## Precalculus (6th Edition) Blitzer

The partial fraction is $\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$.
Let us consider the decomposition as, \begin{align} & \frac{7x-4}{{{x}^{2}}-x-12}=\frac{7x-4}{\left( x+3 \right)\left( x-4 \right)} \\ & =\frac{A}{\left( x+3 \right)}+\frac{B}{\left( x-4 \right)} \end{align} Then, multiply $\left( x+3 \right)\left( x-4 \right)$ on both sides: \begin{align} & \left( x+3 \right)\left( x-4 \right)\frac{7x-4}{\left( x+3 \right)\left( x-4 \right)}=\left( x+3 \right)\left( x-4 \right)\frac{A}{\left( x+3 \right)}+\left( x+3 \right)\left( x-4 \right)\frac{B}{\left( x-4 \right)} \\ & 7x-4=\left( x-4 \right)A+\left( x+3 \right)B \\ & =Ax-4A+Bx+3B \\ & 7x-4=x\left( A+B \right)-4A+3B \end{align} …… (I) Now, compare the coefficient of equation (I): $A+B=7$ …… (II) $-4A+3B=-4$ …… (III) Then, multiply equation (II) by 4 and add with equation (III): \begin{align} & 7B=24 \\ & B=\frac{24}{7} \end{align} Now, put the value of B in equation (II): \begin{align} & A+\frac{24}{7}=7 \\ & A=7-\frac{24}{7} \\ & =\frac{49-24}{7} \\ & =\frac{25}{7} \end{align} Therefore, $\frac{7x-4}{{{x}^{2}}-x-12}=\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$ Thus, the partial fraction of the given expression is $\frac{25}{7\left( x+3 \right)}+\frac{24}{7\left( x-4 \right)}$.