Problem 4 - Precalculus

Question

In Exercises \(1-6\), find only the form needed to begin the process of partial fraction decomposition. Do not create the system of linear equations or attempt to find the actual decomposition. $$ \frac{a x^{2}+b x+c}{x^{3}(5 x+9)\left(3 x^{2}+7 x+9\right)} $$

Step-by-Step Solution

Answer

Start with: \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{5x+9} + \frac{Ex+F}{3x^2+7x+9} \).

Step 1: Identify the Type of Denominator Factors

The given fraction has a denominator with a combination of linear and quadratic factors. Specifically, it includes a repeated linear factor, \(x^3\), a linear factor, \(5x+9\), and an irreducible quadratic factor, \(3x^2+7x+9\).

Step 2: Assign Partial Fraction Forms

Since the denominator contains different types of factors, assign a partial fraction to each. For the repeated linear factor \(x^3\), use: \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} \). For the linear factor \(5x+9\), use: \( \frac{D}{5x+9} \). For the irreducible quadratic factor \(3x^2+7x+9\), use: \( \frac{Ex+F}{3x^2+7x+9} \).

Step 3: Write the Partial Fraction Decomposition Form

Combine these elements to write the partial fraction form. The expression for the partial fraction decomposition is: \[\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{5x+9} + \frac{Ex+F}{3x^2+7x+9} \]

Key Concepts

Linear Factor

In partial fraction decomposition, understanding the role of linear factors is essential. A linear factor is simply a polynomial of degree one, expressed in the form of \(ax + b\). For example, considering the linear factor from the original exercise, \(5x + 9\), we assign a separate term in our partial fraction decomposition based on this linear factor.

When dealing with linear factors in the denominator, we use a simple constant, say \(D\), as the numerator for this term in the decomposition:

• \(\frac{D}{5x + 9}\)

These linear terms in the decomposition reflect how simple linear equations can influence the breakdown of fractions into simpler parts. Thus, focusing on linear factors helps simplify complex rational expressions.

Quadratic Factor

Quadratic factors add another layer of complexity. A quadratic factor is a polynomial of degree two, like \(3x^2 + 7x + 9\) in our exercise. Such factors can't be factored further into linear terms if they are irreducible over the real numbers, which is the case here.

When assigning a term to this sort of factor in the decomposition, it's necessary to use a linear term in the numerator, such as \(Ex+F\). This gives us the fraction:

• \(\frac{Ex+F}{3x^2 + 7x + 9}\)

This approach ensures that the degrees perfectly match the original fractions, and allows solving for multiple unknowns \(E\) and \(F\), making it essential for a correct decomposition.

Denominator Factors

Identifying the factors of the denominator is the first crucial step in partial fraction decomposition. In our exercise, the denominator is \(x^3(5x+9)(3x^2+7x+9)\), which is composed of different factor types:

• Repeated linear factor \(x^3\)
• Linear factor \(5x + 9\)
• Irreducible quadratic factor \(3x^2 + 7x + 9\)

Each of these types dictates the form of the numerators we will use in our partial fractions. Recognizing and categorizing these denominator factors is fundamental to generating the form of the decomposition accurately. Therefore, one should always start by identifying these components to lay the groundwork for subsequent steps.

Repeated Linear Factor

A repeated linear factor is simply a linear factor raised to a power greater than one. In the exercise, \(x^3\) serves as the repeated linear factor, where \(x\) is repeated thrice.

In partial fraction decomposition, the process for repeated linear factors requires assigning multiple terms, with descending powers in the denominators:

• \(\frac{A}{x}\)
• \(\frac{B}{x^2}\)
• \(\frac{C}{x^3}\)

This breakdown helps in matching the structure of the repeated linear term within the original fraction. Here, \(A\), \(B\), and \(C\) are the constants that we will solve for later, accommodating the repetition in the factor. Executing this decomposition accurately is key to resolving more complicated rational equations.