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: 2

Answer

The simplified partial fraction expansion is $\frac{5x+7}{\left( x-1 \right)\left( x+3 \right)}=\frac{A}{\left( x-1 \right)}+\frac{B}{\left( x+3 \right)}$

Work Step by Step

The provided rational expression is as follows: $\frac{5x+7}{\left( x-1 \right)\left( x+3 \right)}$ Now, solving the expression as given below: We set up the partial fraction expansion with unknown constants coefficients and then write a constant coefficients over each of the two distinct algebraic linear factors in the denominator of the expression. Then, decompose the fractional part as follows: $\frac{5x+7}{\left( x-1 \right)\left( x+3 \right)}=\frac{A}{\left( x-1 \right)}+\frac{B}{\left( x+3 \right)}$ Thus, $\frac{A}{\left( x-1 \right)}+\frac{B}{\left( x+3 \right)}$ is a partial fraction expansion of the rational expression $\frac{5x+7}{\left( x-1 \right)\left( x+3 \right)}$ with constants $A$ and $B$.
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