Which values for b make the polynomial 16z2 + bz + 81 a perfect square trinomial?

As (a+b) ^2 = a^2 + 2*a*b + b^2 ... (i)

here we have

16z^2 + bz + 81 which can be written as (4z) ^2 + bz + 9^2

from (i) we can see that in order to be perfect square bz should be

bz = 2 * (4z) * (9)

bz = 72z

dividing z on both sides we get

b=72