Problem 16 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{x+6}{x(x+3)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{2}{x} - \frac{1}{x+3} \).

Step 1: Recognize the Denominator

The given rational function is \( \frac{x+6}{x(x+3)} \). We need to decompose this into partial fractions. First, recognize that the denominator \( x(x+3) \) can be expressed as two linear factors: \( x \) and \( x+3 \).

Step 2: Set Up the Partial Fraction Form

For each factor of the denominator, assign a constant to represent the numerator of each partial fraction. Thus, decompose the expression into \( \frac{A}{x} + \frac{B}{x+3} \), where \( A \) and \( B \) are constants we need to determine.

Step 3: Combine the Partial Fractions

Express the sum \( \frac{A}{x} + \frac{B}{x+3} \) with a common denominator, \( x(x+3) \). It becomes: \( \frac{A(x+3) + Bx}{x(x+3)} \).

Step 4: Setup Equation to Solve for Constants

Since \( \frac{x+6}{x(x+3)} = \frac{A(x+3) + Bx}{x(x+3)} \), we equate the numerators to get the equation: \[ A(x+3) + Bx = x + 6 \].

Step 5: Expand and Simplify

Expand the left side: \( Ax + 3A + Bx = x + 6 \). Combine like terms to get \( (A+B)x + 3A = x + 6 \).

Step 6: Solve for Constants A and B

Equate the coefficients from both sides of the equation: 1. For the \(x\) terms: \( A + B = 1 \)2. For the constant terms: \( 3A = 6 \).From \(3A = 6\), solve for \(A\) to get \(A = 2\). Substitute \(A = 2\) into \(A + B = 1\) to find \(B = 1 - 2 = -1\).

Step 7: Write The Partial Fraction Decomposition

Now that we have \( A = 2 \) and \( B = -1 \), substitute back into the partial fractions: \[ \frac{x+6}{x(x+3)} = \frac{2}{x} - \frac{1}{x+3} \].

Key Concepts

Rational Functions

A rational function is a mathematical expression of the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials, and \( Q(x) eq 0 \). Rational functions are similar to fractions but deal with polynomials instead of integers. These functions are widely used in algebra and calculus to model real-world problems and analyze mathematical behavior.
Understanding rational functions is crucial when you need to simplify complex expressions or find solutions to equations. In partial fraction decomposition, analyzing the form of the rational function allows you to decompose it into simpler, more manageable parts. This procedure is essential, especially when solving integrals or when trying to find roots of polynomial equations.

Linear Factors

In the process of partial fraction decomposition, a key step is recognizing the linear factors of the denominator. A linear factor is any expression of the form \( x - a \) or \( x + a \), where \( a \) is a constant.
For the given rational function \( \frac{x+6}{x(x+3)} \), the denominator \( x(x+3) \) is already factored into two linear factors: \( x \) and \( x+3 \). Identifying linear factors is essential because each one corresponds to a separate term in the partial fraction decomposition.

• Linear factors simplify the mathematical procedure.
• Each linear factor leads to an equation with constants that need solving.

Understanding how to identify and use linear factors is a cornerstone of mastering partial fraction decomposition.

Constants in Partial Fractions

When dealing with partial fraction decomposition, each linear factor in the denominator translates to a constant in the numerator of its respective term. In practice, this means if you have a denominator comprised of linear factors like \( x \) and \( x+3 \), your partial fraction decomposition will take the form \( \frac{A}{x} + \frac{B}{x+3} \).
Constants \( A \) and \( B \) are unknowns that we need to solve for. They're vital because determining their values transforms the complex rational function into simpler, easily integrable terms. Solving for these constants typically involves:

• Eliminating the denominators by equating polynomials on both sides of the expresssion.
• Using algebraic techniques to pay special attention to coefficients.

Finding the correct values of these constants is the ultimate goal in partial fraction decomposition.

Denominator Analysis

Denominator analysis is the first critical step in partial fraction decomposition. It involves identifying the factors of the denominator polynomial and helps set up the rest of the decomposition correctly. In the expression \( \frac{x+6}{x(x+3)} \), the denominator \( x(x+3) \) reveals itself as a product of two linear factors.
Why is this step important? Because the structure of the denominator dictates how the rational function can be broken down into partial fractions. The analysis tells us:

• How many terms the partial fraction decomposition will have.
• What kind of numerators we should initially assign to these terms.

This methodical breakdown ensures that you're equipped to solve for the constants efficiently, paving the way for more straightforward solutions in calculus and algebra.