## Solving the Equation (x+1)(x+3)(x-4)(x-6) + 24 = 0

This problem involves a quartic equation, meaning it has a highest power of 4. While we can't always solve quartic equations directly, this particular problem has a clever solution that simplifies the equation significantly.

### The Key Observation

Notice that the first part of the equation resembles the expansion of a quadratic equation. Let's manipulate it a bit:

**(x+1)(x+3)(x-4)(x-6) + 24 = 0**

We can rewrite this as:

**[(x+1)(x-6)][(x+3)(x-4)] + 24 = 0**

Now, expand each pair of brackets:

**(x² - 5x - 6)(x² - x - 12) + 24 = 0**

### The Substitution Trick

Here's the clever part: Let's substitute **y = x² - x**. This substitution transforms the equation into a quadratic:

**(y - 6)(y - 12) + 24 = 0**

Expanding this gives us:

**y² - 18y + 72 + 24 = 0**

**y² - 18y + 96 = 0**

### Solving the Quadratic

Now, we have a simple quadratic equation that we can solve using the quadratic formula or factoring:

**(y - 12)(y - 8) = 0**

This gives us two solutions for *y*:

**y = 12****y = 8**

### Back to the Original Variable

Remember, we made the substitution **y = x² - x**. Now, we need to substitute back to find the solutions for *x*:

**Case 1: y = 12**

**x² - x = 12**

**x² - x - 12 = 0**

**(x - 4)(x + 3) = 0**

This gives us two solutions: **x = 4** and **x = -3**

**Case 2: y = 8**

**x² - x = 8**

**x² - x - 8 = 0**

This equation doesn't factor easily. We can solve it using the quadratic formula:

**x = [1 ± √(1² - 4(1)(-8))] / 2(1)**

**x = [1 ± √33] / 2**

This gives us two more solutions: **x = (1 + √33) / 2** and **x = (1 - √33) / 2**

### Final Solutions

Therefore, the equation (x+1)(x+3)(x-4)(x-6) + 24 = 0 has **four solutions**:

**x = 4****x = -3****x = (1 + √33) / 2****x = (1 - √33) / 2**