## Precalculus (6th Edition) Blitzer

In order to find the partial function decomposition of the rational fraction with distinct linear factors in the denominator, the expression is equated to the expression $\frac{A}{ax+b}+\frac{B}{cx+d}$ and then the values of A and B are calculated by comparing the terms of the equations.
In order to find the partial function decomposition of the rational fraction with distinct linear factors in the denominator, the expression is equated to the expression $\frac{A}{ax+b}+\frac{B}{cx+d}$ and then the values of A and B are calculated by comparing the equations. Let us take an example: Consider $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}$ And the partial fraction decomposition of the rational fraction is as follows: $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=\frac{A}{x+2}+\frac{B}{x+4}$ And multiply both sides by $\left( x+2 \right)\left( x+4 \right)$: \begin{align} & 2x=A\left( x+4 \right)+B\left( x+2 \right) \\ & 2x=Ax+4A+Bx+2B \\ & 2x=\left( A+B \right)x+4A+2B \end{align} And equate the equation: $A+B=2$ (I) $4A+2B=0$ (II) Solving the above equation, From (I), $A=2-B$, put in equation (II) : \begin{align} & 4\left( 2-B \right)+2B=0 \\ & 8-4B+2B=0 \\ & 8-2B=0 \\ & 2B=8 \end{align} $B=4$ Putting the value of $B=4$ in equation (I) \begin{align} & A+4=2 \\ & A=2-4 \\ & A=-2 \end{align} This implies, $A=-2,B=4$ For these values of $A$ and $B$ in $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=\frac{A}{x+2}+\frac{B}{x+4}$. Thus, $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=-\frac{2}{x+2}+\frac{4}{x+4}$.