factoring poles / zeros: off by constant gain compared with textbook (From Schaum's DSP outline, 2nd edition, problem 5.32)

Book says factor it and extract H(z) from the factored product:

$$
H(z)H(z^{-1})= \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1}  }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^-1 }
$$

ok.. no problem... so i convert to the form used by the book for nearly every problem:

$$
H(z)H(z^{-1})=  \left( \frac{z^{-1}}{z^{-1}} \right) \left( \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1}  }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^-1 } \right)
$$

$$
H(z)H(z^{-1})= \left( \frac{ - \frac{1}{2}  + \frac{5}{4}z^{-1} - \frac{1}{2}z^{-2}  }{ \frac{1}{3}  + \frac{10}{9}z^{-1}-  \frac{1}{3}z^{-2} } \right)
$$

Then I apply quadractic formula to determine the zeros and poles:

$$zeros = \left\{ \frac{1}{2}, 2 \right\}$$
$$poles = \left\{ \frac{1}{3}, 3 \right\}$$

Then I rewrite $H(z)H^{-1}(z)$ in factored form:

$$
H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-2z^{-1})}{\left(1-\frac{1}{3}z^{-1}\right)(1-3z^{-1})}
$$

At this point I realize that I have a different answer from the book which says, the answer at this point should be:

$$
H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-\frac{1}{2}z)}{\left(1-\frac{1}{3}z^{-1}\right)\left(1-\frac{1}{3}z\right)}
$$

I think... well... ok.. it should be equivalent to my version... i can rewrite my equation to get it to match the book.. so i try that:

$$
H(z)H(z^{-1})=\frac{\left(1-\frac{1}{2}z^{-1}\right)(-2z^{-1})(1-\frac{1}{2}z)}{\left(1-\frac{1}{3}z^{-1}\right)(-3z^{-1})(1-\frac{1}{3}z)}
$$

$$
H(z)H(z^{-1})=\left(\frac{2}{3}\right)\frac{\left(1-\frac{1}{2}z^{-1}\right)(1-\frac{1}{2}z)}{\left(1-\frac{1}{3}z^{-1}\right)(1-\frac{1}{3}z)}
$$

At this point i'm scratching my head wondering why it doesn't match because its off by a different gain?  Wondering why the gain doesn't matter when picking out H(z) from the factorization of $H(z)H(z^{-1})$?  Is this good practice to write it the other way around? which way is correct?
Spectral factorization is not unique, unless you specify further constraints.

So let your given $H(z)H(z^-1)$ be (taking from your derivation) :

$$
\begin{align}
H(z)H(z^{-1})&amp;= \frac{ \frac{5}{4} - \frac{1}{2}z - \frac{1}{2}z^{-1}  }{ \frac{10}{9}- \frac{1}{3}z - \frac{1}{3}z^-1 } \tag{1}\\ 
&amp;= \frac{\left(1-\frac{1}{2}z^{-1}\right)(1-2z^{-1})}{\left(1-\frac{1}{3}z^{-1}\right)(1-3z^{-1})}
\end{align}
$$

Then all of the following are valid possibilities that satisf Eq.(1)

$$
H_1(z) =\frac{1-\frac{1}{2}z^{-1}}{1-\frac{1}{3}z^{-1}}
$$

$$
H_2(z) =\frac{1-2z^{-1}}{1-3z^{-1}}
$$

$$
H_3(z) =\frac{1-\frac{1}{2}z^{-1}}{1-3z^{-1}}
$$

$$
H_4(z) =\frac{1-2z^{-1}}{1-\frac{1}{3}z^{-1}}
$$

So you must have further constraints that make one of them as a unique answer. Such as minimum-phase, maximum phase, causal, stable, bothe etc...