Finding minimum-phase/allpass of transfer function I am trying to find the minimum-phase system of the transfer function $H(z) = \frac{2 + 3.125z^{-2}}{1-0.9z^{-1}+0.81z^{-2}}$. 

I know I need to find and remove the allpass (basically reflect the two zeroes on the imaginary axis inside the unit circle?), and I've gotten so far as this: $H(z) = 2\frac{1 + \frac{3.125}{2}z^{-2}}{1-0.9z^{-1}+0.81z^{-2}} = 2\frac{1 + \frac{3.125}{2}z^{-2}}{1-0.9z^{-1}+0.81z^{-2}} \frac{\frac{3.125}{2} + z^{-2}}{1 + \frac{3.125}{2}z^{-2}}=,$ 

and then I'm not really sure where to go next. Is my thinking correct thus far? What step should I do next?

Thank you. 

Edit:
This is the kind of thing I'm trying to do, as per the example in my book:
The allpass part must have its zeros outside the unit circle, so it gets the zeros of your system, with the corresponding poles inside the unit circle. You get the correct pole locations by simply swapping the coefficients of the numerator polynomial:

$$H_a(z)=\frac{2+3.125z^{-2}}{3.125+2z^{-2}}\tag{1}$$

The minimum phase part is obtained by taking the system's poles and cancelling the poles of the allpass part:

$$H_m(z)=\frac{3.125+2z^{-2}}{1-0.9z^{-1}+0.81z^{-2}}\tag{2}$$

The transfer function of the original system is given by

$$H(z)=H_a(z)\cdot H_m(z)\tag{3}$$