(x-3)(x-4)(x-7)(x-8)=60

4 min read Jun 17, 2024

Solving the Equation: (x-3)(x-4)(x-7)(x-8) = 60

This equation presents a challenge, as it involves a product of four factors set equal to a constant. Here's a breakdown of how to approach solving it:

Understanding the Problem

Polynomial Equation: The equation represents a polynomial of degree 4.
Factorization: The left side is already factored, which is a good starting point.
Constant: The right side is a constant, making it a bit harder to directly manipulate the factors.

Solving Strategies

1. Expansion and Standard Form

Expand: You could expand the left side to get a standard polynomial equation. This would involve multiplying out all the factors.
Solve: The resulting polynomial could be solved using techniques like the Rational Root Theorem, factoring, or numerical methods. This approach can be tedious and might involve finding complex solutions.

2. Trial and Error with Integer Solutions

Integer Roots: Since the equation has integer coefficients, it's possible it has integer solutions.
Trial and Error: Try substituting consecutive integers for 'x' and see if any satisfy the equation.
Observation: Notice that the factors are consecutive integers with a difference of 1. This might lead to a pattern you can exploit.

3. Graphing Approach

Visualize: You can graph the function y = (x-3)(x-4)(x-7)(x-8) - 60.
Intersections: The x-intercepts of the graph represent the solutions to the equation.
Limitations: Graphing tools might not give precise solutions, but can provide an estimate.

Example: Finding Integer Solutions

Let's try the trial and error approach with integer solutions:

x = 6: (6-3)(6-4)(6-7)(6-8) = 3 * 2 * -1 * -2 = 12 ≠ 60
x = 5: (5-3)(5-4)(5-7)(5-8) = 2 * 1 * -2 * -3 = 12 ≠ 60
x = 9: (9-3)(9-4)(9-7)(9-8) = 6 * 5 * 2 * 1 = 60

We found that x = 9 is a solution!

Conclusion

Solving this equation involves a combination of strategies. While expanding the equation can lead to a solution, using the factored form and trial and error with integer solutions might be the most efficient method in this case.