Problem 50 - College Algebra Essentials

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 fractions decomposition of the rational expression \(\frac{1}{{x^{2}-a x-b x+a b}}\) is \(\frac{1}{a-b}\frac{1}{{x-a}} -\frac{1}{b-a}\frac{1}{{x-b}}\).

Step 1: Factorize the denominator

First, rewrite the denominator of the rational expression in a simpler form. Here, \(x^{2}-a x-b x+a b\) can be rewritten as \(x^{2}-x(a+b)+ab\). It is recognizable as a quadratic equation, and can be factorized further into \(x-a)(x-b)\). Hence, the rational expression becomes \(\frac{1}{{(x-a)(x-b)}}\).

Step 2: Decompose the expression into partial fractions

Next, decompose the rational expression into smaller fractions, where each is called a partial fraction. If \(x-a\) and \(x-b\) are distinct real roots for the quadratic equation in the denominator, then the partial fraction decomposition for \(\frac{1}{{(x-a)(x-b)}}\) is \(\frac{A}{{x-a}} + \frac{B}{{x-b}}\).

Step 3: Find the coefficients A and B

To find the coefficients \(A\) and \(B\), equate the numerator of the partial fractions to the numerator of the original expression, before factoring out imbedded constants. The equation will yield a system of linear equations which can be solved for \(A\) and \(B\). Here, as the numerator of the original fraction is 1, the equations are \(1 = A(x-b) + B(x-a)\). This expression is true for all \(x\). So each coefficient can be found by considering suitable values of \(x\). If \(x=a\), \(1 = A(a-b)\) and hence \(A=\frac{1}{a-b}\). If \(x=b\), \(1 = B(b-a)\) and hence \(B=-\frac{1}{b-a}\).

Step 4: Write down the partial fraction decomposition

Final step is to write down the answer. The partial fraction decomposition of the given fraction is hence \(\frac{1}{a-b}\frac{1}{{x-a}} -\frac{1}{b-a}\frac{1}{{x-b}}\).

Key Concepts

Understanding Rational Expressions

Rational expressions can be thought of as fractions where both the numerator and the denominator are polynomials. Much like fractions, rational expressions can be simplified, added, subtracted, multiplied, and divided. Simplifying complex rational expressions often involves the technique of partial fraction decomposition, which breaks them down into simpler fractions that are easier to work with, especially when it comes to integration or finding limits in calculus.

Continuous success with rational expressions requires a deep understanding of polynomial identities and factorization. Breaking down complex denominators, as seen in our example problem, allows us to reconfigure the expression into a sum or difference of simpler fractions, each with a linear denominator, which are much more manageable for further mathematical operations. This eventually assists in revealing the deeper characteristics of the expression.

Factorizing Polynomials

Factorizing polynomials is a key skill for working with rational expressions, and it involves expressing a polynomial as the product of its factors. Identifying and applying different factorization techniques such as grouping, square of a binomial, difference of squares, sum and difference of cubes, and others is essential.

In the provided exercise, the denominator is a quadratic polynomial, which can typically be factored into a product of two first-degree binomials. Unraveling the polynomial into its factorized form enables the application of partial fraction decomposition by setting the stage for identifying the partial fractions and determining their coefficients.

Systems of Linear Equations

In the context of partial fraction decomposition, systems of linear equations come into play when solving for the unknown coefficients in the partial fractions. After decomposing the expression, we get an equation in which the coefficients are the only unknowns. To isolate these coefficients, we commonly use methods like substitution, elimination, or even matrix operations such as row reduction.

In simpler cases, as shown in the textbook example, the system may consist of just one or two equations that can be quickly solved by strategically choosing values for the variable that simplify the equation. This is a cornerstone of algebra and is applied widely across various fields of mathematics and science.