Problem 12 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

\(\frac{2x}{(x-1)(x+1)} = \frac{1}{x-1} + \frac{1}{x+1}\)

Step 1: Identify the Denominator

The denominator of the rational function is \((x-1)(x+1)\). Notice that it consists of linear factors, \(x-1\) and \(x+1\). This will guide us into setting up our partial fractions.

Step 2: Set Up the Partial Fraction Decomposition

We set the original function equal to a sum of two fractions: \(\frac{2x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}\). Here, \(A\) and \(B\) are constants that we need to find.

Step 3: Clear the Denominator

Multiply both sides by \((x-1)(x+1)\) to clear the denominators: \[2x = A(x+1) + B(x-1)\] This equation must hold for all \(x\).

Step 4: Expand and Simplify

Expand the expression on the right-hand side: \[ 2x = Ax + A + Bx - B\]Combine like terms: \[ 2x = (A+B)x + (A-B)\]

Step 5: Solve for Coefficients

Match coefficients from both sides of the equation. For the \(x\) terms: \[ 2 = A + B \]For the constant terms: \[ 0 = A - B \]

Step 6: Calculate A and B

Solve the system of equations:1. \( A + B = 2 \)2. \( A - B = 0 \)Adding these equations, we get:\[ 2A = 2 \] \( A = 1 \)Substitute \(A = 1\) into the first equation: \[ 1 + B = 2 \] \( B = 1 \)

Step 7: Write the Partial Fraction Decomposition

With \(A = 1\) and \(B = 1\), the partial fraction decomposition is: \[ \frac{2x}{(x-1)(x+1)} = \frac{1}{x-1} + \frac{1}{x+1} \]

Key Concepts

Rational Function

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. In our example, we have \[ \frac{2x}{(x-1)(x+1)} \] The numerator is simply \(2x\), and the denominator is the product of two linear factors, \((x-1)(x+1)\).

• Rational functions arise in many areas of mathematics, including calculus and algebra, because they can model a variety of real-world situations.
• One of their notable features is their behavior at their roots or points of undefined values, which often leads to vertical asymptotes in graphing.

Understanding rational functions is essential, as they provide the base for various processes such as integration by partial fraction decomposition, which is used to simplify complex fractions.

Linear Factors

Linear factors are components that can be multiplied to give a polynomial. In the denominator of our rational function \((x-1)(x+1)\), we observe two linear factors.

• A linear factor is generally of the form \(ax + b\), where \(a\) and \(b\) are constants. In this case, both \(x-1\) and \(x+1\) are linear factors.
• This implies that both \(x-1\) and \(x+1\) are simple roots of the denominator where each factor equates to zero.

Recognizing these factors simplifies the approach for partial fraction decomposition, allowing us to set up simpler fractions to work with. This method of simplifying rational expressions is invaluable, especially when performing calculus operations like differentiation and integration.

Coefficients

Coefficients are the numerical parts of the terms in an expression or equation. To find the partial fraction decomposition of our rational function, we introduced coefficients \(A\) and \(B\) in the equation:\[ \frac{2x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \]

• The coefficients \(A\) and \(B\) are values we solve for in order to express the fraction as a sum of simpler fractions.
• They are crucial for ensuring that the two sides of the equation remain equivalent.

To determine these coefficients, we clear the denominators by multiplying through the equation, leading us to a system of equations which guide us in finding the correct values.

System of Equations

A system of equations is a collection of equations that are solved together. In the context of partial fraction decomposition, we used a system of equations to resolve the coefficients \(A\) and \(B\). From our setup:\[ 2x = A(x+1) + B(x-1) \]we expanded to \[ 2x = (A+B)x + (A-B) \]

• The system of equations is derived by equating coefficients of like terms from both sides of the equation. For the problem in question, we had:
• \( A + B = 2 \)
• \( A - B = 0 \)

By solving these two linear equations simultaneously, we found that \(A = 1\) and \(B = 1\). Systems of equations are fundamental in algebra and used extensively to find unknown variables when decomposing fractions or solving applied problems in different fields of study.