(x+1)(x+2)(x+3)(x+4)-8=0

3 min read Jun 16, 2024

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.

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.

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.

Following the above steps, we can find that the solutions to the equation (x + 1)(x + 2)(x + 3)(x + 4) - 8 = 0 are:

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 x-intercepts of the graph of the corresponding function.