Problem 28 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{x^{2}}{(x-1)^{2}(x+1)^{2}}$$

Step-by-Step Solution

Answer

After finding the constants, the partial fraction decomposition of the function \( \frac{x^{2}}{(x-1)^{2}(x+1)^{2}} \) will be \( \frac{A}{(x-1)} + \frac{B}{(x-1)^{2}} + \frac{C}{(x+1)} + \frac{D}{(x+1)^{2}} \) with the specific numerical values for \(A, B, C,\) and \(D\).

Step 1: Express as sum of simpler fractions

Given the function \( \frac{x^{2}}{(x-1)^{2}(x+1)^{2}} \), the first step is to express this as a sum of simpler fractions. This can be accomplished by expressing the denominator as \((x-1)^{2}(x+1)^{2}\). We will assume that we can rewrite the function as: \( \frac{x^{2}}{(x-1)^{2}(x+1)^{2}} = \frac{A}{(x-1)^{1}} + \frac{B}{(x-1)^{2}} + \frac{C}{(x+1)^{1}} + \frac{D}{(x+1)^{2}} \).

Step 2: Clear the denominator

To make calculations of the constants easier, it's best to clear the denominator by multiplying both sides by \((x-1)^{2}(x+1)^{2}\), which results in \(x^2 = A(x-1)(x+1)^2 + B(x+1)^2 + C(x-1)^2(x+1) + D(x-1)^2\). This is a polynomial equation that holds true for all \(x\) in the domain of the original function.

Step 3: Substitute values to solve for constants

By choosing suitable values of \(x\) we can simplify the equation to solve for each variable in turn. One strategy is to set \(x\) to the values where each term cancels out. In this case, choose \(x = 1\) and \(x = -1\) to solve for \(B\) and \(D\) first, then choose another value, \(x = 0\) or \(x=2\) to form a system of two equations that can be solved for \(A\) and \(C\). Perform the operations to solve for each constant.

Key Concepts

Rational expression

A rational expression is essentially a fraction where the numerator and the denominator are both polynomials. The term "rational" comes from the word "ratio," indicating a division relationship between two entities. In math, these expressions are significant because they often exhibit complex behaviors, including vertical asymptotes or holes in their graphs.

For example, in the problem provided, the rational expression is \(\frac{x^{2}}{(x-1)^{2}(x+1)^{2}}\). Here, the numerator is a simple polynomial \(x^2\), and the denominator is a product of polynomials \((x-1)^2(x+1)^2\). Rational expressions can often be simplified, especially if their polynomials have common factors. However, in partial fraction decomposition, our goal is to express them as a sum of simpler fractions instead, simplifying their analysis.

Denominators

In any fraction, the denominator is located at the bottom and describes the number of parts the whole is divided into. In rational expressions, the denominator can be complex and usually consists of polynomial terms. The denominator is crucial as it determines the values that make the expression undefined (where the denominator becomes zero).

In partial fraction decomposition, a rational expression with a complex denominator like \((x-1)^{2}(x+1)^{2}\) is split into simpler fractions, each having a portion of the original denominator. Each of these simpler fractions typically has denominators like \((x-1)\), \((x+1)\), \((x-1)^{2}\), \((x+1)^{2}\), and so forth, depending on the original expression. By working with simpler fractions, solving or integrating the expression becomes much easier.

Polynomial equation

A polynomial equation is an equality that involves polynomials, which are expressions containing variables raised to whole number exponents. These equations form the backbone of algebra and calculus as they provide a method to solve for unknown variables. Polynomial equations are crucial in partial fraction decomposition for finding the unknown constants.

During decomposition, after setting up the series of simpler fractions, we multiply through by the common denominator to clear it, generating a polynomial equation. For our example, this leads to:

• \[x^2 = A(x-1)(x+1)^2 + B(x+1)^2 + C(x-1)^2(x+1) + D(x-1)^2\]

Solving this equation using specific values for \(x\) allows us to determine the constants \(A, B, C,\) and \(D\) that make the identity true for all \(x\).

Constants

Constants in the context of algebra and decomposition refer to specific, fixed values that aren't dependent on variables. They play an important role when expressing a more complex rational expression in a simpler form through partial fraction decomposition. Constants are usually represented by letters such as \(A, B, C,\) and \(D\).

To find these constants, we substitute specific values for the variable \(x\) to simplify and solve the polynomial equation obtained after clearing the denominator. By carefully choosing values for \(x\), where most terms cancel out, we can solve the resulting system of equations. For instance, substituting \(x = 1\) and \(x = -1\) directly simplifies the process of finding some constants, since these values will nullify other terms, isolating the constants you want to calculate.