Question

Resolve \(\frac{1}{x^{2}-9}\) into partial fractions. (1) \(\frac{1}{3(x-3)}-\frac{1}{3(x+3)}\) (2) \(\frac{1}{2(x-3)}-\frac{3}{2(x+3)}\) (3) \(\frac{1}{6(x-3)}-\frac{1}{6(x+3)}\) (4) \(\frac{1}{6(x-3)}+\frac{1}{6(x+3)}\)

Short answer

Answer: The correct partial fraction decomposition of \(\frac{1}{x^2 - 9}\) is \(\frac{1}{6(x-3)}-\frac{1}{6(x+3)}\).

Step-by-step solution

Factor the denominator

First, factor the denominator of the given fraction: \(x^2 - 9 = (x - 3)(x + 3)\).

Set up the partial fraction decomposition

Now, set up the partial fraction decomposition: \(\frac{1}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3}\), where A and B are constants that we need to determine.

Clear the denominators

To determine the constants A and B, we can clear the denominators by multiplying both sides of the equation by the common denominator, which is \((x - 3)(x + 3)\). This gives us \(1 = A(x + 3) + B(x - 3)\).

Solve for the coefficients

Now, we need to solve for A and B. We can do this by substituting an x-value that makes one of the terms in the equation equal to zero. (1) Substitute \(x = 3\): We get \(1= A(6)\). Then \(A = \frac{1}{6}\). (2) Substitute \(x = -3\): We get \(1 = B(-6)\). Then \(B = -\frac{1}{6}\).

Write the partial fractions

Now that we have the values for A and B, we can write the partial fraction decomposition: \(\frac{1}{x^2 - 9} = \frac{\frac{1}{6}}{x - 3} - \frac{\frac{1}{6}}{x + 3}\), which can be simplified to: \(\frac{1}{x^2 - 9} = \frac{1}{6(x - 3)} - \frac{1}{6(x + 3)}\). The correct decomposition is Option (3): \(\frac{1}{6(x-3)}-\frac{1}{6(x+3)}\).

Key concepts

Algebra

In algebra, we encounter various types of equations and expressions. A crucial area here is breaking down complex expressions into simpler parts. When given a rational expression involving polynomials, partial fraction decomposition is a useful method.
This technique is especially helpful when solving integration problems or simplifying expressions for easier manipulation.

• Algebraic manipulation involves operations like addition, subtraction, multiplication, and division of algebraic expressions.
• Understanding these operations lays the groundwork for working effectively with partial fractions.

With strong algebra basics, handling rational expressions becomes manageable, paving the way for more advanced problem-solving scenarios.

Rational Expressions

Rational expressions are fractions where the numerator and the denominator are polynomials. They are treated similarly to regular fractions with constants, but involve variables like x. Manipulating these expressions can simplify complex algebraic tasks.

• In our problem, \(\frac{1}{x^2 - 9}\) is a rational expression. The numerator is 1 (a polynomial of degree 0), and the denominator is a polynomial of degree 2.
• By breaking down the denominator into factors, we prepare the expression for partial fraction decomposition.
• The pre-factorization form can be cumbersome to work with, but transforming it into simpler partial fractions makes it easier to analyze or integrate later on.

Mastering the basics of rational expressions is integral to solving higher-level algebra problems such as integration or equations involving fractions.

Denominator Factoring

Factoring is the process of breaking down a complex expression into a product of simpler expressions. In our exercise, we focus on factoring the denominator of the rational expression.
To resolve the expression \(\frac{1}{x^2 - 9}\), we first recognize that the denominator \(x^2 - 9\) is a difference of squares. This can be factored into \( (x - 3)(x + 3) \).

• The difference of squares is represented by \(a^2 - b^2 = (a - b)(a + b)\).
• In this case, \(a = x\) and \(b = 3\), thus swift factorization simplifies our expression.

This step is crucial as it directly allows us to set up the partial fractions, preparing us for the subsequent step of solving the coefficients.

Coefficient Comparison

Once the denominator is factored and the form of the partial fractions is established, the next step is to determine the values of the unknown coefficients. For our expression, these coefficients are denoted by A and B.
We set up the equation: \(\frac{1}{(x - 3)(x + 3)} = \frac{A}{x - 3} + \frac{B}{x + 3}\).
To find A and B, we clear the denominators:

• Multiply both sides by the common denominator \( (x - 3)(x + 3) \) to eliminate the fractions.
• This simplifies the equation to \( 1 = A(x + 3) + B(x - 3) \), a linear equation in terms of x.

Choosing suitable values for x, specifically \(x = 3\) and \(x = -3\), allows us to simplify and solve for A and B independently, as they are constructed to zero out one of the terms each in the equation:
(1) For \(x = 3\), \(A = \frac{1}{6}\)
(2) For \(x = -3\), \(B = -\frac{1}{6}\)
This method ensures a swift resolution, even when dealing with more complex expressions.