## Precalculus (6th Edition)

$\color{blue}{\dfrac{y+3}{y+4}}$
Use the rule $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \cdot \dfrac{d}{c}$ to obtain: $=\dfrac{y^2+y-2}{y^2+3y-4} \cdot \dfrac{y^2+4y+3}{y^2+3y+2}$ Factor each polynomial, then cancel the common factors to obtain: $\require{cancel} \\=\dfrac{(y+2)(y-1)}{(y+4)(y-1)} \cdot \dfrac{(y+1)(y+3)}{(y+1)(y+2)} \\=\dfrac{\cancel{(y+2)}\cancel{(y-1)}}{(y+4)\cancel{(y-1)}} \cdot \dfrac{\cancel{(y+1)}(y+3)}{\cancel{(y+1)}\cancel{(y+2)}} \\=\color{blue}{\dfrac{y+3}{y+4}}$