## 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.