91Ó°ÊÓ

Given rational expression is

$\frac{7x+3}{x^3-2x^2-3x}$

Rearrange the rational expression

$\frac{7x+3}{x^3-2x^2-3x}=\frac{7x+3}{x(x-3)(x+1)}$

partial fraction decomposition of a rational expression

role="math" localid="1647550724172" $$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{7x+3}{x(x-3)(x+1)}=\frac{A}{x}+\frac{B}{(x-3)}+\frac{C}{(x+1)}⋯\left(i\right) \\ \frac{7x+3}{x(x-3)(x+1)}=\frac{A(x-3)(x+1)}{x(x-3)(x+1)}+\frac{Bx(x+1)}{x(x-3)(x+1)}+\frac{Cx(x-3)}{x(x-3)(x+1)} \\ 7x+3=A(x-3)(x+1)+Bx(x+1)+Cx(x-3) \\ \left(0\right)x^2+7x+3=(A+B+C)x^2+(-2A+B-3C)x-3A \end{gathered}$$

Compare the constants in equation ii

$$\begin{gathered} 3=-3A \\ A=-1 \end{gathered}$$

Compare the coefficient of $x^2$in equation ii and Substitute the expression for A

$$\begin{gathered} 0=A+B+C \\ 0=-1+B+C \\ B=1-C \end{gathered}$$

Compare the coefficient of x in equation ii and Substitute the value of A and the expression for B

role="math" localid="1647550664149" $$\begin{gathered} 7=-2A+B-3C \\ 7=-2(-1)+(1-C)-3C \\ C=-1 \end{gathered}$$

so

$$\begin{gathered} B=1-(-1) \\ B=2 \end{gathered}$$

Substitute the value of A, B,and C in the equation i

$$\begin{gathered} \frac{7x+3}{x(x-3)(x+1)}=\frac{A}{x}+\frac{B}{(x-3)}+\frac{C}{(x+1)} \\ \frac{7x+3}{x(x-3)(x+1)}=\frac{-1}{x}+\frac{2}{(x-3)}+\frac{-1}{(x+1)} \\ \frac{7x+3}{x^3-2x^2-3x}=\frac{-1}{x}+\frac{2}{(x-3)}+\frac{-1}{(x+1)} \end{gathered}$$

So the partial fraction decomposition of a rational expression is$\frac{7x+3}{x^3-2x^2-3x}=\frac{-1}{x}+\frac{2}{(x-3)}+\frac{-1}{(x+1)}$