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 842: 60


The statement makes sense.

Work Step by Step

Let us consider an example as follows: $\frac{3}{x}+\frac{1}{x+2}$ Firstly, check the common denominator of the fraction. $x\left( x+2 \right)$ Next to operate the partial fraction decomposition: $\frac{3\left( x+2 \right)+1}{x\left( x+2 \right)}=\frac{3x+7}{x\left( x+2 \right)}$ Writing the fraction with one of the polynomial factors for each of the denominator. $\frac{3x+7}{x\left( x+2 \right)}=\frac{A}{x}+\frac{B}{x+2}$ And by multiplying both sides by the common factors $x\left( x+2 \right)$. $\begin{align} & 3x+7=A\left( x+2 \right)+Bx \\ & 3x+7=Ax+2A+Bx \\ & 3x+7=\left( A+B \right)x+2A \\ \end{align}$ And equating the equation: $\begin{align} & A+B=3 \\ & 2A=7 \\ \end{align}$ Then, solving the above equation for $A$ and $B$. $A=\frac{7}{2},B=-\frac{1}{2}$ Put these values in $\frac{3x+7}{x\left( x+2 \right)}=\frac{A}{x}+\frac{B}{x+2}$ $\frac{3x+7}{x\left( x+2 \right)}=\frac{7}{2x}-\frac{1}{2\left( x+2 \right)}$. Thus, the statement makes sense.
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