Problem 3 - A Graphical Approach to College Algebra

Question

Find the partial fraction decomposition for each rational expression. $$\frac{4 x+2}{(x+2)(2 x-1)}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{6}{5(x+2)} + \frac{8}{5(2x-1)} \).

Step 1: Identify the Denominator

The first step in partial fraction decomposition is to express the denominator of the given rational expression in terms of its factors. Here, the denominator is already factored as \((x+2)(2x-1)\). Therefore, no further factoring is necessary at this point.

Step 2: Set Up the Partial Fraction Form

Since the denominator \((x+2)(2x-1)\) consists of two distinct linear factors, the partial fraction decomposition will take the form:\[\frac{A}{x+2} + \frac{B}{2x-1}\]where \(A\) and \(B\) are constants to be determined.

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \((x+2)(2x-1)\) to eliminate the denominator:\[4x + 2 = A(2x-1) + B(x+2)\]

Step 4: Expand and Simplify

Expand the right side of the equation:\[A(2x - 1) + B(x + 2) = 2Ax - A + Bx + 2B\] which simplifies to:\[(2A + B)x + (-A + 2B)\]

Step 5: Set Up the System of Equations

Equate the coefficients from the equation on both sides:1. \(2A + B = 4\)2. \(-A + 2B = 2\)This system of equations will help us solve for \(A\) and \(B\).

Step 6: Solve for A and B

First, solve the first equation for \(B\):\[B = 4 - 2A\]Substitute \(B\) in the second equation:\[-A + 2(4 - 2A) = 2\]This simplifies to:\[-A + 8 - 4A = 2\]\[-5A + 8 = 2\]Solving for \(A\), we get:\[-5A = 2 - 8\]\[-5A = -6\]\[A = \frac{6}{5}\]Now substitute \(A\) back to find \(B\):\[B = 4 - 2\left(\frac{6}{5}\right)\]\[B = 4 - \frac{12}{5}\]\[B = \frac{20}{5} - \frac{12}{5}\]\[B = \frac{8}{5}\]

Step 7: Write the Decomposition

Using the values of \(A\) and \(B\), the partial fraction decomposition is:\[ \frac{4x+2}{(x+2)(2x-1)} = \frac{6}{5(x+2)} + \frac{8}{5(2x-1)} \]

Key Concepts

Rational Expression

A rational expression is simply a fraction where both the numerator and the denominator are polynomials. Rational expressions are similar to fractions involving numbers, but instead, they use algebraic expressions. The key component of working with rational expressions is understanding how to handle the algebraic terms in both the numerator and denominator.
In the given exercise, the rational expression is \( \frac{4x+2}{(x+2)(2x-1)} \). This means that the numerator is the polynomial \( 4x + 2 \) and the denominator consists of the factors \((x+2)(2x-1)\). Recognizing that you're working with polynomials allows you to apply algebraic techniques, such as factoring and expansion, to simplify or manipulate these expressions.
Importantly, when working with rational expressions, any values that make the denominator zero must be excluded from the domain, since division by zero is undefined.

Linear Factors

In algebra, linear factors are expressions in the form \( ax + b \), where the power of \( x \) is 1. This means it's a first-degree polynomial. Understanding linear factors is crucial for partial fraction decomposition because they represent simple forms that can't be factored further.
For the rational expression in the exercise \( \frac{4x + 2}{(x+2)(2x-1)} \), the denominator is already broken down into linear factors \((x+2)\) and \((2x-1)\). Each of these linear factors corresponds to a separate term in the partial fraction decomposition.
When you have a denominator composed of distinct linear factors, it simplifies the process because each can be dealt with individually. The decomposition involves expressing the original rational expression as a sum of fractions, each with a linear factor as its denominator.

System of Equations

The system of equations arises in partial fraction decomposition when we need to find constants that make the decomposition valid. Once we set up the partial fraction form of the expression, the next step is solving for these unknown constants.
In our exercise, the partial fraction form \( \frac{A}{x+2} + \frac{B}{2x-1} \) leads to an equation \( 4x + 2 = A(2x-1) + B(x+2) \). When expanded and simplified, this equation results in two separate equations:

• \( 2A + B = 4 \)
• \( -A + 2B = 2 \)

Solving this system of linear equations gives us the values of the constants \( A \) and \( B \). The method typically involves substitution or elimination to isolate one variable and solve for the others.
Understanding how to solve a system of equations is a fundamental skill in algebra, making it easier to handle tasks like partial fraction decomposition efficiently.

Algebraic Expansion

Algebraic expansion involves multiplying out expressions to eliminate parentheses and simplify terms. This process is integral to finding partial fraction decompositions because it helps set up a system of equations by equating coefficients.
In our problem, after setting up the partial fraction form \( A(2x-1) + B(x+2) \), the right-hand side needs to be expanded to eliminate brackets:

• First, \( A(2x-1) \) expands to \( 2Ax - A \)
• Then, \( B(x+2) \) expands to \( Bx + 2B \)

When combined, this results in \( (2A + B)x + (-A + 2B) \). This expanded form allows you to directly compare coefficients with the left side of the equation, \( 4x + 2 \), leading to a system of linear equations that can be solved for \( A \) and \( B \).
By practicing algebraic expansion, you develop a deeper understanding of polynomial relationships and manipulation, critical skills in advanced algebra topics like partial fraction decomposition.