## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 7 - Section 7.3 - Partial Fractions - Exercise Set - Page 841: 6

#### Answer

The partial fraction expansion is $\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{A}{x-4}+\frac{Bx+C}{{{x}^{2}}+5}$.

#### Work Step by Step

The provided rational expression is as given below: $\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}$ 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. Then, decompose the fractional part as follows: $\frac{5{{x}^{2}}-9x+19}{\left( x-5 \right)\left( {{x}^{2}}+5 \right)}=\frac{A}{x-4}+\frac{Bx+C}{{{x}^{2}}+5}$ Thus, $\frac{A}{x-4}+\frac{Bx+C}{{{x}^{2}}+5}$ is a partial fraction expansion of the rational expression $\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}$with constants $A$,$B$ and$C$.

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