Problem 16 - Algebra and Trigonometry

Question

\(13-44=\) Find the partial fraction decomposition of the rational function. $$ \frac{x+6}{x(x+3)} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Denominator

The given rational function is \( \frac{x+6}{x(x+3)} \). The denominator \( x(x+3) \) indicates that we should decompose it into two fractions, one with \( x \) and another with \( x+3 \) as the denominators.

Step 2: Setup the Partial Fractions

Assume the partial fraction decomposition as \[ \frac{x+6}{x(x+3)} = \frac{A}{x} + \frac{B}{x+3} \]Here, \( A \) and \( B \) are constants we need to find.

Step 3: Clear the Denominator

Multiply both sides of the equation by the common denominator \( x(x+3) \) to eliminate the fractions:\[ x + 6 = A(x+3) + Bx \].

Step 4: Expand and Collect Like Terms

Expand the equation:\[ x + 6 = Ax + 3A + Bx \]Combine like terms on the right side:\[ x + 6 = (A + B)x + 3A \].

Step 5: Equate Coefficients

Equate the coefficients from both sides:1. For \( x \): \( 1 = A + B \)2. For the constant term: \( 6 = 3A \)

Step 6: Solve the System of Equations

From the second equation, solve for \( A \):\[ 6 = 3A \Rightarrow A = 2 \]Substitute \( A = 2 \) in the first equation:\[ 1 = 2 + B \Rightarrow B = -1 \]

Step 7: Write the Partial Fraction Decomposition

Substitute \( A = 2 \) and \( B = -1 \) into the partial fractions:\[ \frac{x+6}{x(x+3)} = \frac{2}{x} - \frac{1}{x+3} \].

Key Concepts

Rational Functions

A rational function is essentially a fraction where both the numerator and the denominator are polynomials. These types of functions are crucial in understanding various mathematical models and equations. In our given example, the rational function is \( \frac{x+6}{x(x+3)} \). It exhibits a combination of terms in the numerator and is a product of terms in the denominator. Dealing with rational functions often involves simplifying them or decomposing them into simpler parts for easier manipulation and understanding.

Partial fraction decomposition is a technique used for breaking down complex rational functions into simpler fractions. This is particularly useful for integration and solving equations. By analyzing the denominator, we identify the distinct terms, which guide us in setting up the partial fractions. Each part of the denominator, like \( x \) and \( x+3 \), signifies a separate fraction component. Recognizing this assists us in managing even intricate functions comfortably.

• Polynomials in the numerator and denominator define rational functions.
• Decomposition aids in breaking down the equation for easier application of calculus skills.
• Understanding the structure is key to simplifying mathematical problems.

Algebra

Algebra forms the backbone of manipulating and solving rational functions. It involves performing operations and rearranging expressions to reach an equation's solution. In our step-by-step solution, algebraic techniques are utilized extensively; for example, setting up equations by equating coefficients and solving for unknown constants like \( A \) and \( B \).

Clearing the denominator is a classic algebraic step. By multiplying through by the denominator \( x(x+3) \), we transition from a fraction to a polynomial equation. This makes it easier to apply algebraic methods such as expansion, rearrangement, and isolating terms. Throughout these processes, algebraic critical thinking becomes essential. Properly expanding and combining like terms helps in accurately setting up the necessary equations.

• Algebra helps in organizing complex expressions into simpler components.
• Techniques like clearing fractions aid in transforming the problem into a manageable form.
• Strategic thinking is crucial for solving by aligning expressions correctly.

Coefficient Comparison

The method of coefficient comparison is a powerful algebraic technique particularly relevant in partial fraction decomposition. Once the fraction is expressed as the sum of its components, matching the coefficients of like terms on both sides of the equation becomes the next logical step. This rules-based logic ensures that the solution satisfies the original equation across all terms.

In our solution, after clearing the fractions and expanding the expressions, we compare the coefficients for \( x \) and the constant term. This method involves methodically equating the terms on both sides. By equating the \( x \) coefficient from \( Ax + Bx \) with the coefficient of \( x \) from \( x+6 \), and similarly for the constant term, we obtain a system of equations allowing us to solve for \( A \) and \( B \).

Using coefficient comparison provides clarity and a systematic approach to solving complex expressions. It is a step-by-step validation process that ensures all aspects of the original equation are satisfied by the derived components.

• This approach is vital for validating the decomposition process.
• Accurate comparison allows for solving the unknown constants.
• Keeps the process methodical and ensures no aspect is overlooked.