(vii) X^(4)-2x^(3)-x^(2)-2x+1=0

Jasonbradley

3 min read

·

Jun 16, 2024

31

Share

Solving the Quartic Equation: x⁴ - 2x³ - x² - 2x + 1 = 0

This article delves into the solution of the quartic equation x⁴ - 2x³ - x² - 2x + 1 = 0. We'll explore the strategies and techniques involved in finding its roots.

Understanding Quartic Equations

Quartic equations, like the one we're tackling, are polynomial equations with the highest power of the variable being four. They generally have the form:

ax⁴ + bx³ + cx² + dx + e = 0

where a, b, c, d, and e are constants and a ≠ 0.

Finding Solutions to Quartic Equations

Unfortunately, there's no single, universal formula for solving all quartic equations directly like the quadratic formula for second-degree equations. However, we can use various methods:

1. Factoring:

• By Grouping: Look for common factors among the terms. In our case, we can group the terms: (x⁴ - x²) + (-2x³ - 2x) + 1 = 0 x²(x² - 1) - 2x(x² - 1) + 1 = 0 (x² - 1)(x² - 2x + 1) = 0 (x + 1)(x - 1)(x - 1)² = 0 (x + 1)(x - 1)³ = 0 This gives us the solutions x = -1 and x = 1 (with multiplicity 3).

2. Rational Root Theorem:

• This theorem helps find potential rational roots. It states that if a rational number p/q is a root of the equation, then p must be a factor of the constant term (e in the general form) and q must be a factor of the leading coefficient (a).
• Applying this to our equation:
  • Factors of the constant term (1): ±1
  • Factors of the leading coefficient (1): ±1
  • Potential rational roots: ±1
• We've already found these roots using factoring.

3. Numerical Methods:

• When factoring isn't straightforward, numerical methods like the Newton-Raphson method or bisection method can approximate the roots to a desired accuracy.

Conclusion

The quartic equation x⁴ - 2x³ - x² - 2x + 1 = 0 can be solved effectively through factoring. Understanding the different approaches to solving quartic equations equips us to tackle more complex polynomial equations.