Precalculus (6th Edition) Blitzer

When a rational expression contains a power of a factor in the denominator, be sure to set up the partial fraction decomposition to allow for every natural number power of that factor less than or equal to the power. Let us take an example, $\frac{2x+1}{{{\left( x-5 \right)}^{2}}{{x}^{3}}}=\frac{A}{\left( x-3 \right)}+\frac{B}{{{\left( x-5 \right)}^{2}}}+\frac{C}{x}+\frac{D}{{{x}^{2}}}+\frac{E}{{{x}^{3}}}$ …… (I) Although ${{\left( x-5 \right)}^{2}}$ and ${{x}^{2}}$ are quadratic, they are still expressed as powers of linear factors, $x-5$ and $x$ Therefore, the numbers are constant. From equation (I): $2x+1=A\left( x-5 \right){{x}^{3}}+B{{x}^{2}}+c{{\left( x-5 \right)}^{2}}{{x}^{2}}+D{{\left( x-5 \right)}^{2}}x+E{{\left( x-5 \right)}^{2}}$ Put $x=0$ Then, $E=\frac{1}{25}$ Put $x=5$ Then, \begin{align} & 11=B{{\left( 25 \right)}^{2}} \\ & B=\frac{11}{25} \\ \end{align} Similarly get the values of A, B,C, D by putting in the value of x. Thus, the partial decomposition of the rational expression contains the power of a prime cubic factor in the denominator.