Problem 50 - Algebra and Trigonometry

Question

write the partial fraction decomposition of each rational expression. $$\frac{1}{x^{2}-a x-b x+a b} \quad(a \neq b)$$

Step-by-Step Solution

Answer

The partial fraction decomposition of \( \frac{1}{x^{2}-a x-b x+a b} \) is \( \frac{1}{a-b}\frac{1}{x-a} - \frac{1}{a-b}\frac{1}{x-b}\)

Step 1: Factoring the Denominator

Factorize the denominator of the expression which gives: \(x^2 - ax - bx + ab\) = \(x^2 - (a+b)x + ab\) = \(x-a)(x-b)\). Thus the expression becomes \( \frac{1}{(x-a)(x-b)}\).

Step 2: Writing Down the Partial Fractions

Write the partial fractions for the expression such that each term in the denominator has a corresponding fraction. Considering that the denominator is a product of linear terms, a format of A/(x-a) + B/(x-b) can be used. Thus, \( \frac{1}{(x-a)(x-b)}\) = \( \frac{A}{x-a} + \frac{B}{x-b}\).

Step 3: Setting the Fractions Equal

Set the two fractions as equal and multiply both sides by the common denominator, which obtains: 1 = A(x - b) + B(x - a). Given that this equation must hold for all x, it can be solved for the constants A and B by substituting suitable values. That can be done with ease by first setting x=a, which eliminates B, and then setting x=b, which eliminates A.

Step 4: Solving for A

Substitute x=a into the equation, giving: 1 = A(a - b). Thus, solving for A gets \( A = \frac{1}{a - b}\).

Step 5: Solving for B

Secondly, substitute x=b into the equation, which gives: 1 = B(b - a). Solving for B gives \( B = \frac{-1}{a - b}\). As a result, note that B is the negative of A, which should be expected since (b - a) = -(a - b).

Key Concepts

Factoring

Factoring is the process of breaking down an expression into simpler multiple expressions that, when multiplied together, give back the original expression. It is an essential step in partial fraction decomposition.

In our exercise, we were given the denominator \(x^2 - ax - bx + ab\). The first step involves factoring this quadratic expression. Quadratics often follow patterns that help in factoring:

• We notice that \(x^2 - ax - bx + ab\) can be reorganized as \(x^2 - (a+b)x + ab\).
• This is a typical form of a factorable quadratic expression, \((x-m)(x-n)\), where \(m + n = -(a+b)\) and \(m \cdot n = ab\).

In our case, factoring directly gives us \((x-a)(x-b)\). Factoring simplifies the expression, making partial fraction decomposition possible.

Linear Terms

Linear terms are expressions where the variable is raised to the first power. They appear as single powers of \(x\), such as \(x-a\) or \(x-b\).

In partial fraction decomposition, the linear terms from the factored denominator become the basis for each fraction. With our example, after factoring, we get linear terms \((x-a)\) and \((x-b)\).

Each linear term forms the denominator of a new fraction in the partial fractions sum. The idea is to express the original rational expression as a sum of simpler fractions, each with a linear term in the denominator: \(\frac{A}{x-a}\) and \(\frac{B}{x-b}\).

This step transforms the complicated rational expression into something more manageable, readying it for integration or further manipulation.

Denominator

The denominator is everything below the fraction bar in a rational expression. It defines the limitations of the expression and is crucial in operations like decomposition.

In our given problem, the denominator \(x^2 - ax - bx + ab\) needed factoring to progress with partial fraction decomposition.

Factoring reveals the true structure of the denominator; once factored as \((x-a)(x-b)\), it is clear that the denominator consists of two linear factors. These factors indicate the expressions that will become part of the partial fractions.

Understanding the role of the denominator introduces us to important mathematical concepts like restriction points, where the expression is undefined (e.g., \(x=a\) or \(x=b\)). Recognizing these helps avoid errors in solving and applying the expression.

Constants

Constants act as fixed values that complete an equation but are not influenced by the variable part of the expression. They are critical in forming the equations needed to solve for the unknowns in partial fraction decomposition.

In our problem, once the partial fractions \(\frac{A}{x-a} + \frac{B}{x-b}\) are identified, the next step involves determining the constants \(A\) and \(B\).

We set the equation equal to the original expression's numerator by clearing the denominators: \(1 = A(x-b) + B(x-a)\). By substituting convenient values of \(x\) (specifically \(x=a\) and \(x=b\)), we eliminate one constant at a time to find:

• \(A = \frac{1}{a-b}\)
• \(B = \frac{-1}{a-b}\).

These constants complete the decomposition, providing weights for each of the partial fractions.