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: 8


The partial fraction function is $\frac{7}{\left( {{x}^{2}}+7 \right)}-\frac{46}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$.

Work Step by Step

Let us consider the provided expression, $y=\frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$ …… (I) Then, find the partial fraction as given below, $\begin{align} & \frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}=\frac{A}{\left( {{x}^{2}}+7 \right)}+\frac{C}{{{\left( {{x}^{2}}+7 \right)}^{2}}} \\ & {{\left( {{x}^{2}}+7 \right)}^{2}}\frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}={{\left( {{x}^{2}}+7 \right)}^{2}}\left( \frac{A}{\left( {{x}^{2}}+7 \right)}+\frac{C}{{{\left( {{x}^{2}}+7 \right)}^{2}}} \right) \\ & 7{{x}^{2}}-9x+3=A\left( {{x}^{2}}+7 \right)+C \\ & 7{{x}^{2}}-9x+3=A{{x}^{2}}+7A+C \end{align}$ …… (II) Now, compare the coefficient of the equation (II) of ${{x}^{2}},{{x}^{1}}$ and ${{x}^{0}}$: $A=7$ …… (III) $7A+C=3$ …… (IV) And put the value of equation (III) into equation (IV), $\begin{align} & 7\left( 7 \right)+C=3 \\ & C=3-49 \\ & =-46 \end{align}$ Therefore, $\frac{7{{x}^{2}}-9x+3}{{{\left( {{x}^{2}}+7 \right)}^{2}}}=\frac{7}{\left( {{x}^{2}}+7 \right)}-\frac{46}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$ Thus, the partial fraction of the given expression is $\frac{7}{\left( {{x}^{2}}+7 \right)}-\frac{46}{{{\left( {{x}^{2}}+7 \right)}^{2}}}$.
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