(x%2B1)(x%2B3)(x%2B5)%2B15.jpeg "(x-1)(x+1)(x+3)(x+5)+15")
Factoring and Solving the Expression (x-1)(x+1)(x+3)(x+5)+15
This article will explore the factorization and solution of the expression (x-1)(x+1)(x+3)(x+5)+15. We will delve into the techniques used to simplify this expression and find its roots.
Understanding the Expression
The given expression is a polynomial with four factors multiplied together and a constant term added. It appears complex, but we can simplify it using a strategic approach.
Factoring the Expression
- Group the factors: Observe that the first four terms are a product of pairs: (x-1)(x+1) and (x+3)(x+5).
- Expand the pairs:
- (x-1)(x+1) = x² - 1 (using the difference of squares pattern)
- (x+3)(x+5) = x² + 8x + 15
- Substitute and simplify: Substitute the expanded forms back into the original expression: (x² - 1)(x² + 8x + 15) + 15 = x⁴ + 8x³ + 14x² - 8x - 15 + 15 = x⁴ + 8x³ + 14x² - 8x
- Factor out common factors: Notice that all terms have 'x' in common: = x(x³ + 8x² + 14x - 8)
Finding the Roots
To find the roots of the expression, we need to solve the equation: x(x³ + 8x² + 14x - 8) = 0
This equation has one obvious root: x = 0. To find the remaining roots, we need to solve the cubic equation:
x³ + 8x² + 14x - 8 = 0
Unfortunately, there is no general formula to solve cubic equations like this one. We can use numerical methods, graphing calculators, or factorization techniques (if applicable) to approximate the remaining roots.
Conclusion
While the initial expression (x-1)(x+1)(x+3)(x+5)+15 appears complex, it can be simplified by factoring and rearranging terms. We found one obvious root, x=0, and identified a cubic equation to solve for the remaining roots. Finding these roots requires numerical methods or specialized techniques. This exercise demonstrates how algebraic manipulation and factorization techniques can be used to analyze and simplify seemingly complex expressions.