Problem 9 - Precalculus: Mathematical for Calculus

Question

Write the form of the partial fraction decomposition of the function (as in Example 4). Do not determine the numerical values of the coefficients. $$\frac{x^{3}+x+1}{x(2 x-5)^{3}\left(x^{2}+2 x+5\right)^{2}}$$

Step-by-Step Solution

Answer

Partial fraction decomposition: \(\frac{A}{x} + \frac{B}{2x-5} + \frac{C}{(2x-5)^2} + \frac{D}{(2x-5)^3} + \frac{Ex+F}{x^2+2x+5} + \frac{Gx+H}{(x^2+2x+5)^2}\).

Step 1: Identify the Denominator

The denominator is given by the product of the factors: \(x\), \((2x-5)^3\), and \((x^2+2x+5)^2\). Each factor will be associated with a term in the partial fraction decomposition.

Step 2: Set Up Partial Fractions for Linear Factors

For the linear factor \(x\), include a term \(\frac{A}{x}\) in the decomposition. For the factor \((2x-5)^3\), include terms \(\frac{B}{2x-5}\), \(\frac{C}{(2x-5)^2}\), and \(\frac{D}{(2x-5)^3}\) due to its repeating nature.

Step 3: Set Up Partial Fractions for the Irreducible Quadratic Factor

For the irreducible quadratic factor \((x^2+2x+5)^2\), include terms \(\frac{Ex+F}{x^2+2x+5}\) and \(\frac{Gx+H}{(x^2+2x+5)^2}\). Each term involves a linear expression since the quadratic cannot be factored further.

Step 4: Combine the Terms

Combine all the identified terms from the previous steps to form the partial fraction decomposition: \(\frac{A}{x} + \frac{B}{2x-5} + \frac{C}{(2x-5)^2} + \frac{D}{(2x-5)^3} + \frac{Ex+F}{x^2+2x+5} + \frac{Gx+H}{(x^2+2x+5)^2}\).

Key Concepts

Linear Factors

In partial fraction decomposition, linear factors are the simplest to work with. A linear factor is a polynomial factor of degree one, typically seen as something like \( ax + b \). In our exercise, we encountered the linear factor \( x \).
To address linear factors in partial fraction decomposition, you typically include a term for each linear factor. For \( x \), we simply add a term of the form \( \frac{A}{x} \). Here, \( A \) is a constant coefficient that would be determined if we were solving for specific values, but in this exercise, we won't find it.
When you have linear factors raised to a power, such as \((2x-5)^3\), it becomes slightly more complex. Each power level requires a separate fraction:

• \( \frac{B}{2x-5} \) for the factor itself,
• \( \frac{C}{(2x-5)^2} \) for the square,
• \( \frac{D}{(2x-5)^3} \) for the cube.

This approach ensures each part of the original denominator is represented in the partial fraction decomposition, which allows us to decompose the expression fully.

Quadratic Factors

A quadratic factor is a polynomial of degree two. These factors can sometimes be reduced to simpler, linear factors if they have real roots. However, not all quadratic factors can be easily reduced.
In the exercise, we dealt with \((x^2 + 2x + 5)\). Since this expression does not factor into linear terms with real coefficients, it stands as a quadratic factor.
For each irreducible quadratic factor in the denominator, you include a partial fraction with a linear expression in the numerator. For example, due to \((x^2 + 2x + 5)^2\), you need to include:

• \( \frac{Ex+F}{x^2+2x+5} \)
• \( \frac{Gx+H}{(x^2+2x+5)^2} \)

The linear numerators \( Ex+F \) and \( Gx+H \) help capture any potential linear behavior present in the original rational function. This form accommodates parts of the function where a simple constant won't suffice.

Irreducible Quadratic

An irreducible quadratic factor is a second-degree polynomial that cannot be factored further into linear terms over the real numbers. Its roots are complex. In the exercise, \( x^2 + 2x + 5 \) serves as an example of an irreducible quadratic factor.
When dealing with irreducible quadratics in partial fraction decomposition, each one requires a fraction that uses a linear expression in the numerator. This is because the irreducible nature indicates more complexity in the factor. For the term \( x^2 + 2x + 5 \), our decomposition will look like this:

• \( \frac{Ex+F}{x^2+2x+5} \)
• \( \frac{Gx+H}{(x^2+2x+5)^2} \)

The linear terms \( Ex+F \) and \( Gx+H \) ensure we can account for first-degree components that might be present in the numerator of the original fraction. These are essential steps to provide a complete decomposition.