In Problems 11–20, determine the partial fraction expansion for the given rational function.$\frac{{\mathbf{s}}^{\mathbf{2}}\mathbf{-}\mathbf{26}\mathbf{s}\mathbf{-}\mathbf{47}}{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{5}\mathbf{)}}$

• The partial fraction expansion of a rational fraction (that is, a fraction with both a polynomial numerator and a polynomial denominator) is an algebraic procedure that involves expressing the fraction as a sum of a polynomial plus one or more fractions with a simpler denominator.
• The partial fraction decomposition is important because it gives methods for a variety of rational function computations, such as explicit antiderivative computations, Taylor series expansions, inverse Z-transforms, and inverse Laplace transforms.

The given rational function is $\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}$

Rewrite $\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}$as a sum of partial fractions as:

$\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}=\frac{\mathrm{A}}{\mathrm{s}-1}+\frac{\mathrm{B}}{\mathrm{s}+2}+\frac{\mathrm{C}}{\mathrm{s}+5}$

Multiply both sides by the LCD$(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)$as follows:

${\mathrm{s}}^2-26\mathrm{s}-47=\mathrm{A}(\mathrm{s}+2)(\mathrm{s}+5)+\mathrm{B}(\mathrm{s}-1)(\mathrm{s}+5)+\mathrm{C}(\mathrm{s}-1)(\mathrm{s}+2)$

Find the constants as:

For$\mathrm{s}=1,{\left(1\right)}^2-26\left(1\right)-47=\mathrm{A}\left(3\right)\left(6\right)\Rightarrow \mathrm{A}=-4$.

For$\mathrm{s}=-2,{(-2)}^2-26(-2)-47=\mathrm{B}(-3)\left(3\right)\Rightarrow \mathrm{B}=-1$.

For $\mathrm{s}=-5,{(-5)}^2-26(-5)-47=\mathrm{C}(-6)(-3)\Rightarrow \mathrm{C}=6$.

Substitute the value of constants into $\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}=\frac{\mathrm{A}}{\mathrm{s}-1}+\frac{\mathrm{B}}{\mathrm{s}+2}+\frac{\mathrm{C}}{\mathrm{s}+5}$ as follows:

$\begin{array}{c}\frac{{\mathrm{s}}^2-26\mathrm{s}-47}{(\mathrm{s}-1)(\mathrm{s}+2)(\mathrm{s}+5)}=\frac{-4}{\mathrm{s}-1}+\frac{-1}{\mathrm{s}+2}+\frac{6}{\mathrm{s}+5}\\ =\frac{6}{\mathrm{s}+5}-\frac{1}{\mathrm{s}+2}-\frac{4}{\mathrm{s}-1}\end{array}$

Therefore, the partial fraction expansion for the given rational function is

$\frac{6}{\mathrm{s}+5}-\frac{1}{\mathrm{s}+2}-\frac{4}{\mathrm{s}-1}$