Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 42


The simplified partial fraction expansion is $\frac{\left( -2 \right)}{3\left( x-1 \right)}+\frac{2x+13}{3\left( {{x}^{2}}+x+1 \right)}$.

Work Step by Step

The provided partial fraction expression is as follows: $\frac{3x-5}{{{x}^{3}}-1}$ Simplify this as given below: $\begin{align} & \frac{3x-5}{{{x}^{3}}-1}=\frac{3x-5}{{{x}^{3}}-{{1}^{3}}} \\ & =\frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \end{align}$ Step 1: Set up the partial fraction expansion with unknown constant coefficients and then write a constant coefficient over each of the two distinct algebraic linear factors in the denominator of the expression. $\frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1}$ Step 2: Multiply both sides of the equation by the expression $\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$ and consider the least common denominator. $\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)\times \left( \frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)} \right)=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)\times \left( \frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1} \right)$ Then, multiply and simplify as follows: $\begin{align} & 3x-5=\left( {{x}^{2}}+x+1 \right)A+\left( Bx+C \right)\left( x-1 \right) \\ & 3x-5=\left( {{x}^{2}}+x+1 \right)A+\left( Bx+C \right)\left( x-1 \right) \\ & =A\left( {{x}^{2}}+x+1 \right)+B\left( {{x}^{2}}-x \right)+C\left( x-1 \right) \\ & ={{x}^{2}}\left( A+B \right)+x\left( A-B+C \right)+A-C \end{align}$ Step 3: And equating the coefficients of like powers of $ x $ and of the constant terms. Then, we get the system of linear equations with the unknown values of $ A,B $ and $ C $. $ A+B=0$ (I) $ A-B+C=3$ (II) $ A-C=-5$ (III) Step 5: Now solve the system for $ A $, $ B $, and $ C $, Eliminate $ C $ by adding equations (II) and (III) as given below: $\left( A-B+C \right)+A-C=3-5$ $2A-B=-2$ (IV) Eliminate $ B $ by adding equations (I) and (IV) as given below: $\begin{align} & \left( A+B \right)+2A-B=0-2 \\ & 3A=-2 \\ & A=-\frac{2}{3} \end{align}$ Substituting the values of $ A $ in equation (I) and simplify as given below: $\begin{align} & A+B=0 \\ & -\frac{2}{3}+B=0 \\ & B=\frac{2}{3} \end{align}$ Similarly, put the values of $ A $ and $ B $ in equation (II), and find the value of $ C $ as given below: $\begin{align} & A-B+C=3 \\ & -\frac{2}{3}-\frac{2}{3}+C=3 \\ & C=3+\frac{4}{3} \\ & C=\frac{13}{3} \end{align}$ Step 5: By replacing the values of $ A $, $ B $, and $ C $, write the partial function expression as given below: $\begin{align} & \frac{3x-5}{\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+x+1} \\ & =\frac{-\frac{2}{3}}{x-1}+\frac{\frac{2}{3}x+\frac{13}{3}}{{{x}^{2}}+x+1} \\ & =-\frac{2}{3\left( x-1 \right)}+\frac{2x+13}{3\left( {{x}^{2}}+x+1 \right)} \end{align}$ Thus, $-\frac{2}{3\left( x-1 \right)}+\frac{2x+13}{3\left( {{x}^{2}}+x+1 \right)}$ is the required partial fraction expansion of the rational expression $\frac{3x-5}{{{x}^{3}}-1}$.
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