## Precalculus (6th Edition) Blitzer

It is provided that $\frac{A}{\left( x+1 \right)}+\frac{B}{\left( {{x}^{2}}+4 \right)}$ is the partial fraction decomposition of the algebraic expression $\frac{1}{\left( x+1 \right)\left( {{x}^{2}}+4 \right)}$ into a series of smaller components. We know that the partial fraction decomposition is used for rational functions. So, it is the process by which a rational function can be written as a sum of a polynomial and fractions. And the partial fraction decomposition associated with $\frac{p{{x}^{2}}+qx+r}{\left( x-a \right)\left( {{x}^{2}}+bx+c \right)},a\ne b$, and $\left( {{x}^{2}}+bx+c \right)$ is an irreducible polynomial that cannot be factorized into linear factors: $\frac{A}{\left( x-a \right)}+\frac{Bx+C}{\left( {{x}^{2}}+bx+c \right)}$ Therefore, $\frac{A}{\left( x+1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+4 \right)}$ is the final simplified algebraic expression, Thus, the stated partial fraction decomposition $\frac{A}{\left( x+1 \right)}+\frac{B}{\left( {{x}^{2}}+4 \right)}$ of $\frac{1}{\left( x+1 \right)\left( {{x}^{2}}+4 \right)}$ is incorrect and the correct partial fraction decomposition is as given below: $\frac{1}{\left( x+1 \right)\left( {{x}^{2}}+4 \right)}$ $=$ $\frac{A}{\left( x+1 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+4 \right)}$.