Answer
$\displaystyle \frac{y+9}{y-1}, \qquad y\neq 1,2$
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:$\quad$
$y^{2}+7y-18$=$\quad$... we find factors $-9$ and $+2$.
$=(y+9)(y-2)$
Denominator, factored:
$y^{2}-3y+2$=$\quad$... we find factors $-2$ and $-1.$
$=(y-2)(y-1)$
Numbers to be excluded from the domain are numbers that yield 0 in the denominator:
$y\neq 1,2$
Expression = $\displaystyle \frac{(y+9)(y-2)}{(y-2)(y-1)}$=$\qquad$... cancel $(y-2)$
= $\displaystyle \frac{y+9}{y-1}, \qquad y\neq 1,2$
