## Precalculus (6th Edition) Blitzer

The partial fraction is $\frac{3}{x}-\frac{1}{\left( x+1 \right)}+\frac{2}{\left( x-5 \right)}$.
$\frac{4{{x}^{2}}-5x-15}{x\left( x+1 \right)\left( x-5 \right)}=\frac{A}{x}+\frac{B}{\left( x+1 \right)}+\frac{C}{\left( x-5 \right)}$ 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 $\frac{3}{x}-\frac{1}{\left( x+1 \right)}+\frac{2}{\left( x-5 \right)}$.