## Solving the Equation (x+1)(x-1)(x+3)(x+5)+7=0

This equation appears complicated at first glance, but with a few algebraic manipulations and the use of the Rational Root Theorem, we can find its solutions. Let's break down the process:

### Expanding the Equation

First, we expand the product of the first four factors:

(x+1)(x-1)(x+3)(x+5) = (x² - 1)(x² + 8x + 15)

Expanding further, we get:

(x² - 1)(x² + 8x + 15) = x⁴ + 8x³ + 14x² + 8x - 15

Now, the equation becomes:

x⁴ + 8x³ + 14x² + 8x - 15 + 7 = 0

Which simplifies to:

x⁴ + 8x³ + 14x² + 8x - 8 = 0

### Rational Root Theorem

The **Rational Root Theorem** states that if a polynomial equation has integer coefficients, then any rational root of the equation must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

In our equation, the constant term is -8 and the leading coefficient is 1. Therefore, the possible rational roots are:

**p:**±1, ±2, ±4, ±8**q:**±1

This gives us the following possible rational roots: ±1, ±2, ±4, ±8.

### Finding the Roots

We can test these possible roots by substituting them into the equation. We find that **x = -2** is a root of the equation. This means (x + 2) is a factor of the polynomial.

We can use polynomial long division or synthetic division to divide the polynomial by (x + 2):

(x⁴ + 8x³ + 14x² + 8x - 8) / (x + 2) = x³ + 6x² + 2x - 4

Now we have the equation:

(x + 2)(x³ + 6x² + 2x - 4) = 0

We can repeat the process of finding rational roots for the cubic polynomial. The possible rational roots are: ±1, ±2, ±4. We find that **x = -2** is a root again. This means (x + 2) is a factor of the cubic polynomial.

Dividing the cubic polynomial by (x + 2), we get:

(x³ + 6x² + 2x - 4) / (x + 2) = x² + 4x - 2

Now we have:

(x + 2)² (x² + 4x - 2) = 0

The quadratic factor can be solved using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

where a = 1, b = 4, and c = -2. This gives us:

x = (-4 ± √(4² - 4(1)(-2))) / 2(1)

x = (-4 ± √(24)) / 2

x = (-4 ± 2√6) / 2

x = -2 ± √6

### Solutions

Therefore, the solutions to the equation (x+1)(x-1)(x+3)(x+5)+7=0 are:

**x = -2 (double root)****x = -2 + √6****x = -2 - √6**

We have successfully solved the equation by using algebraic manipulation and the Rational Root Theorem.