Problem 11 - Algebra and Trigonometry

Question

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: Determine the Denominators for Partial Fractions

The given rational function is \( \frac{2}{(x-1)(x+1)} \). Notice that the denominator is already factored as \( (x-1)(x+1) \). Each factor is a simple linear factor.

Step 2: Set Up the Partial Fraction Decomposition Form

For a rational function where the denominator is the product of distinct linear factors, like \( (x-1)(x+1) \), we can express it as a sum of partial fractions. Set it up as: \( \frac{A}{x-1} + \frac{B}{x+1} \).

Step 3: Combine the Fractions

Combine the partial fractions to a single fraction: \( \frac{A}{x-1} + \frac{B}{x+1} = \frac{A(x+1)+B(x-1)}{(x-1)(x+1)} \). The goal is for this to equal the original function's numerator when combined.

Step 4: Create and Simplify the Equation

Next, equate the numerators because the denominators are already equal: \( A(x+1) + B(x-1) = 2 \). Expand this to: \( Ax + A + Bx - B = 2 \). This simplifies to \( (A+B)x + (A-B) = 2 \).

Step 5: Solve the System of Equations

Equate the coefficients with those on the right-hand side: \( A + B = 0 \) and \( A - B = 2 \). Solve the equations simultaneously. Adding them gives \( 2A = 2 \), so \( A = 1 \). Substituting \( A = 1 \) into \( A + B = 0 \) gives \( 1 + B = 0 \), thus \( B = -1 \).

Step 6: Write the Decomposition

Hence, the partial fraction decomposition of \( \frac{2}{(x-1)(x+1)} \) is \( \frac{1}{x-1} - \frac{1}{x+1} \).

Key Concepts

Rational Functions

Rational functions form the basis of partial fraction decomposition. They are mathematical expressions represented as a fraction of two polynomials:

• the numerator is a polynomial of any degree,
• the denominator is a polynomial of degree one or higher.

In our example, \[ \frac{2}{(x-1)(x+1)} \]the numerator is the number 2, a polynomial of degree 0, while the denominator is the product of two linear polynomials, \((x-1)(x+1)\). Rational functions are key in calculus, allowing us to integrate more complex expressions by breaking them down into simpler parts. They can resemble division between numbers but involve more complex factors. Whenever dealing with a rational function, it's essential to understand components like the distinct factors in the denominator, aiding in further processes like partial fraction decomposition.

Linear Factors

Linear factors are the building blocks when decomposing rational functions into partial fractions. A linear factor is simply a polynomial of degree 1, usually expressed as \( (x-a) \).In our exercise, the rational function's denominator has two linear factors:

• \( x-1 \)
• \( x+1 \)

What makes linear factors advantageous in partial fraction decomposition is their simplicity. They provide a straightforward path towards expressing any rational function in a separate sum of fractions. Each linear factor in the denominator becomes the denominator of a partial fraction term, where the numerators in these fractions are constants that we solve for. Recognizing and working with distinct linear factors simplifies the process, ensuring a smoother decomposition.

Algebraic Equations

In solving partial fraction decompositions, algebraic equations come into play when finding the unknown constants in the numerators. Once the fraction has been set up by using linear factors, as in:\[ \frac{A}{x-1} + \frac{B}{x+1} = \frac{A(x+1)+B(x-1)}{(x-1)(x+1)} \]we equate the numerators:\[ A(x+1) + B(x-1) = 2 \]To find the values of \( A \) and \( B \), we simplify and rearrange this equation:

• Expand to: \( Ax + A + Bx - B \)
• Combine like terms: \( A+B \) and \( A-B \)
• Compare both sides: resulting in equations \( A + B = 0 \) and \( A - B = 2 \)

Solving these simultaneous equations involves fundamental algebraic skills. We identify the constants by making both sides of the equality the same. Thus, algebraic manipulation provides the values necessary for the partial fraction breakdown: \( A = 1 \), \( B = -1 \), resulting in the final decomposition.