(x-3)(x%2B5)(x%2B7)-297%3D0.jpeg "(x-1)(x-3)(x+5)(x+7)-297=0")
Solving the Quartic Equation: (x-1)(x-3)(x+5)(x+7) - 297 = 0
This article explores the process of solving the quartic equation (x-1)(x-3)(x+5)(x+7) - 297 = 0.
Expanding and Simplifying
The first step involves expanding the equation by multiplying the factors:
- (x-1)(x-3) = x² - 4x + 3
- (x+5)(x+7) = x² + 12x + 35
Now, we need to multiply these two quadratic expressions:
- (x² - 4x + 3)(x² + 12x + 35) = x⁴ + 8x³ - 29x² - 126x + 105
Finally, we can subtract 297 from both sides to get the simplified quartic equation:
- x⁴ + 8x³ - 29x² - 126x - 192 = 0
Factoring the Equation
Our aim is to factor the quartic equation into simpler factors. This is where a little observation and manipulation are key:
- Grouping Terms: Rearrange the terms to facilitate factoring:
- (x⁴ - 192) + (8x³ - 29x²) - 126x = 0
- Difference of Squares: Recognize the first term as a difference of squares:
- (x² - √192)(x² + √192) + x²(8x - 29) - 126x = 0
- Factor out Common Factors: Factor out common factors in the second and third terms:
- (x² - √192)(x² + √192) + x²(8x - 29) - 63(2x) = 0
- Further Grouping: Group the terms to find a common factor:
- (x² - √192)(x² + √192) + (x² - 63)(8x - 29) = 0
Now, we have factored the equation into two parts, each of which can be further simplified:
- (x² - √192)(x² + √192) = (x - √√192)(x + √√192)(x - i√√192)(x + i√√192)
- (x² - 63)(8x - 29) = (x - √63)(x + √63)(8x - 29)
Final Solution
By combining the factors, we arrive at the complete factored form of the equation:
- (x - √√192)(x + √√192)(x - i√√192)(x + i√√192)(x - √63)(x + √63)(8x - 29) = 0
Therefore, the solutions to the equation are:
- x = √√192
- x = -√√192
- x = i√√192
- x = -i√√192
- x = √63
- x = -√63
- x = 29/8
These are the seven distinct roots of the quartic equation.
Conclusion
Solving the equation (x-1)(x-3)(x+5)(x+7) - 297 = 0 involved a series of steps: expanding, simplifying, factoring, and isolating the roots. The solution reveals seven distinct roots, including real and complex solutions. This process illustrates the powerful tools of algebra and factorization in solving polynomial equations.