Problem 15 - Precalculus: Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

\(\frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4}\)

Step 1 - Understand the Numerator and Denominator

The given rational function is \( \frac{5}{(x-1)(x+4)} \). The numerator is the constant 5, and the denominator is the product of two linear factors: \((x-1)(x+4)\). For partial fraction decomposition, we need to express this as a sum of fractions.

Step 2 - Decompose into Partial Fractions

Since the denominator \((x-1)(x+4)\) is a product of linear factors, we can decompose the fraction as follows: \[ \frac{5}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4} \] where \(A\) and \(B\) are constants to be determined.

Step 3 - Clear the Denominator

Multiply through by the common denominator \((x-1)(x+4)\) to eliminate the fractions: \[ 5 = A(x+4) + B(x-1) \] This equation is now a polynomial in \(x\) which we'll solve to find \(A\) and \(B\).

Step 4 - Solve for Constants A and B

To determine \(A\) and \(B\), expand and compare the coefficients. Start by plugging in different values of \(x\).- Substitute \(x = 1\): \[ 5 = A(1+4) + B(1-1) \] \[ 5 = 5A \] so, \( A = 1 \).- Substitute \(x = -4\): \[ 5 = A(-4+4) + B(-4-1) \] \[ 5 = -5B \] so, \( B = -1 \).

Step 5 - Write the Final Decomposition

Substitute \(A = 1\) and \(B = -1\) back into the partial fraction decomposition to get:\[ \frac{5}{(x-1)(x+4)} = \frac{1}{x-1} - \frac{1}{x+4} \] This is the partial fraction decomposition.

Key Concepts

Rational Functions

Rational functions are expressions formed by the division of two polynomials. The general form of a rational function is written as \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). In the problem you provided, the rational function is \( \frac{5}{(x-1)(x+4)} \). This means the polynomial in the numerator is just the constant 5, while the denominator is a product of two linear polynomials, \( (x-1) \) and \( (x+4) \).
Understanding rational functions is crucial because their behavior is governed by the properties of both the numerator and the denominator. The zeros of the numerator determine the roots of the function, while the zeros of the denominator (excluding them from the domain of the function) create asymptotes, which are lines that the graph of the function approaches but never touches. By breaking down complex rational expressions into simpler parts through partial fraction decomposition, we can integrate, differentiate, or otherwise analyze these functions more easily.

Linear Factors

Linear factors are critical components when dealing with the decomposition of rational functions. A linear factor is typically an expression of the form \( ax + b \), where \( a \) and \( b \) are constants and \( x \) is the variable. Separating complex polynomials into their linear components is a common technique in algebra to simplify various operations.
In our original problem, the denominator \( (x-1)(x+4) \) is already factored into linear components. This characteristic allows us to apply partial fraction decomposition effectively. Each linear factor in the denominator corresponds to a separate term in the partial fraction decomposition. For instance, \( \frac{5}{(x-1)(x+4)} \) is decomposed into the sum \( \frac{A}{x-1} + \frac{B}{x+4} \).

• This method involves predicting that each linear factor will contribute a simple rational term.
• The constants \( A \) and \( B \) are determined through solving algebraic equations formed by eliminating the denominators.

By decomposing into these units, you can more easily evaluate or manipulate the function, which is particularly useful in calculus and engineering.

Algebraic Equations

Algebraic equations play an integral role in finding the constants when performing partial fraction decomposition. After splitting the rational function using linear factors, the challenge shifts to solving algebraic equations to isolate the constants of each decomposed fraction.
In the given exercise, after the partial fraction setup \( \frac{5}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4} \), we clear the denominators by multiplying through by \((x-1)(x+4)\). This process yields a linear equation in the form \( 5 = A(x + 4) + B(x - 1) \). We can solve this equation by substituting specific values of \( x \) that simplify the equation:

• By substituting \( x = 1 \), you find that \( A = 1 \).
• By substituting \( x = -4 \), you determine \( B = -1 \).

These values of \( A \) and \( B \) are substituted back into the decomposed fractions, hence completing the process of partial fraction decomposition. This method demonstrates how algebraic equations are pivotal in breaking down and simplifying rational functions into manageable parts through partial fractions.