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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. $$ \frac{a x^{2}+b x+c}{x^{3}(5 x+9)\left(3 x^{2}+7 x+9\right)} $$

Short Answer

Expert verified
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 by step solution

01

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, \(3x^2+7x+9\).
02

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: \( \frac{Ex+F}{3x^2+7x+9} \).
03

Write the Partial Fraction Decomposition Form

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Consider the following scenario. In the small village of Pedimaxus in the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, \(90 \%\) of those who subscribe to the Pedimaxus Tribune want to keep getting it, but \(10 \%\) want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, \(80 \%\) want to continue with it and \(20 \%\) want switch to the Tribune. We can express this situation using matrices. Specifically, let \(X\) be the 'state matrix' given by $$ X=\left[\begin{array}{l} T \\ P \end{array}\right] $$ where \(T\) is the number of people who get the Tribune and \(P\) is the number of people who get the Picayune in a given week. Let \(Q\) be the 'transition matrix' given by $$ Q=\left[\begin{array}{ll} 0.90 & 0.20 \\ 0.10 & 0.80 \end{array}\right] $$ such that \(Q X\) will be the state matrix for the next week. Show that \(S Y=X_{s}\) for any matrix \(Y\) of the form $$ Y=\left[\begin{array}{r} y \\ 150-y \end{array}\right] $$ This means that no matter how the distribution starts in Pedimaxus, if \(Q\) is applied often enough, we always end up with 100 people getting the Tribune and 50 people getting the Picayune.

In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{-2 x^{2}+20 x-68}{x^{3}+4 x^{2}+4 x+16} $$

Consider the following scenario. In the small village of Pedimaxus in the country of Sasquatchia, all 150 residents get one of the two local newspapers. Market research has shown that in any given week, \(90 \%\) of those who subscribe to the Pedimaxus Tribune want to keep getting it, but \(10 \%\) want to switch to the Sasquatchia Picayune. Of those who receive the Picayune, \(80 \%\) want to continue with it and \(20 \%\) want switch to the Tribune. We can express this situation using matrices. Specifically, let \(X\) be the 'state matrix' given by $$ X=\left[\begin{array}{l} T \\ P \end{array}\right] $$ where \(T\) is the number of people who get the Tribune and \(P\) is the number of people who get the Picayune in a given week. Let \(Q\) be the 'transition matrix' given by $$ Q=\left[\begin{array}{ll} 0.90 & 0.20 \\ 0.10 & 0.80 \end{array}\right] $$ such that \(Q X\) will be the state matrix for the next week. If you didn't see the pattern, we'll help you out. Let $$ X_{s}=\left[\begin{array}{r} 100 \\ 50 \end{array}\right]. $$ Show that \(Q X_{s}=X_{s}\) This is called the steady state because the number of people who get each paper didn't change for the next week. Show that \(Q^{n} X \rightarrow X_{s}\) as \(n \rightarrow \infty\).

Find the inverse of the given matrix using their determinants and adjoints. $$ B=\left[\begin{array}{rr} 12 & -7 \\ -5 & 3 \end{array}\right] $$

Let \(A\) be an arbitrary invertible \(3 \times 3\) matrix. (a) Show that \(\operatorname{det}\left(I_{3}\right)=1\). (See footnote \(^{8}\) below.) (b) Using the facts that \(A A^{-1}=I_{3}\) and \(\operatorname{det}\left(A A^{-1}\right)=\operatorname{det}(A) \operatorname{det}\left(A^{-1}\right),\) show that $$\operatorname{det}\left(A^{-1}\right)=\frac{1}{\operatorname{det}(A)}$$

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