/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 In Exercises 7 - 18 , find the p... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{-x^{2}+15}{4 x^{4}+40 x^{2}+36} $$

Short Answer

Expert verified
The partial fraction decomposition is \(-\frac{1}{2(x^2 + 3)} + \frac{18}{(2x^2 + 6)^2}\).

Step by step solution

01

Factor the Denominator

The denominator is \(4x^4 + 40x^2 + 36\). Notice that this is a quadratic in terms of \(x^2\). We can rewrite it as \(4(x^2)^2 + 40x^2 + 36\). To factor, use substitution: let \(u = x^2\), then it becomes \(4u^2 + 40u + 36\). Factor it to get \((2u + 6)(2u + 6)\) or \((2x^2 + 6)^2\).
02

Simplify the Denominator

Since \((2x^2 + 6)^2\) can be factored further by factoring \(2\) out giving us \(4(x^2 + 3)^2\), the denominator is now expressed as \((2(x^2 + 3))^2\).
03

Set Up the Partial Fraction Decomposition

Given \(\frac{-x^2 + 15}{4x^4 + 40x^2 + 36}\), the partial fraction decomposition takes the form \(\frac{Ax + B}{2x^2 + 6} + \frac{Cx + D}{(2x^2 + 6)^2}\). This is determined by the repeated factor \(2x^2 + 6\) in the denominator.
04

Clear the Denominators

Multiply through by the common denominator \((2x^2 + 6)^2\) to eliminate the fractions, resulting in the equation: \(-x^2 + 15 = (Ax + B)(2x^2 + 6) + (Cx + D)\).
05

Expand and Combine Terms

Expand the right-hand side: \((Ax + B)(2x^2 + 6) = Ax(2x^2) + Ax(6) + B(2x^2) + B(6) = 2Ax^3 + 6Ax + 2Bx^2 + 6B\). Add \(Cx + D\) to get the complete expansion: \(2Ax^3 + 2Bx^2 + (6A + C)x + (6B + D)\).
06

Solve for Coefficients

Since the left side is \(-x^2 + 15\), match coefficients with the right: for \(-x^2\), we get \(2B = -1\); for constant \(15\), \(6B + D = 15\). There are no other \(x^3\) or \(x^1\) terms, so \(A = 0\) and \(C = 0\). Solve these simultaneous equations to find \(B = -\frac{1}{2}\) and \(D = 18\).
07

Write the Partial Fraction Decomposition

With the coefficients determined, the partial fraction decomposition is: \(\frac{0x - \frac{1}{2}}{2x^2 + 6} + \frac{0x + 18}{(2x^2 + 6)^2}\), which simplifies to \(-\frac{1}{2(x^2 + 3)} + \frac{18}{(2x^2 + 6)^2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rational Expressions
A rational expression is simply a fraction in which both the numerator and denominator are polynomials. Polynomials are algebraic expressions made up of variables and constants connected by addition, subtraction, and multiplication. In the realm of rational expressions, these can be as simple as linear equations or as complex as quadratics and higher-degree polynomials.

The rational expression given, \( \frac{-x^{2}+15}{4x^{4}+40x^{2}+36} \), requires us to manipulate polynomials in both the numerator and the denominator. Our task involves finding a way to break down this complex expression into simpler components through partial fraction decomposition. This is a technique used to rewrite the expression so each part is easier to analyze and integrate if needed.

Understanding the structure and behavior of rational expressions is crucial for solving many algebraic problems. It allows us to perform operations such as addition, subtraction, and integration on complex polynomial expressions.
Factoring
Factoring is a mathematical process of breaking down a complex expression into simpler parts, called factors, which when multiplied together will yield the original expression. It's like solving a puzzle where the pieces come together to form the bigger picture.

For the denominator \(4x^4 + 40x^2 + 36\), factoring helps us identify repetitive components or simpler expressions within a complex polynomial. To start factoring, note that it takes the form of a quadratic in terms of \(x^2\), which can be written as \(4(x^2)^2 + 40x^2 + 36\). By letting \(u = x^2\), the expression simplifies to \(4u^2 + 40u + 36\).

Factoring the quadratic \(u\) gives us \((2u + 6)(2u + 6)\), which eventually results in \((2(x^2 + 3))^2\) when plugged back. This step is crucial for simplifying the decomposition. Factoring simplifies the expression for use in partial fraction decomposition, making it easier to work with later on.
Denominator
In a rational expression, the denominator is the polynomial that sits beneath the fraction bar. It's important because it determines the nature and constraints of the rational expression.

For the expression \(\frac{-x^2 + 15}{4x^4 + 40x^2 + 36}\), the denominator \(4x^4 + 40x^2 + 36\) underwent a transformation through factoring to become \((2(x^2 + 3))^2\). This is a more manageable form revealing that the denominator has a repeated factor. Such repetition directly impacts the form of the partial fraction decomposition.

When dealing with rational expressions, the denominator plays a crucial role in finding limits, asymptotes, and discontinuities. Incorrect handling of the denominator can lead to mistakes in the solution process, particularly in partial fraction decomposition and integration.
Coefficients
Coefficients are the numerical or constant parts of a term that are multiplied by the variable(s) in an algebraic expression. They are fundamental in algebraic manipulation and form the essence of solving equations.

In the context of the given partial fraction decomposition solution, coefficients were identified by expanding the expression and matching terms with corresponding powers of \(x\) in the untouched polynomial. For example, identifying zero coefficients for\(x^3\) and \(x\) helped ensure accuracy in matching it with the expression \(-x^2 + 15\).

Determining each coefficient ensures that the decomposition accurately reflects the original expression. It requires solving simultaneous equations, often employing strategies like comparison of coefficients, resulting in terms like \(2B = -1\) and \(6B + D = 15\).

This step highlights how each coefficient contributes to the overall form of the partial fraction. The proper identification and calculation of coefficients are vital to successfully completing partial fraction decomposition.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A local buffet charges \(\$ 7.50\) per person for the basie buffet and \(\$ 9.25\) for the deluxe buffet (which includes crab legs.) If 27 diners went out to eat and the total bill was \(\$ 227.00\) before taxes, how many chose the basic buffet and how many chose the deluxe buffet?

In Exercises 7 - 18 , find the partial fraction decomposition of the following rational expressions. $$ \frac{2 x^{2}+3 x+14}{\left(x^{2}+2 x+9\right)\left(x^{2}+x+5\right)} $$

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 the conditions do not change from week to week, then \(Q\) remains the same and we have what's known as a Stochastic Process \({ }^{10}\) because Week \(n\) 's numbers are found by computing \(Q^{n} X .\) Choose a few values of \(n\) and, with the help of your classmates and calculator, find out how many people get each paper for that week. You should start to see a pattern as \(n \rightarrow \infty\).

Solve the given system of nonlinear equations. Sketch the graph of both equations on the same set of axes to verify the solution set. $$ \left\\{\begin{aligned} x^{2}+y^{2} &=16 \\ 16 x^{2}+4 y^{2} &=64 \end{aligned}\right. $$

With the help of your classmates, find at least two different row echelon forms for the matrix $$ \left[\begin{array}{rr|r} 1 & 2 & 3 \\ 4 & 12 & 8 \end{array}\right]. $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.