Problem 4 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

The partial fraction decomposition is \(\frac{-1}{2(x+1)} + \frac{3}{2(x-1)}\).

Step 1: Identify the form of the partial fraction decomposition

In this expression, the denominator \((x+1)(x-1)\) consists of two distinct linear factors. Thus, the partial fraction decomposition will be of the form \(\frac{A}{x+1} + \frac{B}{x-1}\).

Step 2: Multiply through by the common denominator

Multiply both sides by the common denominator \((x+1)(x-1)\) to eliminate the fractions. This gives: \[x + 2 = A(x-1) + B(x+1)\]

Step 3: Expand the right side

Distribute the terms on the right side:\[x + 2 = Ax - A + Bx + B\] This simplifies to: \[x + 2 = (A + B)x + (B - A)\]

Step 4: Compare coefficients

Equate the coefficients from both sides of the equation to find A and B. From \(x\) terms: \[A + B = 1\]From constant terms: \[B - A = 2\]

Step 5: Solve the system of equations

We have the system of equations: 1. \(A + B = 1\) 2. \(B - A = 2\) Add these two equations to eliminate \(A\): \[A + B + B - A = 1 + 2 \]This simplifies to \(2B = 3\), so \(B = \frac{3}{2}\). Substitute back into the first equation: \[A + \frac{3}{2} = 1\]Thus, \(A = 1 - \frac{3}{2} = -\frac{1}{2}\).

Step 6: Write the partial fraction decomposition

Substitute the values of \(A\) and \(B\) back into the partial fraction: \[\frac{x+2}{(x+1)(x-1)} = \frac{-\frac{1}{2}}{x+1} + \frac{\frac{3}{2}}{x-1}\] This can be simplified to: \[\frac{x+2}{(x+1)(x-1)} = \frac{-1}{2(x+1)} + \frac{3}{2(x-1)}\]

Key Concepts

Rational Expression

A rational expression is simply a fraction where both the numerator and the denominator are polynomials. It extends the concept of fractions to algebraic expressions. Consider \[\frac{x+2}{(x+1)(x-1)}\]. Here, the numerator is \(x + 2\) and the denominator is the product of two polynomials, \(x+1\) and \(x-1\). This makes it a rational expression.

Rational expressions can have variables in both the numerator and the denominator, unlike regular fractions that only have numeric terms. Understanding how to manipulate these expressions is crucial in algebra, especially when simplifying them, finding common denominators, or performing operations like addition and subtraction.

To further simplify or decompose these expressions, we often use techniques like partial fraction decomposition, especially when solving calculus problems or integrating rational functions.

It's important to remember that rational expressions are undefined when the denominator equals zero. This concept leads to restrictions in the domain of these functions, often appearing in the form of vertical asymptotes in graphs.

Linear Factors

Linear factors are expressions of the form \(ax + b\), where both \(a\) and \(b\) are constants. They represent lines when graphed, hence the name 'linear'.

In the partial fraction decomposition of a rational expression, identifying linear factors is crucial because they determine the form of the decomposition. For the expression \(\frac{x+2}{(x+1)(x-1)}\), the denominator is already factored into two distinct linear factors: \((x+1)\) and \((x-1)\).

Each linear factor corresponds to a separate fraction in the decomposition, for which constants \(A\), \(B\), etc., are solved:

• \(\frac{A}{x+1}\)
• \(\frac{B}{x-1}\)

Understanding linear factors is not only fundamental in partial fraction decomposition but also in other areas like solving quadratic equations, graphing linear equations, and analyzing polynomial functions.

The unique characteristics of each factor, such as their roots, influence how the rational expression behaves or is simplified.

System of Equations

A system of equations consists of two or more equations with the same set of variables. Solving a system means finding values for the variables that satisfy all equations simultaneously.

In partial fraction decomposition, we often set up a system of equations to determine the unknown constants in the decomposed fractions. Let's consider the equations from the example:

• \(A + B = 1\)
• \(B - A = 2\)

This system arises when we compare the coefficients from both sides of the expanded equation, derived after equating and combining terms.

We solve this system using substitution or elimination methods. For these equations, adding them eliminates \(A\), allowing us to solve for \(B\). Then, substituting back to find \(A\) gives us the complete solution.

This technique not only applies to partial fraction decomposition but is also widely used in fields like physics, economics, and engineering, wherever multiple conditions are imposed on a set of variables.