Solving the Equation: (x+1)(x+2)(x+3)(x+4)  8 = 0
This equation looks quite intimidating at first glance, but with a bit of algebraic manipulation and ingenuity, we can find its solutions.
Simplifying the Equation

Expand the first four terms: We can expand the product of the first four terms using the distributive property (or by using the FOIL method repeatedly). This will give us a polynomial of degree 4.

Combine terms: After expanding, we can combine like terms to get a simplified polynomial equation.
Finding the Solutions

Looking for patterns: Observe that the constant term in the simplified equation is 8. This suggests that we might be able to factor the equation.

Trying a guess: Let's try substituting x = 1 into the original equation. We get: (1 + 1)(1 + 2)(1 + 3)(1 + 4)  8 = 0. This indicates that x = 1 is a solution to the equation.

Using the Factor Theorem: Since x = 1 is a solution, we know that (x + 1) is a factor of the polynomial. We can use polynomial long division or synthetic division to divide the polynomial by (x + 1) and get a cubic polynomial.

Repeating the process: We can repeat the process of finding solutions and factoring for the resulting cubic polynomial. We might find another solution (and another factor), which would leave us with a quadratic equation.

Solving the quadratic: We can then use the quadratic formula to find the remaining solutions.
The Solutions
Following the above steps, we can find that the solutions to the equation (x + 1)(x + 2)(x + 3)(x + 4)  8 = 0 are:
 x = 1
 x = 4
 x = 2 + √2
 x = 2  √2
Key Points
 This equation is a quartic equation (degree 4).
 It has four solutions (real and/or complex).
 The solutions can be found by factoring, polynomial division, and the quadratic formula.
 The solutions to this equation represent the xintercepts of the graph of the corresponding function.