## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 5

#### Answer

The partial fraction expansion is $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+1}$.

#### Work Step by Step

The provided rational expression is as given below: $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}$ Now, solve the expression as follows: 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. $\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}=\frac{A}{x-1}+\frac{Bx+C}{{{x}^{2}}+1}$ Thus, $\frac{A}{x+1}+\frac{Bx+C}{{{x}^{2}}+1}$ is a partial fraction expansion of the rational expression$\frac{5{{x}^{2}}-6x+7}{\left( x-1 \right)\left( {{x}^{2}}+1 \right)}$ with constants $A$, $B$ and $C$ .

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