## Factoring the expression (x+2)(x+3)(x+4)(x+5)-8

The expression (x+2)(x+3)(x+4)(x+5)-8 might look intimidating at first glance, but we can factor it in a clever way to reveal its hidden beauty.

### Step 1: Recognizing the Pattern

Observe that the first part of the expression is a product of four consecutive terms. This hints at a pattern that can be exploited for factorization.

### Step 2: Introducing a "Middle Term"

Let's introduce a new variable, say 'y', to simplify our expression. Let:

**y = (x+2)(x+5)**

Notice that if we multiply out the terms of (x+3)(x+4), we get:

**(x+3)(x+4) = x² + 7x + 12**

Now, we can rewrite the original expression in terms of 'y':

**(x+2)(x+3)(x+4)(x+5) - 8 = y(x² + 7x + 12) - 8**

### Step 3: Completing the Square

Let's focus on the quadratic term (x² + 7x + 12). We can manipulate it by completing the square. To do this:

- Take half of the coefficient of the x term (7/2) and square it (49/4).
- Add and subtract this value inside the parentheses:

**(x² + 7x + 12) = (x² + 7x + 49/4 - 49/4 + 12)**

- The first three terms now form a perfect square trinomial:

**(x² + 7x + 49/4 - 49/4 + 12) = (x + 7/2)² - 1/4**

### Step 4: Substituting and Factoring

Substitute this back into our expression:

**y(x² + 7x + 12) - 8 = y((x + 7/2)² - 1/4) - 8**

Now, we can factor out a common factor of 'y':

**y((x + 7/2)² - 1/4) - 8 = y(x + 7/2)² - y/4 - 8**

We can rewrite the last two terms to get a difference of squares:

**y(x + 7/2)² - y/4 - 8 = y(x + 7/2)² - (1/4)(y + 32)**

Finally, we can factor this expression as a difference of squares:

**y(x + 7/2)² - (1/4)(y + 32) = [√(y)(x + 7/2) + √(1/4)(y + 32)][√(y)(x + 7/2) - √(1/4)(y + 32)]**

### Step 5: Replacing 'y'

Remember that **y = (x+2)(x+5)**. Substitute this back into the factored expression to get the final result:

**[√((x+2)(x+5))(x + 7/2) + √(1/4)((x+2)(x+5) + 32)][√((x+2)(x+5))(x + 7/2) - √(1/4)((x+2)(x+5) + 32)]**

### Conclusion

This factorization might seem intricate, but it demonstrates the power of recognizing patterns, completing the square, and utilizing variable substitutions to simplify complex expressions.