## Factoring (x+1)(x-1)(x-3)(x-5)+12

This problem involves factoring a polynomial expression with a specific structure. We can use a clever manipulation to simplify the factorization process. Here's how:

**1. Rearranging the terms:**

Notice that the first four terms are the product of four consecutive differences: (x+1)(x-1)(x-3)(x-5). Let's rearrange the terms:

(x+1)(x-1)(x-3)(x-5) + 12 = [(x+1)(x-5)][(x-1)(x-3)] + 12

**2. Expanding and simplifying:**

Now, expand the products within the brackets:

[(x+1)(x-5)][(x-1)(x-3)] + 12 = (x² - 4x - 5)(x² - 4x + 3) + 12

**3. Substitution:**

To make the expression easier to work with, let's substitute **y = x² - 4x**. This gives us:

(x² - 4x - 5)(x² - 4x + 3) + 12 = (y - 5)(y + 3) + 12

**4. Factoring the quadratic:**

Now we have a simple quadratic expression:

(y - 5)(y + 3) + 12 = y² - 2y - 15 + 12 = y² - 2y - 3

This quadratic factors easily:

y² - 2y - 3 = (y - 3)(y + 1)

**5. Back-substituting:**

Substitute **y = x² - 4x** back into the equation:

(y - 3)(y + 1) = (x² - 4x - 3)(x² - 4x + 1)

**Therefore, the factored form of (x+1)(x-1)(x-3)(x-5)+12 is (x² - 4x - 3)(x² - 4x + 1).**