x = -3\) are the critical points.

### Step 5: Test intervals between the critical points.
The critical points divide the number line into three intervals: 
1. \(x < -3\)
2. \(-3 < x < 5\)
3. \(x > 5\)

Now, test a value from each interval:

- For \(x < -3\) (e.g., \(x = -4\)):
  \[
  (x - 5)(x + 3) = (-4 - 5)(-4 + 3) = (-9)(-1) = 9 > 0
  \]
- For \(-3 < x < 5\) (e.g., \(x = 0\)):
  \[
  (x - 5)(x + 3) = (0 - 5)(0 + 3) = (-5)(3) = -15 < 0
  \]
- For \(x > 5\) (e.g., \(x = 6\)):
  \[
  (x - 5)(x + 3) = (6 - 5)(6 + 3) = (1)(9) = 9 > 0
  \]

### Step 6: Write the solution.
The inequality \((x - 5)(x + 3) > 0\) holds when \(x < -3\) or \(x > 5\).

Thus, the solution is:
\[
x \in (-\infty, -3) \cup (5, \infty)
\]

Would you like more details on any part of this solution or have further questions?

Here are five related questions to expand your understanding:
1. How would you solve \((x+1)(x-3) \geq 12\)?
2. What is the graphical representation of the solution to this inequality?
3. How do you solve quadratic inequalities in general?
4. What are some techniques to factor complex quadratics?
5. How does solving inequalities differ from solving equations?

**Tip:** Always test values from each interval around the critical points to determine where an inequality holds true.

Discover how to solve the quadratic inequality (x+1)(x-3) > 12 with a step-by-step explanation. This guide covers factoring, critical points, and sign analysis, perfect for students in Grades 9-11.

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