Answer
The partial fraction decomposition of the rational expression is
$\frac{7{{x}^{2}}-7x+23}{\left( x-3 \right)\left( {{x}^{2}}+4 \right)}=\frac{5}{x-3}+\frac{2x-1}{{{x}^{2}}+4}$.
Work Step by Step
By putting the constant $ A $ over the linear factor $ x-3$ and a linear expression $ Bx+C $ over the quadratic factor:
$\frac{7{{x}^{2}}-7x+23}{\left( x-3 \right)\left( {{x}^{2}}+4 \right)}=\frac{A}{x-3}+\frac{Bx+C}{{{x}^{2}}+4}$
And multiply both sides by $\left( x-3 \right)\left( {{x}^{2}}+4 \right)$ as shown below:
$\begin{align}
& 7{{x}^{2}}-7x+23=A\left( {{x}^{2}}+4 \right)+\left( Bx+C \right)\left( x-3 \right) \\
& 7{{x}^{2}}-7x+23=A{{x}^{2}}+4A+B{{x}^{2}}-3Bx+Cx-3C \\
& 7{{x}^{2}}-7x+23=\left( A+B \right){{x}^{2}}+\left( C-3B \right)x+\left( 4A-3C \right)
\end{align}$
By equating the terms of equal powers, we get:
$ A+B=7$ (I)
$ C-3B=-7$ (II)
$4A-3C=23$ (III)
And from equation (II),
$ B=\frac{1}{3}\left( C+7 \right)$ (IV)
Put the value of B from equation (IV) in equation (I):
$\begin{align}
& A+\frac{1}{3}\left( C+7 \right)=7 \\
& 3A+C+7=21 \\
& 3A+C=14
\end{align}$
Multiplying both sides by 3 as given below:
$9A+3C=42$ (V)
Add equations (III) and (V):
$\begin{align}
& 13A=65 \\
& A=5
\end{align}$
From equation (I), compute the value of B:
$\begin{align}
& B=7-5 \\
& =2
\end{align}$
From equation (II), compute the value of C:
$\begin{align}
& C=3B-7 \\
& =6-7 \\
& =-1
\end{align}$
So, the values are $ A=5,B=2,C=-1$.
Put the values of $ A,\,\,B,\text{ and }C $ in the given equation, and define the partial fraction decomposition:
$\frac{A}{x-3}+\frac{Bx+C}{{{x}^{2}}+4}=\frac{5}{x-3}+\frac{2x-1}{{{x}^{2}}+4}$
Thus, the partial fraction decomposition of the rational expression is
$\frac{7{{x}^{2}}-7x+23}{\left( x-3 \right)\left( {{x}^{2}}+4 \right)}=\frac{5}{x-3}+\frac{2x-1}{{{x}^{2}}+4}$.
