/*! 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 16 Find the partial fraction decomp... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the partial fraction decomposition for each rational expression. \(=\) $$\frac{6 x^{5}+7 x^{4}-x^{2}+2 x}{3 x^{2}+2 x-1}$$

Short Answer

Expert verified
Set up partial fractions and solve for coefficients. Result: \[\frac{A}{3x - 1} + \frac{B}{x + 1}\].

Step by step solution

01

Factor the Denominator

First, factor the denominator of the rational expression. Given the denominator is a quadratic polynomial, factor it: \[3x^2 + 2x - 1\].
02

Setup Partial Fraction Decomposition

Since the denominator factors as \[ (3x - 1)(x + 1) \], set up the partial fraction decomposition form as: \[\frac{A}{3x - 1} + \frac{B}{x + 1} \].
03

Clear the Denominator

Multiply both sides of the equation by the original denominator \[(3x^2 + 2x - 1)\] to clear the fractions: \[ 6x^5 + 7x^4 - x^2 + 2x = A(x + 1) + B(3x - 1) \].
04

Expand and Collect Like Terms

Expand the right side of the equation and collect like terms: \[ A(x + 1) + B(3x - 1) = Ax + A + 3Bx - B \].
05

Solve for Coefficients

Match the coefficients of corresponding powers of x on both sides of the equation to form a system of linear equations to solve for A and B.

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Key Concepts

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

Factor the Denominator
When working with partial fraction decomposition, the first step is to factor the denominator of the rational expression. This is crucial because it breaks down the problem into simpler parts.
In our exercise, we started with the rational expression \[\frac{6x^{5}+7x^{4}-x^{2}+2x}{3x^{2}+2x-1}\].

The denominator is \[3x^2 + 2x - 1\], which is a quadratic polynomial. To factor it, we look for two numbers that multiply to \(-3 \times 1 = -3\) and add to \2\. These numbers are 3 and -1. Then, we rewrite the quadratic as \[(3x - 1)(x + 1)\].

This factoring step makes the decomposition process much more manageable, as it allows for the rational expression to be split into simpler fractions.
Rational Expression
A rational expression is a fraction where both the numerator and the denominator are polynomials. In our problem, \[\frac{6x^{5}+7x^{4}-x^{2}+2x}{3x^{2}+2x-1}\] is the rational expression we are working with.

Identifying the numerator and the denominator is essential in partial fraction decomposition, as the goal is to express the rational expression as a sum of simpler fractions. By decomposing it into simpler parts, we simplify the process of integrating or differentiating these expressions.

For our specific problem, once the denominator has been factored into \[(3x - 1)(x + 1)\], it allows us to rewrite the rational expression in a form that is easier to manage, specifically \[\frac{A}{3x - 1} + \frac{B}{x + 1}\]. This form helps isolate coefficients that can be solved via algebraic methods.
System of Linear Equations
After setting up the partial fraction decomposition, the next step involves clearing the denominator by multiplying both sides of the equation by the original denominator. This results in an equation without fractions:
\[ 6x^5 + 7x^4 - x^2 + 2x = A(x + 1) + B(3x - 1) \].

By expanding and collecting like terms, you will obtain:
\[ Ax + A + 3Bx - B \].

At this point, you form a system of linear equations by matching the coefficients of corresponding powers of x from both sides of the equation. This gives you a set of equations to solve for the unknowns A and B.

Solving this system will provide the values for A and B, completing the partial fraction decomposition. For our example, equating the coefficients for each power of x allows us to solve for these variables systematically, ensuring accuracy in the decomposition process. This method of using a system of linear equations is crucial for exact solutions in partial fraction problems.

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