Problem 13 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{2}{(x-1)(x+1)} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Objective

Our objective is to write the rational function \(\frac{2}{(x-1)(x+1)}\) as a sum of simpler fractions, known as partial fractions.

Step 2: Set Up Partial Fractions

Assume the expression \(\frac{2}{(x-1)(x+1)}\) can be decomposed into the form \(\frac{A}{x-1} + \frac{B}{x+1}\), where \(A\) and \(B\) are constants to be determined.

Step 3: Create a Common Denominator

Write the equation: \[ \frac{2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \] Multiply each term by the common denominator \((x-1)(x+1)\) to clear the fractions: \[ 2 = A(x+1) + B(x-1) \]

Step 4: Expand and Combine Like Terms

Expand the right side: \(A(x+1) = Ax + A\) and \(B(x-1) = Bx - B\). Combine them to get: \(Ax + A + Bx - B = (A + B)x + (A - B)\). Set up the equation: \(2 = (A + B)x + (A - B)\).

Step 5: Equate Coefficients

Since there are no \(x\) terms on the left side, \(A + B = 0\). Since the constant is 2, \(A - B = 2\). Now, solve these equations simultaneously.

Step 6: Solve the Equations

From \(A + B = 0\), we have \(B = -A\). Substitute this into \(A - B = 2\): \(A - (-A) = 2\), so \(2A = 2\). Thus, \(A = 1\). Substitute \(A = 1\) back into \(B = -A\) to find \(B = -1\).

Step 7: Write the Decomposition

Substitute \(A = 1\) and \(B = -1\) back into the partial fractions: \[ \frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1} \].

Key Concepts

Rational Function

A rational function is a type of mathematical expression that is made by dividing one polynomial by another. Just like fractions in arithmetic, these expressions are indicated by two separate polynomial functions: a numerator and a denominator. In our exercise, the rational function is represented as \[\frac{2}{(x-1)(x+1)}\]which means the numerator is 2, and the denominator is \[ (x-1)(x+1). \]In the wider scope, rational functions are pivotal because they embody the idea of division in polynomials, enabling complex expressions to be simplified or analyzed more conveniently. They play a crucial role in calculus and algebra for solving equations, doing curve sketching, and analyzing asymptotic behavior. Without a good understanding of rational functions, moving forward with advanced algebraic concepts can be challenging.

Common Denominator

The concept of a common denominator is essential when dealing with fractions, including rational functions. The purpose of finding a common denominator is to create equivalent fractions that can easily be added or subtracted. In our example, we're decomposing the rational function \[\frac{2}{(x-1)(x+1)}\]into partial fractions:\[\frac{A}{x-1} + \frac{B}{x+1}\]To do this, we multiply each term by the common denominator \[(x-1)(x+1)\] to eliminate the fractions:\[2 = A(x+1) + B(x-1).\] By clearing the fractions, it simplifies the equation and helps in setting up a straightforward path to solve for unknown coefficients like \(A\) and \(B\). The common denominator is a vital step not only in partial fraction decomposition but also in many arithmetic operations involving fractions.

Equate Coefficients

Equating coefficients is a technique used to find the constants when decomposing a rational function. Once you multiply both sides by the common denominator, as in:\[2 = A(x+1) + B(x-1),\]you expand and combine like terms to form a polynomial equation. This results in \[(A + B)x + (A - B).\]To solve for \(A\) and \(B\), you equate the coefficients of corresponding terms on both sides of the equation. Since there was no \(x\) term in \(2\), we assume:

• \(A + B = 0\)
• \(A - B = 2\)

By equating coefficients, problems which initially seem complicated become simple algebra. For example, these two equations lead to the conclusion \(A = 1\) and \(B = -1\). This step is not only crucial in partial fraction decomposition but also in many areas of algebra and calculus, where clarity in understanding variable relationships is required.