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: 41


The simplified partial fraction expansion is $\frac{3}{x-2}+\frac{x-1}{{{x}^{2}}+2x+4}$.

Work Step by Step

The provided rational fraction expression is as follows: $\frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-8}$ Then, demonstrating the steps as given below: $\begin{align} & \frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-8}=\frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-{{2}^{3}}} \\ & =\frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)} \end{align}$ 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. $\frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)}=\frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+4}$ Step 2: Multiply both sides of the equation by the expression $\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)$, considering the least common denominator. $\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\times \left( \frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)} \right)=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\times \left( \frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+4} \right)$ Then, solving both sides of the equation, we get: $4{{x}^{2}}+3x+14=\left( {{x}^{2}}+2x+4 \right)A+\left( Bx+C \right)\left( x-2 \right)$ $\begin{align} & 4{{x}^{2}}+3x+14=\left( {{x}^{2}}+2x+4 \right)A+\left( Bx+C \right)\left( x-2 \right) \\ & =A\left( {{x}^{2}}+2x+4 \right)+B\left( {{x}^{2}}-2x \right)+C\left( x-2 \right) \\ & ={{x}^{2}}\left( A+B \right)+x\left( 2A-2B+C \right)+4A-2C \end{align}$ Step 3: Then equating the coefficients of like powers of $ x $ and of the constant terms, we get the system of linear equations with the unknowns values of $ A $, $ B $, $ C $ and $ D $. $ A+B=4$ (I) $2A-2B+C=3$ (II) $4A-2C=14$ (III) Step 4: Solve the system for $ A $, $ B $, $ C $ and $ D $. Eliminate $ C $ from the equation by multiplying by 2 in equation (II) and then adding equations (II) and (III) as given below: $\begin{align} & \left( 4A-4B+2C \right)+4A-2C=6+14 \\ & 8A-4B=20 \end{align}$ $2A-B=5$ (IV) Also, eliminate $ B $ from the equation by adding equations (I) and (IV) as given below: $\begin{align} & \left( A+B \right)+2A-B=4+5 \\ & 3A=9 \\ & A=3 \end{align}$ Substitute the values of $ A $ in equation (I) and simplify as given below: $\begin{align} & A+B=4 \\ & 3+B=4 \\ & B=4-3 \\ & B=1 \end{align}$ Similarly, put the values of $ A $ and $ B $ in equation (II) and find the value of $ C $ as given below: $\begin{align} & 2A-2B+C=3 \\ & 2\times 3-2\times 1+C=3 \\ & C=3+2-6 \\ & C=-1 \end{align}$ Step 5: Replace the values of $ A $, $ B $, and, $ C $, writing the partial function expression as given below: $\begin{align} & \frac{4{{x}^{2}}+3x+14}{\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)}=\frac{A}{x-2}+\frac{Bx+C}{{{x}^{2}}+2x+4} \\ & =\frac{3}{x-2}+\frac{1\times x-1}{{{x}^{2}}+2x+4} \\ & =\frac{3}{x-2}+\frac{x-1}{{{x}^{2}}+2x+4} \end{align}$ Thus, $\frac{3}{x-2}+\frac{x-1}{{{x}^{2}}+2x+4}$ is the required partial fraction expansion of $\frac{4{{x}^{2}}+3x+14}{{{x}^{3}}-8}$.
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