Answer
$\displaystyle \frac{y-5}{y+4}, \qquad y\neq-4, -1$
Work Step by Step
Factoring $x^{2}+bx+c$,
we search for two factors of c (m and n) such that m+n=b.
If they exist, $x^{2}+bx+c =(x+m)(x+n)$
Numerator, factored:$\quad$
$y^{2}-4y-5$=$\quad$... we find factors $-5$ and $+1,$
$=(y-5)(y+1)$
Denominator, factored:
$y^{2}+5y+4$=$\quad$... we find factors $+4$ and $+1.$
$=(y+4)(y+1)$
Numbers to be excluded from the domain are numbers that yield 0 in the denominator:
$y\neq-4, -1$
Expression = $\displaystyle \frac{(y-5)(y+1)}{(y+4)(y+1)}$=$\qquad$... cancel $(y+1)$
= $\displaystyle \frac{y-5}{y+4}, \qquad y\neq-4, -1$
