Find all the poles and multiplicities of

a) $\displaystyle \left(\frac{4z+6}{z^2+6z+9}\right)^3$

b) $\displaystyle \frac{6z}{z^2+5z+6}+\frac{1}{z+2}$

Attempt:

a) ok so $\displaystyle \left(\frac{4z+6}{z^2+6z+9}\right)^3$

$\displaystyle = \left(\frac{2z+3}{(z+3)^2}\right)^3$

Would that mean pole at z= -3 of order 2 or order 6?

b) $\displaystyle \frac{6z}{z^2+5z+6}+\frac{1}{z+2}$

$\displaystyle = \frac{6z}{(z+2)(z+3)}+ \frac{1}{z+2}$

So there is a pole at z=-3 of order 1 but is the pole at z=-2 of order 1 or order 2?