Problem 35 - Precalculus: Mathematical for Calculus

Question

Find the partial fraction decomposition of the rational function. $$\frac{x-3}{x^{3}+3 x}$$

Step-by-Step Solution

Answer

The partial fraction decomposition is \( \frac{-1}{x} + \frac{x + 1}{x^2 + 3} \).

Step 1: Factor the Denominator

First, we need to factor the denominator of the given rational function. The expression is \( x^3 + 3x \). We can factor out an \( x \) from the terms to get: \( x(x^2 + 3) \). Therefore, the factorization of the denominator is \( x(x^2 + 3) \).

Step 2: Set Up the Partial Fractions

Since the denominator has a linear factor \( x \) and an irreducible quadratic factor \( x^2 + 3 \), the partial fraction decomposition will be of the form:\[\frac{x-3}{x(x^2 + 3)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 3}\]where \( A \), \( B \), and \( C \) are constants to be determined.

Step 3: Clear the Fractions

Multiply both sides of the equation by the common denominator \( x(x^2 + 3) \) to clear the fractions:\[x - 3 = A(x^2 + 3) + (Bx + C)x\]Simplify the expression on the right.

Step 4: Expand and Collect Like Terms

Expand the expression:\[x - 3 = Ax^2 + 3A + Bx^2 + Cx\]Combine like terms:\[x - 3 = (A + B)x^2 + Cx + 3A\]

Step 5: Set Up a System of Equations

By equating the coefficients of the powers of \( x \) from both sides of the equation, we get:- For \( x^2 \): \( A + B = 0 \)- For \( x \): \( C = 1 \)- Constant term: \( 3A = -3 \)

Step 6: Solve the System of Equations

From \( 3A = -3 \), solve for \( A \):\( A = -1 \)Substitute \( A = -1 \) into \( A + B = 0 \):\(-1 + B = 0 \Rightarrow B = 1 \)We already know \( C = 1 \).

Step 7: Write the Partial Fraction Decomposition

Using \( A = -1 \), \( B = 1 \), and \( C = 1 \), substitute back into the partial fractions:\[\frac{x-3}{x(x^2 + 3)} = \frac{-1}{x} + \frac{x + 1}{x^2 + 3}\]

Key Concepts

Rational Functions

Rational functions are functions represented as the ratio of two polynomials. They take the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials and \( Q(x) eq 0 \). These functions are significant in various fields, including calculus and algebra, due to their varied applications.Understanding rational functions is key to performing operations like addition, subtraction, integration, and especially decomposition through partial fractions. The given rational function in our problem, \( \frac{x-3}{x^{3}+3x} \), is a classic example. Recognizing the setup of rational functions allows us to apply suitable algebraic techniques, such as factoring and decomposing, to simplify expressions and solve related equations.

Factorization

Factorization is the process of breaking down expressions into a product of simpler parts, often called factors. In the context of rational functions, especially in partial fraction decomposition, factorization of the denominator is crucial.In our exercise, we factor the denominator \( x^3 + 3x \). First, we observe that both terms share a common factor of \( x \). We factor it out:

• \( x(x^2 + 3) \)

This factorization reveals a linear factor \( x \) and a quadratic factor \( x^2 + 3 \). Recognizing these factors helps us determine the form of the partial fractions. Understanding how to factor polynomial expressions is a fundamental skill in algebra that aids in simplifying calculations and solving equations.

System of Equations

Setting up and solving a system of equations is an essential step in determining the coefficients in partial fraction decomposition. After factoring, we express the rational function in terms of partial fractions with unknown coefficients.For our specific rational function, this involves equating the linear polynomial \( x - 3 \) to the expression created by our partial fraction setup:

• \( x - 3 = A(x^2 + 3) + (Bx + C)x \)

By expanding and collecting terms, we get an equation in which coefficients of like terms must match:

• \( (A + B)x^2 + Cx + 3A \) must equal \( x - 3 \)

From this, set individual equations:

• \( A + B = 0 \)
• \( C = 1 \)
• \( 3A = -3 \)

Solving these equations yields the values for \( A \), \( B \), and \( C \), which are then used in the decomposition.

Polynomial Decomposition

Polynomial decomposition, in terms of partial fraction decomposition, refers to expressing a rational function as a sum of simpler fractions. This decomposition is useful for integration and simplifies the analysis of rational functions by breaking them into more manageable parts.The standard approach requires identifying the factors of the denominator and assigning a typical form to the numerator of each fraction. In our problem:

• Linear factor: \( \frac{A}{x} \)
• Quadratic factor: \( \frac{Bx + C}{x^2 + 3} \)

These forms are set based on the nature of the factors (linear vs. quadratic). The aim of the decomposition is to find values for \( A \), \( B \), and \( C \) such that the original rational function equals the sum of its parts:

• \( \frac{x-3}{x(x^2 + 3)} = \frac{-1}{x} + \frac{x + 1}{x^2 + 3} \)

Decomposition enables easier integration and understanding of the behavior of rational functions across their domains.