Problem 18 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{4 x^{2}-5 x-15}{x(x+1)(x-5)}$$

Step-by-Step Solution

Answer

The solution to the partial fraction decomposition of \(\frac{4 x^{2}-5 x-15}{x(x+1)(x-5)}\) would be the completed \(\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-5}\) equation after we have solved for \(A\), \(B\), and \(C\) by utilizing the system of linear equations.

Step 1: Initial fraction

We're looking to decompose the fraction \(\frac{4 x^{2}-5 x-15}{x(x+1)(x-5)}\) into its partial fractions. This means to rewrite the fraction in terms of simpler fractions.

Step 2: Identify the factors

The denominator of the fraction can be decomposed into three factors, \(x\), \(x+1\), and \(x-5\). So the initial fraction can be written as the sum of the three fractions \(\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-5}\), where \(A\), \(B\), and \(C\) are the parameters to be found.

Step 3: Set up the equations

Next, we will multiply both sides by the denominator of the initial fraction to get rid of the fraction and set up the equations. We will multiply \(x(x+1)(x-5) ( \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-5}) = 4x^2 - 5x - 15\). This allows us to identify coefficients for like terms on each side of the expression so that we can solve for \(A\), \(B\), and \(C\).

Step 4: Solve the system

By comparing coefficients on left and right side of the equation, we end up with the system of linear equations. This system can be solved through substitution or elimination.

Step 5: Confirm the solution

Once we get the values of \(A\), \(B\), and \(C\), we will substitute these parameters into \(\frac{A}{x} + \frac{B}{x+1} + \frac{C}{x-5}\). This will give us the partial fraction decomposition of the given rational expression.

Key Concepts

Rational Expressions

Rational expressions are like fractions, but instead of integers, you have polynomials in the numerator and the denominator. Take for example, the expression \(\frac{4x^{2}-5x-15}{x(x+1)(x-5)}\), which is the focus of our exercise. To understand these expressions, it's crucial to grasp that they follow the same rules as regular fractions. They can be simplified, expanded, and, notably, decomposed into simpler parts, which is where partial fraction decomposition comes into play.

Partial fraction decomposition aims to break down complex rational expressions into simpler, more manageable fractions that are easier to integrate or differentiate if needed. This task is often encountered in calculus, where managing more straightforward expressions can simplify further operations such as integration. The process involves finding a few constants that will represent the numerators of the 'partial' fractions whose denominators are the factors of the original denominator.

Algebraic Fractions

Algebraic fractions, also known as fractional expressions, are fractions where the numerator or the denominator (or both) include algebraic expressions. Much like the rational expressions, they can seem daunting at first, but they adhere to the same principles as the simple fractions you learn about in primary school.

To work effectively with algebraic fractions, you need to feel comfortable factoring polynomials, finding common denominators, and simplifying expressions. In the context of partial fraction decomposition, an essential skill is rewriting the algebraic fraction as a sum of simpler fractions. In our original exercise, the algebraic fraction contains a quadratic polynomial in the numerator and a cubic polynomial in the denominator. By decomposing this algebraic fraction, we gain a clearer view of its behavior, especially concerning its variables, since the decomposed form involves fractions with linear denominators, easier to interpret and analyze.

System of Linear Equations

At the core of solving a partial fraction decomposition like our initial exercise is the ability to solve a system of linear equations. This is because, after setting up the partial fractions, you'll end up with an equation that, when expanded, gives you several equations relating the constants you need to find.

To solve these systems, you can use various methods such as substitution, elimination, or even matrix operations (in more advanced cases). In the context of partial fraction decomposition, substitution and elimination are most commonly used. They involve rearranging the equations to isolate one variable at a time and progressively solving for each unknown. This is critical because accurately determining these unknowns will allow you to rewrite the original complex fraction as a sum of simpler fractions, vastly simplifying further mathematical operations.