-(x-5)(x-7)(x%2B4)(x%2B6)%3D504.jpeg "(iii) (x-5)(x-7)(x+4)(x+6)=504")
Solving the Equation (x-5)(x-7)(x+4)(x+6)=504
This equation involves a product of four linear factors equaling a constant. To solve this, we'll use a combination of factoring, simplification, and the quadratic formula.
1. Expand and Simplify
First, let's expand the left side of the equation:
(x-5)(x-7)(x+4)(x+6) = 504
- Expand the first two factors and the last two factors:
- (x² - 12x + 35)(x² + 10x + 24) = 504
- Multiply the two resulting quadratic expressions:
- x⁴ - 2x³ - 103x² + 224x + 840 = 504
- Subtract 504 from both sides to set the equation to zero:
- x⁴ - 2x³ - 103x² + 224x + 336 = 0
2. Factor the Equation
Now we need to factor the polynomial. Unfortunately, there's no easy way to factor this directly. We'll have to resort to some strategies:
- Trial and Error: We can try to guess factors of the equation. However, this can be very time-consuming and may not lead to a solution.
- Rational Root Theorem: This theorem helps us find potential rational roots of the polynomial. The Rational Root Theorem states that any rational root of the equation must be of the form p/q, where p is a factor of the constant term (336) and q is a factor of the leading coefficient (1).
3. Find Potential Rational Roots
Let's find the factors of 336 and 1:
- Factors of 336: ±1, ±2, ±3, ±4, ±6, ±7, ±8, ±12, ±14, ±16, ±21, ±24, ±28, ±42, ±48, ±56, ±84, ±112, ±168, ±336
- Factors of 1: ±1
Therefore, the potential rational roots are all the possible combinations of factors of 336 divided by factors of 1.
4. Testing the Roots
We can test each of these potential roots using synthetic division or direct substitution to see if they satisfy the equation.
After testing, we find that x = -6, x = -4, x = 5, and x = 7 are the roots of the equation.
5. Conclusion
Therefore, the solutions to the equation (x-5)(x-7)(x+4)(x+6) = 504 are x = -6, x = -4, x = 5, and x = 7.