Problem 17 - Precalculus: Mathematics for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{12}{x^{2}-9}$$

Step-by-Step Solution

Answer

\(\frac{12}{x^2-9} = \frac{2}{x-3} - \frac{2}{x+3}\).

Step 1: Factor the Denominator

The first step is to factor the denominator of the given rational function. The expression is \(x^2 - 9\), which can be factored as \((x-3)(x+3)\) because it is a difference of squares.

Step 2: Set Up the Partial Fraction Form

Using the factored denominator \((x-3)(x+3)\), express the rational function as a sum of partial fractions. Assume \(\frac{A}{x-3} + \frac{B}{x+3}\) where \(A\) and \(B\) are constants to be determined.

Step 3: Combine the Partial Fractions

Bring both fractions to a common denominator and equate this to the original rational function: \[ \frac{A(x+3) + B(x-3)}{(x-3)(x+3)} = \frac{12}{(x-3)(x+3)} \]This leads to the equation:\[ A(x+3) + B(x-3) = 12 \]

Step 4: Solve for Coefficients A and B

Expand the left side to get: \[ Ax + 3A + Bx - 3B = 12 \]Combine like terms: \[ (A + B)x + (3A - 3B) = 12 \]Since there are no \(x\) terms on the right side, we have:1. \(A + B = 0\)2. \(3A - 3B = 12\)

Step 5: Solve the System of Equations

First, solve equation \(A + B = 0\), which means \(A = -B\). Substitute into equation \(3A - 3B = 12\):\[ 3(-B) - 3B = 12 \]\[ -3B - 3B = 12 \]\[ -6B = 12 \]Divide both sides by -6:\[ B = -2 \]Substituting back into \(A = -B\) gives \(A = 2\).

Step 6: Write the Solution

The partial fraction decomposition of the rational function is:\[ \frac{12}{x^2 - 9} = \frac{2}{x-3} + \frac{-2}{x+3} \]

Key Concepts

Rational Functions

A rational function is a ratio of two polynomials. It's expressed as \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomial functions. The denominator \(Q(x)\) must not be zero, as division by zero is undefined. Rational functions are common in mathematics because they can model a wide variety of relationships between variables.

The goal of rational functions in the context of partial fraction decomposition is to express a complex rational function as a sum of simpler fractions. This simplifies the function and makes integration or other operations easier to perform. For this, the denominator of the rational function is typically factored to break it into simpler partial fractions.

Factoring Polynomials

Factoring polynomials is an essential skill in manipulating expressions and solving equations. When factoring a polynomial, we aim to write it as a product of simpler polynomials. This is particularly useful when handling rational functions.

Take the polynomial \(x^2 - 9\). This expression is a difference of squares, a special type of polynomial that can be factored using the identity \(a^2 - b^2 = (a-b)(a+b)\). Here, \(x^2 - 9\) factors into \((x-3)(x+3)\). This step is crucial in preparing the rational function for partial fraction decomposition, as it separates the polynomial into factors that can be addressed individually.

Solving Systems of Equations

Once the denominator of a rational function is factored and a partial fraction setup is proposed, determining the constants in the numerators involves solving a system of equations. This process consists of two main steps: setting up equations based on equivalent expressions and solving them to find unknown values.

For example, in the equation \(A(x+3) + B(x-3) = 12\), the function is expanded and like terms are grouped, producing a linear system:

• \(A + B = 0\)
• \(3A - 3B = 12\)

Solving this system by substitution or elimination helps find the constants \(A\) and \(B\), which completes the partial fraction decomposition. It's important to solve these systems accurately to ensure the decomposition is correct.

Difference of Squares

The difference of squares is a special polynomial expression of the form \(a^2 - b^2\) that factors into \((a-b)(a+b)\). Recognizing this form is extremely helpful in simplifying polynomials and rational expressions.

In the provided problem, \(x^2 - 9\) is clearly a difference of squares, where \(a = x\) and \(b = 3\). This factors into \((x-3)(x+3)\).

• The ability to identify and factor a difference of squares quickly can simplify many algebraic processes.
• Factoring in this way can also aid in solving equations, as it breaks down complex expressions into more manageable pieces.

In summary, mastering the process of identifying and working with the difference of squares will streamline many algebraic tasks, including partial fraction decomposition.