## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{3}{x}+\frac{2}{\left( x-1 \right)}-\frac{1}{\left( x+3 \right)}$
$\frac{4{{x}^{2}}+13x-9}{x\left( x-1 \right)\left( x+3 \right)}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{\left( x+3 \right)}$ Now, multiply $x\left( x-3 \right)\left( 2x+1 \right)$ on both sides: \begin{align} & x\left( x-1 \right)\left( x+3 \right)\frac{4{{x}^{2}}+13x-9}{x\left( x-1 \right)\left( x+3 \right)}=x\left( x-1 \right)\left( x+3 \right)\frac{A}{x}+x\left( x-1 \right)\left( x+3 \right)\frac{B}{\left( x-1 \right)}+x\left( x-1 \right)\left( x+3 \right)\frac{C}{\left( x+3 \right)} \\ & 4{{x}^{2}}+13x-9=\left( x-1 \right)\left( x+3 \right)A+x\left( x+3 \right)B+x\left( x-1 \right)C \\ & =\left( {{x}^{2}}+3x-x-3 \right)A+B{{x}^{2}}+3Bx+C{{x}^{2}}-Cx \\ & =A{{x}^{2}}+2Ax-3A+B{{x}^{2}}+3Bx+C{{x}^{2}}-Cx \end{align} further simplify then, $4{{x}^{2}}+13x-9={{x}^{2}}\left( A+B+C \right)+x\left( 2A+3B-C \right)-3A$ …… (1) Then, compare the coefficient of equation (1): $A+B+C=4$ …… (2) $2A+3B-C=13$ …… (3) $-3A=-9$ …… (4) $A=3$ …… (5) Now, put the value of A in equation (2): $3+B+C=4$ $B+C=1$ …… (6) Also, put the value of A in equation (3): $2\left( 3 \right)+3B-C=13$ $3B-C=7$ …… (7) Then, add equation (6) and equation (7), then $B=2$ And put the value of B in equation (7): \begin{align} & 3\left( 2 \right)-C=7 \\ & C=-1 \end{align} Therefore, $\frac{4{{x}^{2}}+13x-9}{x\left( x-1 \right)\left( x+3 \right)}=\frac{3}{x}+\frac{2}{\left( x-1 \right)}-\frac{1}{\left( x+3 \right)}$ Thus, the partial fraction of the provided expression is $\frac{3}{x}+\frac{2}{\left( x-1 \right)}-\frac{1}{\left( x+3 \right)}$.