/*! 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 4 Expand the quotients in Exercise... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Expand the quotients in Exercises \(1-8\) by partial fractions. $$\frac{2 x+2}{x^{2}-2 x+1}$$

Short Answer

Expert verified
The partial fraction decomposition is \( \frac{2}{x-1} + \frac{4}{(x-1)^2} \).

Step by step solution

01

Identify the Type of Partial Fraction Decomposition

The given function is a fraction \( \frac{2x+2}{x^2-2x+1} \). First, observe that the denominator \( x^2-2x+1 \) is a perfect square, specifically \( (x-1)^2 \). Since the denominator is a quadratic repeated factor, the partial fraction decomposition will be of the form \( \frac{A}{x-1} + \frac{B}{(x-1)^2} \).
02

Setup the Partial Fraction Equation

Set up the partial fraction equation by writing: \( \frac{2x+2}{x^2-2x+1} = \frac{A}{x-1} + \frac{B}{(x-1)^2} \).
03

Eliminate the Denominator

Multiply both sides by \( (x-1)^2 \) to eliminate the denominators: \( 2x+2 = A(x-1) + B \).
04

Expand and Simplify the Equation

Expand the right side: \( A(x-1) + B = Ax - A + B \). Set this equal to the left side: \( 2x + 2 = Ax - A + B \).
05

Compare Coefficients

By matching the coefficients of corresponding powers of \( x \), we have two equations:1. For \( x \): \( A = 2 \)2. For the constant term: \( -A + B = 2 \).
06

Solve for Constants

We already found \( A = 2 \). Substitute \( A = 2 \) into the second equation: \( -2 + B = 2 \)\( B = 4 \).
07

Write the Final Partial Fraction Decomposition

Substitute \( A \) and \( B \) back into the partial fraction form: \( \frac{2}{x-1} + \frac{4}{(x-1)^2} \).

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.

Quadratic Equation
A quadratic equation is a polynomial equation of degree 2. It has a general form of:\[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Quadratic equations are widely used in various mathematical calculations and can be solved using methods such as factoring, completing the square, or using the quadratic formula.
For the given problem, the denominator \( x^2 - 2x + 1 \) is a quadratic expression. In this particular case, the quadratic expression is unique because it factors into a perfect square, \((x-1)^2\). This factorization helps simplify the problem and guides us to the correct form of partial fraction decomposition.
Perfect Square
A perfect square is a quadratic expression that can be expressed as the square of a binomial. For example, the expression \( (x-1)^2 \) is a perfect square, meaning:\[ (x-1)^2 = x^2 - 2x + 1 \]Identifying a perfect square in partial fraction decomposition is crucial, as it can change the approach you take to find the solution. Recognizing \( x^2 - 2x + 1 \) as \( (x-1)^2 \) simplifies setting up the decomposition since the factor \( x-1 \) is repeated.
Understanding and recognizing perfect squares helps in:
  • Simplifying expressions
  • Reducing algebraic complexity
  • Guiding the correct setup of partial fraction solutions
Coefficient Matching
Coefficient matching is a technique used to solve equations by aligning coefficients of corresponding terms. In the context of partial fraction decomposition, it helps determine the unknowns or constants in the numerator. After expressing the original rational function in its decomposed form, the goal is to align both sides by matching the terms with the same power of \( x \).
In the example solution, multiplying through by \( (x-1)^2 \) results in:\[ 2x + 2 = Ax - A + B \]Here, the coefficients of \( x \) are matched to find \( A \), while constant terms are used to solve for \( B \). From the equation:
  • Term for \( x \): \( A = 2 \)
  • Constant term: \( -A + B = 2 \)
This systematic alignment simplifies solving for unknown constants, providing a clearer path to the final solution.
Rational Functions
A rational function is defined as the quotient of two polynomials. These functions have the form:\[ f(x) = \frac{P(x)}{Q(x)} \]where \( P(x) \) and \( Q(x) \) are polynomial expressions and \( Q(x) eq 0 \).
Rational functions are significant in many applications where relationships between quantities are non-linear.
In the given problem, \( \frac{2x+2}{x^2-2x+1} \) is a rational function, with a numerator \( 2x+2 \) and a denominator \( x^2-2x+1 \). Such functions might present challenges in integration and calculus, paving the way for techniques like partial fraction decomposition. This method facilitates simplification by expressing the function as a sum of simpler fractions, making integration or further analysis more manageable.

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

Failure time The time between failures of a photocopier is exponentially distributed. Half of the copiers at a university require service during the first 2 years of operations. If the university purchased 150 copiers, how many do you expect to require service during the first year of their operation?

The integrals in Exercises \(1-40\) are in no particular order. Evaluate each integral using any algebraic method or trigonometric identity you think is appropriate, and then use a substitution to reduce it to a standard form. $$ \int \frac{x^{2}}{x^{2}+1} d x $$

Lifetime of light bulbs A manufacturer of light bulbs finds that the mean lifetime of a bulb is 1200 hours. Assume the life of a bulb is exponentially distributed. $$ \begin{array}{l}{\text { a. Find the probability that a bulb will last less than its guaranteed }} \\ {\text { lifetime of } 1000 \text { hours. }} \\\ {\text { b. In a batch of light bulbs, what is the expected time until half }} \\\ {\text { the light bulbs in the batch fail? }}\end{array} $$

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{2}^{\infty} \frac{d x}{\sqrt{x^{2}-1}}$$

The instructions for the integrals in Exercises \(1-10\) have two parts, one for the Trapezoidal Rule and one for Simpson's Rule. I. Using the Trapezoidal Rule a. Estimate the integral with \(n=4\) steps and find an upper bound for \(\left|E_{T}\right| .\) b. Evaluate the integral directly and find \(\left|E_{T}\right|\) c. Use the formula (( true value)) \(\times 100\) to express \(\left|E_{T}\right|\) as \right.\right. a percentage of the integral's true value. II. Using Simpson's Rule a. Estimate the integral with \(n=4\) steps and find an upper bound for \(\left|E_{S}\right| .\) b. Evaluate the integral directly and find \(\left|E_{S}\right|\) c. Use the formula ( \(\left|E_{S}\right| /(\) true value) \() \times 100\) to express \(\left|E_{S}\right|\) as a percentage of the integral's true value. $$ \int_{1}^{2} x d x $$

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.