Problem 9 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

\( \frac{4x+2}{(x+2)(2x-1)} = \frac{6/5}{x+2} + \frac{8/5}{2x-1} \)

Step 1: Understand the Problem

We have the rational expression \( \frac{4x+2}{(x+2)(2x-1)} \). This expression needs to be broken down into simpler fractions, each with a denominator that is a factor of \((x+2)(2x-1)\).

Step 2: Set Up the Partial Fraction Form

Since the denominator \((x+2)(2x-1)\) has two linear factors, the decomposition can be set up as \( \frac{A}{x+2} + \frac{B}{2x-1} \), where \(A\) and \(B\) are constants that we need to find.

Step 3: Multiply by the Common Denominator

Multiply through by \((x+2)(2x-1)\) to eliminate the denominators: \[ 4x+2 = A(2x-1) + B(x+2) \]

Step 4: Expand and Simplify the Equation

Expand the right side of the equation: \[ 4x + 2 = (2A)x - A + Bx + 2B \]Combine like terms: \[ 4x + 2 = (2A + B)x + (-A + 2B) \]

Step 5: Create and Solve System of Equations

By equating the coefficients from both sides of the equation, we get the system: 1. \(2A + B = 4\)2. \(-A + 2B = 2\)Solve these equations. From equation 1, express \(B = 4 - 2A\). Substitute in equation 2: \[ -A + 2(4-2A) = 2 \]Simplify and solve for \(A\): \[ -A + 8 - 4A = 2 \] \[ -5A = -6 \] \[ A = \frac{6}{5} \]

Step 6: Find B Using the Value of A

Substitute \(A = \frac{6}{5}\) back into \(B = 4 - 2A\): \[ B = 4 - 2 \left(\frac{6}{5}\right) \] \[ B = 4 - \frac{12}{5} \] \[ B = \frac{20}{5} - \frac{12}{5} \] \[ B = \frac{8}{5} \]

Step 7: Write the Partial Fraction Decomposition

The partial fraction decomposition of \( \frac{4x+2}{(x+2)(2x-1)} \) is \[ \frac{6/5}{x+2} + \frac{8/5}{2x-1} \]

Key Concepts

Rational Expressions

In mathematics, a rational expression is a fancy name for a fraction where both the numerator and the denominator are polynomials. Just like ordinary fractions, rational expressions can be simplified, multiplied, divided, added, or subtracted. It's crucial that the denominator is never zero.
This is because division by zero is undefined in mathematics. For example, the expression \( \frac{4x+2}{(x+2)(2x-1)} \) is a rational expression, where \(4x+2\) is the numerator and \((x+2)(2x-1)\) is the polynomial denominator.

• Simplifying Rational Expressions: To simplify any rational expression, factor both the numerator and the denominator and cancel any common factors if there are any.
• Partial Fraction Decomposition: This is a method that expresses a given rational expression as a sum of simpler fractions. These simpler fractions are often easier to integrate or differentiate.

Understanding the nature of rational expressions is the first step in mastering more complex algebraic manipulations, including partial fraction decomposition.

Linear Factors

Linear factors are expressions of the form \(ax + b\), where \(a\) and \(b\) are constants. In the context of partial fraction decomposition, it's important to identify linear factors of the denominator in a rational expression.

When we encounter a fraction like \( \frac{4x+2}{(x+2)(2x-1)} \), the denominator \((x+2)(2x-1)\) consists of two linear factors \(x+2\) and \(2x-1\). Each term in the partial fraction decomposition will have these linear factors in their denominators.

• Step involved: Recognize the denominator of the expression as a product of linear factors.
• Form of decomposition: Set up the decomposition as \( \frac{A}{x+2} + \frac{B}{2x-1} \), where \(A\) and \(B\) are constants.

This setup is what allows us to solve for the unknowns \(A\) and \(B\) and rewrite the original expression as a sum of simpler fractions.

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. Solving a system of equations involves finding values of the unknowns that satisfy all the equations simultaneously.
In partial fraction decomposition, we use the setup from our decomposition to create such a system. From the expression \( \frac{4x+2}{(x+2)(2x-1)} \), after eliminating the denominator and expanding, you get:

• The expression \(4x + 2 = A(2x-1) + B(x+2)\) simplifies to \( (2A + B)x + (-A + 2B) \).
• By comparing coefficients on both sides, we derive the system of equations:
  1. \(2A + B = 4\)
  2. \(-A + 2B = 2\)

Solving this system involves methods such as substitution or elimination. These steps give us the values for \(A\) and \(B\):

• First, solve one equation for one of the variables.
• Substitute back into the other equation to find the other variable.

This step is pivotal because it allows us to express the original rational expression in its decomposed form, ultimately aiding in further mathematical operations like integration or more complex algebra.