## Factoring (x+1)(x+2)(x+3)(x+4)-15

This problem involves factoring a polynomial expression that appears quite complex at first glance. However, with a few strategic steps, we can simplify it and arrive at a factored form.

### The Trick

The key to factoring this expression lies in recognizing a pattern. Let's try grouping the terms:

**Group 1:**(x+1)(x+4)**Group 2:**(x+2)(x+3)

Now, let's expand these groups:

**Group 1:**x² + 5x + 4**Group 2:**x² + 5x + 6

Notice that both groups share the same middle term (5x). This is crucial for our next step.

### Factoring by Substitution

Let's introduce a new variable, say 'y', to represent the common term (5x+4) in both groups.

- Let
**y = x² + 5x + 4**

Now, our original expression becomes:

**y (y + 2) - 15**

This is much simpler to factor. It's a quadratic equation that we can factor directly:

**y² + 2y - 15 = (y + 5)(y - 3)**

### Substituting Back

Remember, we introduced 'y' as a placeholder. Now, we need to substitute back the original expression for 'y':

**(x² + 5x + 4 + 5)(x² + 5x + 4 - 3)**

Simplifying, we get:

**(x² + 5x + 9)(x² + 5x + 1)**

### Final Factored Form

Therefore, the factored form of (x+1)(x+2)(x+3)(x+4)-15 is:

**(x² + 5x + 9)(x² + 5x + 1)**