Problem 15 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{5}{(x-1)(x+4)} $$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{1}{x-1} - \frac{1}{x+4} \).

Step 1: Identify the Denominator

First, identify the factors of the denominator. The given function is \( \frac{5}{(x-1)(x+4)} \). The denominator consists of the linear factors \( (x-1) \) and \( (x+4) \).

Step 2: Set Up the Partial Fractions

Write the expression as a sum of two partial fractions. Assume: \( \frac{5}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4} \). Your task is to find the constants \( A \) and \( B \).

Step 3: Clear the Denominator

Multiply through by the denominator \( (x-1)(x+4) \) to get rid of the fractions: \( 5 = A(x+4) + B(x-1) \).

Step 4: Expand and Rearrange

Expand the right side of the equation: \( 5 = Ax + 4A + Bx - B \). Combine like terms: \( 5 = (A+B)x + (4A-B) \).

Step 5: Equate Coefficients

Since the left side \( 5 \) is a constant (no x-term), equate the coefficients on both sides: \( A + B = 0 \) and \( 4A - B = 5 \).

Step 6: Solve the System of Equations

Solve for \( A \) and \( B \): 1. From \( A + B = 0 \), we get \( B = -A \).2. Substitute \( B = -A \) into \( 4A - B = 5 \): \( 4A - (-A) = 5 \) or \( 5A = 5 \).3. Solving gives \( A = 1 \) and hence \( B = -1 \).

Step 7: Write the Partial Fraction Decomposition

Substituting \( A \) and \( B \) back into the fractions, we get: \( \frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4} \).

Key Concepts

Rational Functions

Rational functions are expressions formed by dividing one polynomial by another. They take the standard form \( R(x) = \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials. In the context of partial fraction decomposition, understanding rational functions is crucial as it allows us to express complex fractions as sums of simpler fractions. This simplification is particularly beneficial in calculus, especially when integrating.

• **Numerator:** The top part of the fraction, which is \( P(x) \) in the standard form.
• **Denominator:** The bottom part of the fraction, \( Q(x) \), important for determining the factors for decomposition.
• **Partial Fraction Decomposition:** A technique used to express the rational function as a sum of simpler fractions.

In the exercise example, the rational function is given as \( \frac{5}{(x-1)(x+4)} \). The goal is to decompose this complex expression into simpler fractions that are easier to manipulate or integrate. This involves identifying the polynomial in the denominator and using its factors to set up the decomposition.

Denominator Factors

Denominator factors play a crucial role in partial fraction decomposition. To decompose a rational function effectively, it's essential to break down the denominator into its constituent factors. These factors can either be linear or quadratic, and the type of factors impacts the form of the partial fractions.

• **Linear Factors:** In the example, the factors are \( (x-1) \) and \( (x+4) \). Each linear factor leads to a separate term in the partial fraction decomposition.
• **Distinct Linear Factors:** When the linear factors are different, as in \( (x-1) \) and \( (x+4) \), each has its own simple fraction with a unique constant in the numerator.

By identifying these factors, we can set up an equation representing the original rational function as a sum of simpler fractions. This process involves assuming a fractional expression for each factor, such as \( \frac{A}{x-1} + \frac{B}{x+4} \), where \( A \) and \( B \) are constants to be determined.

System of Equations

Once we have set up the partial fractions using the denominator factors, the next step is to determine the constants involved by solving a system of equations. This involves clearing the denominators and equating coefficients to form a system that can be solved algebraically.

• **Clearing the Denominator:** Multiply through by the common denominator to eliminate the fractions. This creates a polynomial equation.
• **Equate Coefficients:** Rearrange the terms and match the coefficients of like terms from both sides of the equation. This process creates a set of equations.
• **Solving the System:** In the example, solving \( A + B = 0 \) and \( 4A - B = 5 \) helps in finding \( A = 1 \) and \( B = -1 \).

The solution of this system gives the values of the constants \( A \) and \( B \). Substituting them back into the partial fraction form provides the complete decomposition. This method simplifies the function into \( \frac{1}{x-1} - \frac{1}{x+4} \). Solving the system accurately ensures that the rational function is correctly expressed in its decomposed form, aiding further calculations or integrations.