/*! 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 25 Find the decomposition of the pa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the decomposition of the partial fraction for the repeating linear factors. \(\frac{5-x}{(x-7)^{2}}\)

Short Answer

Expert verified
Decomposition: \(\frac{-1}{x-7} + \frac{-2}{(x-7)^2}\).

Step by step solution

01

Identify the Partial Fraction Form

The given expression is \(\frac{5-x}{(x-7)^2}\). For a repeated linear factor \((x-a)^2\), the partial fraction decomposition will be \(\frac{A}{x-7} + \frac{B}{(x-7)^2}\).
02

Write the Equation

Set up the equation \(\frac{5-x}{(x-7)^2} = \frac{A}{x-7} + \frac{B}{(x-7)^2}\). This states that the rational expression is equal to the sum of the two partial fractions.
03

Clear the Denominator

Multiply both sides by \((x-7)^2\) to remove the denominator: \(5-x = A(x-7) + B\).
04

Expand and Simplify

Expand the expression on the right-hand side: \(A(x-7) = Ax - 7A\). So we get \(5-x = Ax - 7A + B\).
05

Equate and Solve for Coefficients

Compare coefficients: 1. For the \(x\) terms: \(-1 = A\)2. For the constant terms: \(5 = -7A + B\)Substitute \(A = -1\) into the second equation: \(5 = -7(-1) + B\). This gives \(5 = 7 + B\), so \(B = -2\).
06

Write the Final Decomposition

The decomposition is \(\frac{-1}{x-7} + \frac{-2}{(x-7)^2}\).

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.

Repeated Linear Factors
Repeated linear factors occur when a polynomial denominator in a rational expression includes a factor that is raised to a power greater than one. This is represented as
  • e.g., \((x-a)^n\) where \(n\) is greater than 1.
  • This means the same factor, \((x-a)\), is multiplied by itself \(n\) times.
When dealing with repeated linear factors, partial fraction decomposition requires a unique setup. Each instance of the repeated factor will contribute a separate term to the decomposition, with varying powers of the factor in the denominator.For example, for a factor like \((x-7)^2\), the decomposition form is:
  • \(\frac{A}{x-7}\) for the first power
  • \(\frac{B}{(x-7)^2}\) for the second power
In our original problem, \((x-7)^2\) required terms \(\frac{A}{x-7}\) and \(\frac{B}{(x-7)^2}\) to represent the repeated factors appropriately.This method efficiently handles every level of repetition and ensures the rational expression can be decomposed into simpler parts.
Decomposition of Rational Expressions
Decomposition of rational expressions involves breaking down a complicated rational expression into a sum of simpler fractions.The goal is to rewrite the expression in a way that is easier to work with, especially for integration or solving equations.The expression is split into partial fractions, each with its own distinct denominator.In our exercise, \(\frac{5-x}{(x-7)^2}\), we are working with a repeated linear factor.Using partial fraction decomposition, we set:
  • \(\frac{5-x}{(x-7)^2} = \frac{A}{x-7} + \frac{B}{(x-7)^2}\)
This method allows us to solve for constant coefficients, \(A\) and \(B\), by eliminating the denominators and comparing coefficients.The decomposition includes distinct fractions, each taking a part of the original expression:
  • The breakdown into easier terms means specific coefficients \(A\) and \(B\) need to be calculated to ensure the decomposition equals the original expression.
This step-by-step calculation makes the expressions manageable and simpler for further operations, like integration, where dealing with smaller fractions is advantageous.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and arithmetic operations (addition, subtraction, multiplication, and division).These expressions can represent all sorts of mathematical concepts and problems.In calculus, algebraic expressions are often manipulated into different forms to simplify calculation and analysis.The original expression \(\frac{5-x}{(x-7)^2}\)is an example where algebra is used to restructure for the purpose of simplification.Understanding algebraic expressions involves various operations:
  • Identifying terms involving variables and their coefficients – like recognizing \(-x\) and its coefficient in the expression is \(-1\).
  • Grouping terms methodically by their degree (e.g., linear, quadratic).
  • Applying fundamental operations to simplify or alter the expressions for practical use in solving or integrating.
By mastering these skills, students can efficiently deal with any given rational expression and solve it by partial fraction decomposition or other algebraic methods.Ultimately, algebra forms the backbone that supports more advanced concepts in mathematics, making understanding and working with expressions essential for mathematical proficiency.

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

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. Three roommates shared a package of 12 ice cream bars, but no one remembers who ate how many. If Tom ate twice as many ice cream bars as Joe, and Albert ate three less than Tom, how many ice cream bars did each roommate eat?

For the following exercises, use this scenario: A health-conscious company decides to make a trail mix out of almonds, dried cranberries, and chocolate- covered cashews. The nutritional information for these items is shown in Table 1 . $$ \begin{array}{|c|c|c|c|} \hline & \text { Fat (g) } & \text { Protein (g) } & \text { Carbohydrates (g) } \\ \hline \text { Almonds (10) } & 6 & 2 & 3 \\ \hline \text { Cranberries (10) } & 0.02 & 0 & 8 \\ \hline \text { Cashews (10) } & 7 & 3.5 & 5.5 \\ \hline \end{array} $$ For the "energy-booster" mix, there are 1,000 pieces in the mix, containing \(145 \mathrm{~g}\) of protein and \(625 \mathrm{~g}\) of carbohydrates. If the number of almonds and cashews summed together is equivalent to the amount of cranberries, how many of each item is in the trail mix?

For the following exercises, set up the augmented matrix that describes the situation, and solve for the desired solution. Every day, a cupcake store sells \(5,000\) cupcakes in chocolate and vanilla flavors. If the chocolate flavor is 3 times as popular as the vanilla flavor, how many of each cupcake sell per day?

For the following exercises, find the inverse of the given matrix. $$\left[\begin{array}{cccc}{1} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1} \\\ {0} & {1} & {1} & {0} \\ {0} & {0} & {1} & {1}\end{array}\right]$$

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. Anna, Ashley, and Andrea weigh a combined 370 \(\mathrm{lb}\) . If Andrea weighs 20 \(\mathrm{lb}\) more than Ashley, and Anna weighs 1.5 times as much as Ashley, how much does each girl weigh?

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.