#### Answer

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.

#### Work Step by Step

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, $\frac{2x}{\left( x+2 \right)\left( x+4 \right)}=-\frac{2}{x+2}+\frac{4}{x+4}$.