Factorize: $1-\frac{1}{6}z^{-1}-\frac{1}{3}z^{-2}$ into $(1-az^{-1})(1-bz^{-1})$

Question

Currently stuck with factorizing $1-\frac{1}{6}z^{-1}-\frac{1}{3}z^{-2}$ into $(1-az^{-1})(1-bz^{-1})$

How do I get there? I would like to know how to do it without guessing.

Tried pq, but got stuck with

z1=$-\frac{1}{12}+\sqrt{\frac{13}{36}}$

Thanks for any help. I am really struggling with factorizing in general. The formula is the denumerator of a rational function. So I am basically looking for zeros. Hence the try with pq.

Also applied it to the formula multiplied by $z^2$

Answer 1

Multiply by $6z^2$: you get $$ 6z^2-z-2 $$ The roots are easily seen to be $2/3$ and $-1/2$, so the polynomial factors as $$ 6\Bigl(z-\frac{2}{3}\Bigr)\Bigl(z+\frac{1}{2}\Bigr) $$ Divide back by $6z^2$ to get $$ \Bigl(1-\frac{2}{3}z^{-1}\Bigr)\Bigl(1+\frac{1}{2}z^{-1}\Bigr) $$

Answer 2

There is an error inside the square root, p/2=-1/12 not -1/6