Problem 13 - Precalculus: Mathematical for Calculus

Question

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: Set Up Partial Fractions

Since the denominator of your function \(\frac{5}{(x-1)(x+4)}\) is a product of distinct linear factors, we decompose it as \(\frac{A}{x-1} + \frac{B}{x+4}\), where \(A\) and \(B\) are constants to be determined.

Step 2: Combine the Fractions

The common denominator of both fractions is \((x-1)(x+4)\). Combine the fractions on the right by expressing them over this common denominator: \(\frac{A(x+4) + B(x-1)}{(x-1)(x+4)}\). This should equal the original fraction \(\frac{5}{(x-1)(x+4)}\).

Step 3: Set Numerators Equal

Since the denominators are the same, equate the numerators: \(A(x+4) + B(x-1) = 5\). This gives you \(Ax + 4A + Bx - B = 5\).

Step 4: Simplify and Collect Like Terms

Combine like terms: \((A + B)x + (4A - B) = 5\). Now, equate the coefficients from both sides of the equation: 1. For the \(x\) term, \(A + B = 0\); 2. For the constant term, \(4A - B = 5\).

Step 5: Solve the System of Equations

Solve the system of equations: \(A + B = 0\) and \(4A - B = 5\).First, from \(A + B = 0\), express \(B\) in terms of \(A\): \(B = -A\).Substitute \(B = -A\) into the second equation: \(4A + A = 5\) yields \(5A = 5\), therefore \(A = 1\). Substitute \(A = 1\) back into \(B = -A\) to find \(B = -1\).

Step 6: Write the Decomposition

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

Key Concepts

Rational Functions

Rational functions are expressions that involve ratios of polynomials. They are written in the form \( \frac{P(x)}{Q(x)} \), where \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x)\) is not zero. In the given exercise, the rational function is \( \frac{5}{(x-1)(x+4)} \). This means that the numerator is a constant, 5, and the denominator is a product of two linear factors (more on that below). The objective is to express this rational function as a sum of simpler fractions, making it easier to integrate or differentiate if needed.

This process is known as partial fraction decomposition. It allows us to break down complex fractions into a series of simpler parts that are much easier to handle in calculus or algebra. For instance, transforming \( \frac{5}{(x-1)(x+4)} \) into \( \frac{1}{x-1} - \frac{1}{x+4} \) simplifies operations like integration. Understanding the structure of rational functions and how to manipulate them through decomposition is a fundamental tool in mathematics.

Linear Factors

Linear factors are components of polynomials expressed in their simplest form, typically seen as \( ax + b \). They are called "linear" because they represent straight lines when graphed. In the context of partial fraction decomposition, each term in the denominator of the rational function is a linear factor.

For the function \( \frac{5}{(x-1)(x+4)} \), the linear factors are \( x-1 \) and \( x+4 \). These are distinct and simple, allowing us to decompose the fraction into separate parts. Each linear factor corresponds to a constant over that factor in the partial fractions, for example, \( \frac{A}{x-1} + \frac{B}{x+4} \). Understanding linear factors is crucial because they determine how a given rational function is split for easier manipulation.

• Distinct Linear Factors: Each different factor is treated separately.
• Repeated Linear Factors: Requires special handling (not in this example).

Mastering the concept of linear factors paves the way for efficiently solving and simplifying various algebraic equations.

System of Equations

A system of equations involves multiple equations that are solved together to find the values of unknown variables. In the step-by-step solution for the partial fraction decomposition exercise, a system of linear equations is formed when equating coefficients.

After setting the numerators equal from the decomposition (\(A(x+4) + B(x-1) = 5\)), we collect like terms and equate them, resulting in two equations:

• \( A + B = 0 \)
• \( 4A - B = 5 \)

These equations are solved simultaneously to find the values of \( A \) and \( B \). Solving such a system can be accomplished by methods such as substitution or elimination. In the example, substitution is used by expressing \( B \) in terms of \( A \) from the first equation and substituting it into the second. This allows us to find \( A = 1 \) and \( B = -1 \).

This technique is instrumental in mathematics as it enables breaking down complex problems into simpler, manageable steps, leading to the conceptual clarity necessary for accurate calculations.