Answer
The partial fraction is, $\frac{6}{\left( x-3 \right)}+\frac{3}{\left( x+5 \right)}$
Work Step by Step
We find the partial fraction of the decomposition as given below,
$\begin{align}
& \frac{9x+21}{{{x}^{2}}+2x-15}=\frac{9x+21}{\left( x-3 \right)\left( x+5 \right)} \\
& =\frac{A}{\left( x-3 \right)}+\frac{B}{\left( x+5 \right)}
\end{align}$
Now, multiply $\left( x-3 \right)\left( x+5 \right)$ on both sides:
$\begin{align}
& \left( x-3 \right)\left( x+5 \right)\frac{9x+21}{\left( x-3 \right)\left( x+5 \right)}=\left( x-3 \right)\left( x+5 \right)\frac{A}{\left( x-3 \right)}+\left( x-3 \right)\left( x+5 \right)\frac{B}{\left( x+5 \right)} \\
& 9x+21=\left( x+5 \right)A+\left( x-3 \right)B \\
& =Ax+5A+Bx-3B \\
& 9x+21=x\left( A+B \right)+5A-3B
\end{align}$ …… (1)
Then, compare the coefficient of equation (1):
$A+B=9$ …… (2)
$5A-3B=21$ …… (3)
Then, multiply equation (2) by 3 and add with equation (3),
Then,
$\begin{align}
& A=\frac{48}{8} \\
& =6
\end{align}$
Now, put the value of A in equation (II):
$\begin{align}
& 6+B=9 \\
& B=9-6 \\
& =3
\end{align}$
Therefore,
$\frac{9x+21}{{{x}^{2}}+2x-15}=\frac{6}{\left( x-3 \right)}+\frac{3}{\left( x+5 \right)}$
Thus, the partial fraction of the given expression is $\frac{6}{\left( x-3 \right)}+\frac{3}{\left( x+5 \right)}$.
