Problem 8 - Calculus - AP Edition

Question

Write out the form of the partial fraction decomposition. (Do not find the numerical values of the coefficients.) $$\frac{1-3 x^{4}}{(x-2)\left(x^{2}+1\right)^{2}}$$

Step-by-Step Solution

Answer

\(\frac{A}{x-2} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2}\)

Step 1: Identify the Denominator Factors

The denominator \((x-2)(x^2+1)^2\) consists of linear and quadratic factors. We need to consider decomposition over each factor.

Step 2: Decompose the Linear Factor

For the linear factor \((x-2)\), we write a single fraction with an unknown numerator. The numerator is a constant:\[ \frac{A}{x-2} \]

Step 3: Decompose the Quadratic Factor

For the quadratic factor \((x^2+1)^2\), since it's repeated and irreducible, we need fractions for each power of the factor:\[ \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \]Each numerator is of one degree less than the denominator.

Step 4: Combine the Fractions

Combine all terms from the decompositions of each factor:\[ \frac{A}{x-2} + \frac{Bx+C}{x^2+1} + \frac{Dx+E}{(x^2+1)^2} \]This is the partial fraction decomposition form for the given expression.

Key Concepts

Linear Factors

In the realm of partial fraction decomposition, linear factors are one of the simplest elements to deal with. A linear factor takes the form of \(x-a\), where \(a\) is a constant. For instance, in the given exercise, \(x-2\) is identified as a linear factor.
To decompose a fraction with a linear factor in the denominator, you need to assign a constant numerator to it. This helps simplify the fraction considerably. So, the form of this simple decomposition is \( \frac{A}{x-2} \), where \(A\) is an unknown constant.

Remember that only linear denominators result in the numerator being a constant, as they are of degree one. Linear factors do not involve squares, cubes, or higher powers, making them straightforward in terms of decomposition.
Understanding these factors is crucial for setting up the fraction correctly during partial fraction decomposition.

Quadratic Factors

Quadratic factors are a bit more complex than linear factors. As seen in the exercise, quadratic factors have the form \(ax^2+bx+c\). A classic example here is \((x^2+1)\), which is found in the original expression's denominator.
Since it's an irreducible quadratic, the numerator that corresponds to this quadratic factor must be of one degree less. This means a linear expression of form \(Bx + C\). Thus, when decomposing, the fraction for the quadratic factor looks like this: \( \frac{Bx+C}{x^2+1} \).

Things get more intricate when dealing with repeated quadratic factors, such as \((x^2+1)^2\). For each power of the factor, you must add another term.

• The first term corresponds to \((x^2+1)\) with a numerator of \(Bx+C\).
• The second term corresponds to \((x^2+1)^2\) with a numerator of \(Dx+E\).

The terms \(Dx+E\) and \(Bx+C\) ensure that the decomposition remains accurate across different power levels of the quadratic factor. It's about balancing the equation to fit within the structure of partial fraction decomposition.

Denominator Identification

Identifying the structure of the denominator is the pivotal first step in tackling partial fraction decomposition. In our example, the denominator is \( (x-2)(x^2+1)^2 \).
This part of the exercise involves recognizing both linear and quadratic components within the denominator.

First, identify each distinct factor:

• Linear Factor: \(x-2\)
• Quadratic Factor: \((x^2+1)\) (appearing twice)

The nature of these factors will guide the decomposition process. Linear factors suggest simple fraction forms, while quadratic factors, especially repeated ones, require more complex setups. These setups ensure that each factor's multiplicity and type are correctly accounted for in the partial fractions.
Accurate recognition of the denominator's factors is key to establishing the correct setup for numerator construction and guide further simplification.