Solving the Polynomial Equation x4 - x3 - 4x2 x 1 0: A Step-by-Step Guide

Introduction

This article delves into the process of solving a polynomial equation of degree four: x4 - x3 - 4x2 x 1 0. Polynomial equations, while fundamental in algebra, can be challenging to solve, particularly when the degree is four or higher. However, with systematic methods and a bit of algebraic manipulation, we can determine the roots of such equations. This guide will walk through the steps involved in solving this specific polynomial equation, providing a detailed walkthrough of the solution process.

Manipulating the Equation

First, let's manipulate the given polynomial equation x4 - x3 - 4x2 x 1 0 by dividing both sides by x2.

This yields:

x2 - x - 4 - 2} 0

Next, we rewrite the equation in a more manageable form:

x2 - 42} - x - 0

Further simplification by adding x2 - 2 2 - x - 0 gives us:

x2 2 - x - - 6 0

And then, we group the terms as follows:

(x 1/x) 2 - 6 0

Substitution and Solving the Equation

For simplicity, we substitute t x 1/x. Squaring both sides of the substitution yields t2 x2 2 1/x2, which can be rewritten as:

t2 - 2 x2 1/x2

Thus, the original equation transforms to:

t2 - 2 - t - 4 0

This simplifies to:

t2 - t - 6 0

Factoring the quadratic expression, we get:

(t - 3)(t 2) 0

Therefore, the solutions for t are t 3 and t -2.

Solution in Terms of x

Returning to the original variable x, we solve:

x 1/x 3 and x 1/x -2

For x 1/x 3, we clear denominators and bring all terms to one side:

x2 - 3x 1 0

Solving the quadratic equation using the quadratic formula, we obtain:

x (-b ± √(b2 - 4ac)) / 2a

Here, a1, b-3, c1, thus:

x (3 ± √(9 - 4)) / 2 (3 ± √5) / 2

For the second equation x 1/x -2, we similarly get:

x2 2x 1 0

Factoring this, we have:

(x 12) 0

Thus, the solution is:

x -1

Conclusion

Summarizing the solutions for the polynomial equation x4 - x3 - 4x2 x 1 0, we have:

x (3 √5) / 2, (3 - √5) / 2, -1

The double root at x -1 indicates that this is a repeated root.

This solution demonstrates the power of algebraic manipulations and substitution techniques in solving polynomial equations. These methods are not only useful for specific problems but also for understanding the underlying mathematical principles that govern more complex algebraic equations.