Answer
$ \displaystyle \frac{y-7}{y-5}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ = multiplying with $\displaystyle \frac{Q}{P}$
Rewrite the problem:
$ \displaystyle \frac{y^{2}-4y-21}{y^{2}-10y+25}\cdot\frac{y^{2}-6y+5}{y^{2}+2y-3}=\qquad$ ... factor what we can
... $y^{2}-4y-21=$... factors of $-21$ whose sum is $-4$ ... are $+3$ and $-7.$
$=(y+3)(y-7)$
... $y^{2}-10y+25=$... square of a difference = $(y-5)^{2}$
... $y^{2}-6y+5=$... factors of $+5$ whose sum is $-6$ ... are $-5$ and $-1.$
$=(y-5)(y-1)$
... $y^{2}+2y-3=$... factors of $-3$ whose sum is $+2$ ... are $-1$ and $+3.$
$=(y+3)(y-1)$
Rewrite the problem:
$ = \displaystyle \frac{(y+3)(y-7)}{(y-5)(y-5)}\cdot\frac{(y-5)(y-1)}{(y+3)(y-1)}$=$\qquad$ ... reduce common factors
=$ \displaystyle \frac{(1)(y-7)}{(1)(y-5)}\cdot\frac{(1)(1)}{(1)(1)}$=
=$ \displaystyle \frac{y-7}{y-5}$
