Problem 48 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{a x+b}{x^{2}-c^{2}} \quad(c \neq 0)$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the rational expression \(\frac{a x+b}{x^{2}-c^{2}}\) is \(\frac{A}{x-c} + \frac{B}{x+c}\), where A and B are constants determined by solving the two equations obtained by considering cases \(x=c\) and \(x=-c\).

Step 1: Factorizing the denominator

In the first step, we factorize the denominator which is the difference of two squares. We can use the formula for factoring the difference of squares: \(a^2 - b^2 = (a-b)(a+b)\). Thus, the denominator \(x^{2}-c^{2}\) can be factorized as \((x-c)(x+c)\). So, we rewrite our function as: \(\frac{a x+b}{(x-c)(x+c)}\).

Step 2: Writing the function as Partial Fraction Decomposition

With the denominator factorized, it’s then possible to rewrite the initial fraction as the sum of two fractions, with each of the factors as a denominator. We write the function as: \(\frac{a x+b}{(x-c)(x+c)} = \frac{A}{x-c} + \frac{B}{x+c}\), where A and B are constants to be determined.

Step 3: Resolving for constants A and B

We resolve for constants A and B by multiplying the above equation by \((x-c)(x+c)\) which gives us \(ax + b = A(x+c) + B(x-c)\). In order to find the values of A and B, we can consider the cases \(x=c\) and \(x=-c\). From these cases, we can solve the two equations to find the values of A and B.

Key Concepts

Rational Expressions

Rational expressions are like fractions where the numerator and the denominator are polynomials. In simpler terms, it's a ratio or a division between two polynomials. For example, in the expression \(\frac{ax+b}{x^2-c^2}\), the numerator is \(ax+b\), and the denominator is \(x^2-c^2\). Understanding rational expressions is crucial as they form the basis for partial fraction decomposition. This process allows us to break down complex rational expressions into simpler ones, making them easier to integrate or differentiate in calculus.

Factoring

Factoring is a mathematical process where you express a polynomial as the product of its simplest parts or factors. In the context of the given exercise, we deal with the expression \(x^2-c^2\). This expression can be recognized as a difference of squares, which follows the formula \(a^2-b^2 = (a-b)(a+b)\). Thus, \(x^2-c^2\) can be factored into \((x-c)(x+c)\). Factoring is an essential step in partial fraction decomposition as it simplifies the denominator, allowing us to separate the single fraction into multiple simpler fractions with different denominators.

Difference of Squares

The difference of squares is a specific type of polynomial that can be factored using the formula \(a^2-b^2 = (a-b)(a+b)\). It appears often in algebra and calculus, and recognizing it can greatly simplify your work. In our expression, \(x^2-c^2\), \(x^2\) and \(c^2\) are perfect squares. Applying the difference of squares formula, we rewrite the expression as \((x-c)(x+c)\). Recognizing patterns like the difference of squares allows for efficient factoring, which is essential in simplifying rational expressions for further operations.

Constants Determination

After factoring the denominator, the next step is to determine the constants in the partial fraction decomposition. Once you have expressed the rational fraction as \(\frac{ax+b}{(x-c)(x+c)} = \frac{A}{x-c} + \frac{B}{x+c}\), you need to find the values for constants \(A\) and \(B\). To do this, multiply through by the common denominator \((x-c)(x+c)\) to eliminate the fractions: \(ax + b = A(x+c) + B(x-c)\). By substituting specific values of \(x\), such as \(x=c\) and \(x=-c\), you solve for \(A\) and \(B\) effectively. This determination of constants is a crucial step, as it completes the decomposition process, allowing for easy integration or simplification of the expression.