## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{y+9}{y-1}, \qquad y\neq 1,2$
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 $(y-2)$ = $\displaystyle \frac{y+9}{y-1}, \qquad y\neq 1,2$