## University Calculus: Early Transcendentals (3rd Edition)

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