Partial Fraction Decomposition With Repeated Linear Factors

Partial Fraction Decomposition with Repeated Linear Factors: A Comprehensive Guide

Meta Description: Learn how to perform partial fraction decomposition when dealing with repeated linear factors in the denominator of a rational function. This guide provides a step-by-step approach with clear examples.

Partial fraction decomposition is a crucial technique in calculus, particularly when integrating rational functions. It involves breaking down a complex rational function into simpler fractions that are easier to integrate. While straightforward with distinct linear factors, the process becomes slightly more nuanced when dealing with repeated linear factors in the denominator. This article will guide you through the process, clarifying the approach and providing illustrative examples.

Understanding Repeated Linear Factors

A repeated linear factor in the denominator of a rational function appears more than once. For example, in the expression (3x² + 2x + 1) / (x(x-1)²), the factor (x-1) is repeated. This repetition necessitates a different approach compared to situations with only distinct linear factors.

The Method: A Step-by-Step Guide

The key difference when dealing with repeated linear factors lies in how we represent the partial fractions. For each repeated linear factor (ax + b)^n, we introduce n separate partial fractions, each with a different power of (ax + b) in the denominator. The numerators will be constants.

Let's break down the process:

1. Factor the Denominator Completely: Ensure the denominator is completely factored into linear factors, including repeated ones.

2. Set Up the Partial Fraction Decomposition: For each distinct linear factor (ax + b), include a term of the form A/(ax + b). For each repeated linear factor (ax + b)^n, include n terms: A₁/(ax + b) + A₂/(ax + b)² + ... + Aₙ/(ax + b)ⁿ. The Aᵢ are constants that we need to solve for.

3. Clear the Fractions: Multiply both sides of the equation by the original denominator to eliminate the fractions.

4. Solve for the Constants: This step often involves substituting values of x that simplify the equation. Strategic substitution (e.g., using values that make certain factors zero) can help isolate and solve for the constants. If strategic substitution doesn't yield all constants, you can equate coefficients of like powers of x.

5. Rewrite the Original Function: Substitute the solved constants back into the partial fraction decomposition to obtain the final form.

Example: Partial Fraction Decomposition with Repeated Linear Factor

Let's decompose the rational function: f(x) = (3x² + 2x + 1) / (x(x-1)²)

1. Factorization: The denominator is already factored.

2. Partial Fraction Setup: (3x² + 2x + 1) / (x(x-1)²) = A/x + B/(x-1) + C/(x-1)²

3. Clearing Fractions: Multiplying both sides by x(x-1)² yields: 3x² + 2x + 1 = A(x-1)² + Bx(x-1) + Cx

4. Solving for Constants:

   • Let x = 0: 1 = A(-1)² => A = 1
   • Let x = 1: 6 = C
   • Equating Coefficients: Expanding the equation and comparing coefficients of x², we get: 3 = A + B. Since A = 1, B = 2.

5. Final Decomposition: (3x² + 2x + 1) / (x(x-1)²) = 1/x + 2/(x-1) + 6/(x-1)²

Applications and Further Exploration

Partial fraction decomposition with repeated linear factors finds application in various areas, including:

• Calculus: Simplifying integrals of rational functions.
• Differential Equations: Solving certain types of differential equations.
• Signal Processing: Analyzing and manipulating signals.

Understanding this technique is essential for mastering advanced calculus and its applications in various scientific and engineering fields. Practice with different examples will solidify your understanding and build your confidence in tackling more complex rational functions. Remember to always check your work by combining your partial fractions back into the original expression.