Problem 31 - A Graphical Approach to Precalculus with Limits

Question

Find the partial fraction decomposition for each rational expression. $$\frac{3 x-1}{x\left(2 x^{2}+1\right)^{2}}$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Denominator Terms

The denominator of the expression is \(x(2x^2 + 1)^2\). The first term \(x\) is a linear factor, and \((2x^2 + 1)^2\) is a repeated quadratic factor.

Step 2: Set the Partial Fraction Decomposition Format

Since the denominator has a linear factor \(x\) and a repeated quadratic factor \((2x^2 + 1)^2\), the decomposition format is:\[\frac{3x - 1}{x(2x^2 + 1)^2} = \frac{A}{x} + \frac{Bx + C}{2x^2 + 1} + \frac{Dx + E}{(2x^2 + 1)^2}\]

Step 3: Combine Terms to a Single Fraction

Multiply each term on the right by the appropriate factor to reach a common denominator:\[A(2x^2 + 1)^2 + (Bx + C)x(2x^2 + 1) + (Dx + E)x = 3x - 1\] This simplifies to a single fraction with the same denominator as the original expression.

Step 4: Expand and Simplify

Expand each term on the right side:\[A(4x^4 + 4x^2 + 1) + (Bx + C)(2x^3 + x) + (Dx^2 + Ex)\]Ensure all the terms are grouped according to powers of \(x\).

Step 5: Equate the Coefficients

Set the expanded form equal to the numerator \(3x - 1\). Match the coefficients of corresponding powers of \(x\) from both sides:- Coefficient of \(x^4\): \(4A = 0\)- Coefficient of \(x^3\): \(2B = 0\)- Coefficient of \(x^2\): \(A + B = 0\)- Coefficient of \(x\): \(C = 3\)- Constant: \(A = -1\)

Step 6: Solve the Coefficients

From \(4A = 0\), we know \(A = 0\).From \(2B = 0\), \(B = 0\).From \(C = 3\), \(C = 3\).Thus, the decomposition is:\[\frac{3}{x} - \frac{x}{2x^2 + 1}.\]

Key Concepts

Rational Expressions

Rational expressions are like fractions, but they contain polynomial expressions in both the numerator and the denominator. When dealing with rational expressions, it's important to understand that the numerator and the denominator can each have multiple terms, often combining variables and constants. Dealing with these expressions requires a clear understanding of operations with polynomials.

To simplify or decompose rational expressions, one effective strategy is converting them into simpler parts using partial fraction decomposition. This involves breaking the expression into a sum of simpler, separable fractions which are easier to manipulate and understand. Each component or fraction derived from the original is linked to factors in the denominator. This method allows us to handle complex rational expressions by analyzing their individual simpler components.

Repeated Quadratic Factors

Repeated quadratic factors in a rational expression's denominator imply that a quadratic expression appears multiple times as a factor. For instance, the expression \(2x^2 + 1\) in our exercise is squared, indicating it is a repeated quadratic factor.

To handle repeated quadratic factors, the partial fraction decomposition requires a systematic format to represent each term. For example, with \( (2x^2 + 1)^2\), you must include separate terms for each repetition of the factor: one term for \( (2x^2 + 1) \) and another for \( (2x^2 + 1)^2\). This acknowledges the repetition by assigning a fraction to every power of the quadratic factor:

• The first term is \( \frac{Bx + C}{2x^2 + 1} \).
• The second term is \( \frac{Dx + E}{(2x^2 + 1)^2} \).

The numerators typically involve a linear expression because the factors are quadratic, ensuring the decomposition addresses potential roots and retains polynomial integrity in their combinations.

Linear Factors

Linear factors are expressions in the form of a single variable term added or subtracted by a constant, like \( x \) or \( x - 2 \). They are simpler than quadratic factors, making them easier to decompose when using partial fractions.

In the context of partial fraction decomposition, identifying linear factors such as \( x \) is crucial because it tells us how to structure part of the decomposition. For each linear factor, we assign a fraction of the form \( \frac{A}{x} \), where \( A \) is a constant to be determined.

• These linear factor terms are generally the simplest part of the decomposition.
• It's straightforward because each term corresponds directly to a first-degree polynomial.

Using linear factors provides a foundation on which to build the decomposition, simplifying the overall partial fraction decomposition process, and clarifying the structure needed to solve rational expressions.