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Solving the Equation: x⁴ + 4x⁴ + 4x + 1 = 2

This article delves into the solution of the equation x⁴ + 4x⁴ + 4x + 1 = 2, providing a comprehensive guide through various mathematical approaches. We'll explore simplification techniques, methods for solving polynomial equations, and ultimately, arrive at the solutions for x. Understanding this seemingly simple equation offers a valuable insight into fundamental algebraic concepts and problem-solving strategies.

Simplifying the Equation

Our first step is to simplify the given equation: x⁴ + 4x⁴ + 4x + 1 = 2. Notice that we have like terms involving x⁴. Combining these, we get:

5x⁴ + 4x + 1 = 2

Subtracting 2 from both sides, we transform the equation into a standard polynomial form:

5x⁴ + 4x - 1 = 0

Approaches to Solving Quartic Equations

Solving quartic equations (equations of the fourth degree) can be complex. There's no single, universally easy method. The strategies we'll consider include:

• Factoring: Attempting to factor the polynomial into simpler expressions. This is often the quickest method if it works. However, quartic equations don't always factor easily.
• Rational Root Theorem: This theorem helps identify potential rational roots (roots that are fractions of integers). While it doesn't guarantee finding all roots, it narrows down possibilities.
• Numerical Methods: For quartic equations that are difficult or impossible to solve algebraically, numerical methods (like the Newton-Raphson method) provide approximate solutions. These are iterative methods that refine an initial guess until a solution of sufficient accuracy is reached.
• Ferrari's Method: This is a general method for solving quartic equations that involves transforming the equation into a simpler form and then solving the resulting cubic and quadratic equations. It's a more involved algebraic approach.

Applying the Rational Root Theorem

The Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root p/q (where p and q are integers and q ≠ 0), then p must be a factor of the constant term, and q must be a factor of the leading coefficient.

In our equation, 5x⁴ + 4x - 1 = 0:

• The constant term is -1. Its factors are ±1.
• The leading coefficient is 5. Its factors are ±1, ±5.

Therefore, the possible rational roots are ±1 and ±1/5.

Testing Potential Roots

Let's test these potential roots by substituting them into the equation 5x⁴ + 4x - 1 = 0:

• x = 1: 5(1)⁴ + 4(1) - 1 = 8 ≠ 0
• x = -1: 5(-1)⁴ + 4(-1) - 1 = 0

We've found a root! x = -1 is a solution.

Factoring the Equation (Using the Found Root)

Since x = -1 is a root, (x + 1) is a factor of the polynomial. We can perform polynomial long division or synthetic division to find the other factor.

Performing polynomial long division:

(5x⁴ + 4x - 1) / (x + 1) = 5x³ - 5x² + 5x - 1

Therefore, our equation can be factored as:

(x + 1)(5x³ - 5x² + 5x - 1) = 0

Solving the Cubic Equation

Now we need to solve the cubic equation 5x³ - 5x² + 5x - 1 = 0. This cubic equation doesn't factor easily. We can explore numerical methods or more advanced techniques to find its roots. Let's consider a numerical approach.

Numerical Methods: Newton-Raphson Method (Example)

The Newton-Raphson method is an iterative method for finding successively better approximations to the roots of a real-valued function. The formula is:

x_(n+1) = x_n - f(x_n) / f'(x_n)

where:

• x_n is the current approximation
• x_(n+1) is the next approximation
• f(x_n) is the value of the function at x_n
• f'(x_n) is the derivative of the function at x_n

For our cubic equation, f(x) = 5x³ - 5x² + 5x - 1. Its derivative is f'(x) = 15x² - 10x + 5.

We'd need to choose an initial guess for x and then iterate the Newton-Raphson formula until the solution converges to a desired level of accuracy. This process is best performed using computational tools or software. The result would give us approximate values for the remaining roots.

Further Exploration: Advanced Techniques

For a more rigorous approach to solving the cubic equation 5x³ - 5x² + 5x - 1 = 0, we could employ techniques like Cardano's method, a general algebraic method for solving cubic equations. However, Cardano's method can lead to complex calculations and is beyond the scope of a basic explanation.

Summary of Solutions

In summary:

• We started with the equation x⁴ + 4x⁴ + 4x + 1 = 2.
• Simplified it to 5x⁴ + 4x - 1 = 0.
• Found one rational root: x = -1.
• Factored the equation to (x + 1)(5x³ - 5x² + 5x - 1) = 0.
• Identified that the remaining roots require numerical methods or more advanced techniques (like Cardano's method for the cubic equation) for their precise determination.

The exact solutions for the cubic equation 5x³ - 5x² + 5x - 1 = 0 would require the use of a numerical solver or an advanced algebraic approach. The approximate solutions obtained through numerical methods would then be the remaining roots of the original quartic equation. Understanding the different methods, their strengths and limitations, is crucial for effectively solving polynomial equations of higher degrees. While we have explored several approaches, the practical application often depends on the complexity of the specific equation and the available computational tools.