Problem 5 - Precalculus Mathematics for Calculus

Question

Write the form of the partial fraction decomposition of the function (as in Example 4 ). Do not determine the numerical values of the coefficients. $$\frac{x^{2}-3 x+5}{(x-2)^{2}(x+4)}$$

Step-by-Step Solution

Answer

\(\frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+4}\)

Step 1: Identify the Type of Denominators

The function given is \( \frac{x^{2}-3x+5}{(x-2)^{2}(x+4)} \). The denominator is \((x-2)^2(x+4)\), which includes a repeated linear factor \((x-2)^2\) and a simple linear factor \((x+4)\).

Step 2: Set Up Partial Fractions

For a repeated linear factor \((x-2)^2\), the partial fraction takes the form \( \frac{A}{x-2} + \frac{B}{(x-2)^2} \). For the simple linear factor \((x+4)\), it takes the form \( \frac{C}{x+4} \).

Step 3: Write the Partial Fraction Decomposition

Combine the results from the previous step to write the partial fraction decomposition:\[ \frac{x^{2}-3x+5}{(x-2)^{2}(x+4)} = \frac{A}{x-2} + \frac{B}{(x-2)^2} + \frac{C}{x+4} \] where \(A\), \(B\), and \(C\) are coefficients to be determined.

Key Concepts

Rational Functions

Rational functions are expressions that involve the ratio of two polynomials. In simple terms, they can be viewed as fractions where both the numerator and the denominator are polynomials. For example, the function \( \frac{x^{2}-3x+5}{(x-2)^{2}(x+4)} \) is a rational function.
These functions are essential in calculus and algebra because they model behaviors that are rational, meaning they are not whole numbers, negative numbers, or strings. Rational functions can be used for various real-world applications, such as modeling growth and decay in life sciences, economics, and engineering.
When dealing with rational functions, it is crucial to understand the structure of both the numerator and the denominator. This structure helps identify the types of factors present, whether simple or repeated, which leads into the critical step of breaking them down into partial fractions. This breakdown can simplify integration and provides straightforward solutions in differential equations.

Repeated Linear Factors

Repeated linear factors occur in the denominator of a rational function where a factor is raised to a power greater than one. In the example \((x-2)^2\), the linear factor \((x-2)\) is seen twice in the multiplication of factors. This indicates that (x-2) is a repeated linear factor.
Dealing with repeated linear factors in partial fraction decomposition is slightly more complex than with simple linear factors. The goal is to express the rational function as a sum of simpler fractions. For a factor like \((x-2)^2\), we must consider two partial fractions:

• \(\frac{A}{x-2}\), catering to the first occurrence of the factor.
• \(\frac{B}{(x-2)^2}\), addressing the repeated factor.

Each of these fractions will have a unique coefficient that needs to be found through further calculation or manipulation. Repeated factors ensure all possibilities are accounted for in the fraction breakdown so the overall equation remains accurate.

Simple Linear Factors

Simple linear factors are straightforward because they appear in the denominator of a rational function only once. Their unique presence allows us to focus on a single fraction representation. In the function \( \frac{x^{2}-3x+5}{(x-2)^{2}(x+4)} \), \((x+4)\) is an example of a simple linear factor.
The partial fraction decomposition for a simple linear factor like \((x+4)\) introduces only one term: \(\frac{C}{x+4}\). This singular, uncomplicated fraction overlays on top of the contribution from repeated linear factors, such as \((x-2)^2\).
Identifying and handling simple linear factors first can simplify the overall process of partial fraction decomposition. By noting their singularity, you can establish a foundation for calculating coefficients effectively without unnecessary complexity. Understanding this distinction between simple and repeated factors makes working with rational functions methodical and manageable.