Answer
$1$
Work Step by Step
Factor what we can:
$10z^{2}+13z-3=10z^{2}+15z-2z-3 =$... factor in pairs
$=5z(2z+3)-(2z+3)=(2z+3)(5z-1)$
$2z^{2}-3z-2z+3=$... factor in pairs
$ = z(2z-3)-(2z-3)=(2z-3)(z-1)$
$15z^{2}-28z+5=15z^{2}-25z-3z+5=$... factor in pairs
$=5z(3z-5)-(3z-5)=(3z-5)(5z-1)$
$3z^{2}-8z+5=3z^{2}-3z-5z+5=$... factor in pairs
$=3z(z-1)-5(z-1)=(z-1)(3z-5)$
$25z^{2}-10z+1=(5z)^{2}-2(5z)(1)+(1)^{2}=(5z-1)^{2}$
$4z^{2}-9=(2z)^{2}-3^{2}=(2z-3)(2z+3)$
Rewrite the problem:
$ \displaystyle \frac{(2z+3)(5z-1)}{(z-1)(3z-5)}\cdot\frac{(2z-3)(z-1)}{(5z-1)(5z-1)}\cdot\frac{(3z-5)(5z-1)}{(2z-3)(2z+3)}=\qquad$ ... reduce common factors
$= \displaystyle \frac{(1)(1)}{(z-1)(3z-5)}\cdot\frac{(2z-3)(z-1)}{(1)(5z-1)}\cdot\frac{(3z-5)(5z-1)}{(2z-3)(1)}$
$= \displaystyle \frac{(1)(1)}{(1)(3z-5)}\cdot\frac{(1)(1)}{(1)(5z-1)}\cdot\frac{(3z-5)(5z-1)}{(1)(1)}$
$= \displaystyle \frac{(1)(1)}{(1)(1)}\cdot\frac{(1)(1)}{(1)(1)}\cdot\frac{(1)(1)}{(1)(1)}$
= $1$
