Answer
$$\frac{z}{z^3-z^2-6z}=-\frac{1}{5(z+2)}+\frac{1}{5(z-3)}$$
Work Step by Step
$$\frac{z}{z^3-z^2-6z}=\frac{1}{z^2-z-6}=\frac{1}{(z+2)(z-3)}$$
The partial fraction decomposition then has the following form: $$\frac{z}{z^3-z^2-6z}=\frac{1}{(z+2)(z-3)}=\frac{A}{z+2}+\frac{B}{z-3}$$
To find $A, B$, we clear fractions and get
$$1=A(z-3)+B(z+2)$$ $$1=Az-3A+Bz+2B$$ $$1=(A+B)z+(2B-3A)$$
Equating coefficients of corresponding powers of $z$, we get
- $A+B=0$
- $2B-3A=1$
So $A=-1/5$ and $B=1/5$
Therefore, $$\frac{z}{z^3-z^2-6z}=-\frac{1}{5(z+2)}+\frac{1}{5(z-3)}$$
