Answer
The partial fraction is $\frac{3}{x}-\frac{1}{\left( x+1 \right)}+\frac{2}{\left( x-5 \right)}$.
Work Step by Step
$\frac{4{{x}^{2}}-5x-15}{x\left( x+1 \right)\left( x-5 \right)}=\frac{A}{x}+\frac{B}{\left( x+1 \right)}+\frac{C}{\left( x-5 \right)}$
Now, multiply $x\left( x+1 \right)\left( x-5 \right)$ on both sides:
$\begin{align}
& x\left( x+1 \right)\left( x-5 \right)\frac{4{{x}^{2}}-5x-15}{x\left( x+1 \right)\left( x-5 \right)}=x\left( x+1 \right)\left( x-5 \right)\frac{A}{x}+x\left( x+1 \right)\left( x-5 \right)\frac{B}{\left( x+1 \right)}+x\left( x+1 \right)\left( x-5 \right)\frac{C}{\left( x-5 \right)} \\
& 4{{x}^{2}}-5x-15=\left( x+1 \right)\left( x-5 \right)A+x\left( x-5 \right)B+x\left( x+1 \right)C \\
& =\left( {{x}^{2}}-5x+x-5 \right)A+B{{x}^{2}}-5Bx+C{{x}^{2}}+Cx \\
& =A{{x}^{2}}-4Ax-5A+B{{x}^{2}}-5Bx+C{{x}^{2}}+Cx
\end{align}$ $4{{x}^{2}}-5x-15={{x}^{2}}\left( A+B+C \right)+x\left( -4A-5B+C \right)-5A$ …… (1)
Then, compare the coefficient of equation (1):
$A+B+C=4$ ….. (2)
$-4A-5B+C=-5$ ……(3)
$-5A=-15$ …… (4)
$A=3$ …… (5)
And put the value of A in equation (2) and equation (3):
$3+B+C=4$
$B+C=1$ …… (6)
And,
$\begin{align}
& -4\left( 3 \right)-5B+C=-5 \\
& -12-5B+C=-5
\end{align}$
$-5B+C=7$ …… (7)
Subtract equation (7) from equation (6), then
$B=-1$
Then, put the value of B in equation (6):
$\begin{align}
& -1+C=1 \\
& C=2
\end{align}$
Therefore,
$\frac{4{{x}^{2}}-5x-15}{x\left( x+1 \right)\left( x-5 \right)}=\frac{3}{x}-\frac{1}{\left( x+1 \right)}+\frac{2}{\left( x-5 \right)}$
Thus, the partial fraction of the given expression is $\frac{3}{x}-\frac{1}{\left( x+1 \right)}+\frac{2}{\left( x-5 \right)}$.
