91Ó°ÊÓ

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 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.

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 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.