91Ó°ÊÓ

Given rational expression is

$\frac{x^2-x-8}{(x+1)(x^2+5x+6)}$

Rearrange partial expression

$\frac{x^2-x-8}{(x+1)(x^2+5x+6)}=\frac{x^2-x-8}{(x+1)(x+2)(x+3)}$

partial fraction decomposition of a rational expression

role="math" localid="1647548448526" $$\begin{gathered} \frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{x-a_1}+\frac{A_2}{x-a_2}+⋯+\frac{A_n}{x-a_n} \\ \frac{x^2-x-8}{(x+1)(x+2)(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x+3)}⋯\left(i\right) \\ \frac{x^2-x-8}{(x+1)(x+2)(x+3)}=\frac{A(x+2)(x+3)}{(x+1)(x+2)(x+3)}+\frac{B(x+1)(x+3)}{(x+1)(x+2)(x+3)}+\frac{C(x+1)(x+2)}{(x+1)(x+2)(x+3)} \\ {x}^2-x-8=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)⋯\left(ii\right) \end{gathered}$$

Substitute $x=-1$in equation ii

$$\begin{gathered} 1+1-8=2A+0+0 \\ A=-3 \end{gathered}$$

Substitute$x=-2$ in equation ii

$$\begin{gathered} 4+2-8=0-B+0 \\ B=2 \end{gathered}$$

Substitute $x=-3$in equation ii

$$\begin{gathered} 9+3-8=0+0+2C \\ C=2 \end{gathered}$$

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

$$\begin{gathered} \frac{x^2-x-8}{(x+1)(x+2)(x+3)}=\frac{A}{(x+1)}+\frac{B}{(x+2)}+\frac{C}{(x+3)} \\ \frac{x^2-x-8}{(x+1)(x+2)(x+3)}=\frac{-3}{(x+1)}+\frac{2}{(x+2)}+\frac{2}{(x+3)} \\ \frac{x^2-x-8}{(x+1)(x^2+5x+6)}=\frac{-3}{(x+1)}+\frac{2}{(x+2)}+\frac{2}{(x+3)} \end{gathered}$$

So the partial fraction decomposition of a rational expression is$\frac{x^2-x-8}{(x+1)(x^2+5x+6)}=\frac{-3}{(x+1)}+\frac{2}{(x+2)}+\frac{2}{(x+3)}$