Problem 6 - Precalculus with Calculus Previews

Question

Write out the appropriate form of the partial fraction decomposition of the given rational expression. Do not evaluate the coefficients. $$ \frac{4}{x^{3}\left(x^{2}+1\right)} $$

Step-by-Step Solution

Answer

The decomposition is \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx + E}{x^2 + 1}\).

Step 1: Identify the Denominator Types

First, observe that the denominator is composed of the term \(x^3(x^2 + 1)\), which contains a repeated linear factor \(x^3\) and an irreducible quadratic factor \(x^2 + 1\).

Step 2: Set up the Partial Fraction Decomposition Template

For the repeated linear factor \(x^3\), we decompose into fractions: \(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}\). For the irreducible quadratic \(x^2 + 1\), we decompose into a fraction of the form \(\frac{Dx + E}{x^2 + 1}\).

Step 3: Write the Full Partial Fraction Decomposition

Combine the fractions to write the full decomposition: \[ \frac{4}{x^3(x^2 + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx + E}{x^2 + 1} \]. This represents the structure without solving for the coefficients.

Key Concepts

Repeated Linear Factors

When we talk about repeated linear factors in partial fraction decomposition, we're dealing with a scenario where a factor in the denominator is raised to a power greater than one. In our example, that's the term \( x^3 \). This means that \( x \) is considered a repeated linear factor because it appears three times in the product.
To properly decompose a rational expression with a repeated linear factor like this, we need to account for each occurrence of the factor.
Thus, we break it down into:

• \( \frac{A}{x} \) - for the first appearance of \( x \)
• \( \frac{B}{x^2} \) - capturing the second appearance of \( x \)
• \( \frac{C}{x^3} \) - for the third appearance

This ensures that all contributions of the factor are considered, allowing us to separately handle the influence each power of \( x \) has on the expression.

Irreducible Quadratic Factors

Irreducible quadratic factors, like \( x^2 + 1 \) in our expression, cannot be simplified further by factoring. These are polynomial factors of degree two that do not have real roots.
Since they cannot be reduced using real numbers, we treat them differently in decomposition. The representation involves both a linear term and a constant term in the numerator, which covers all potential outcomes as these factors could impact calculations.
Thus, for an irreducible quadratic factor, our fractional form becomes \( \frac{Dx + E}{x^2 + 1} \).
Here, \( Dx + E \) provides the necessary flexibility for capturing the nuances of any interaction this quadratic factor might have with the rest of the expression.

Rational Expressions

Rational expressions are essentially fractions where both the numerator and the denominator are polynomials. In mathematics, they offer a broad toolbox for expressing relationships and solving equations.
The entire idea behind partial fraction decomposition is to simplify complex rational expressions into simpler, more manageable terms, each of which is easy to integrate, differentiate, or manipulate.
In our exercise, the rational expression \( \frac{4}{x^3(x^2 + 1)} \) benefits from decomposition by translating the single fraction into a sum of individual, simpler fractions.

• Each smaller fraction corresponds to a part of the original denominator.
• This approach provides clarity and easier computational access when solving equations involving these expressions.

Decomposing rational expressions like this is invaluable for understanding the components involved and makes dealing with complex mathematics accessible.