## College Algebra 7th Edition

$\dfrac{x+1}{x-4}$
Use the rule $\dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b} \times \dfrac{d}{c}$ to obtain: $=\dfrac{x^2-2x-15}{x^2-6x+5} \cdot \dfrac{x^2-1}{x^2-x-12}$ Factor each polynomial completely to obtain: $=\dfrac{(x-5)(x+3)}{(x-5)(x-1)} \cdot \dfrac{(x-1)(x+1)}{(x-4)(x+3)}$ Cancel the common factors to obtain: $\require{cancel} \\=\dfrac{\cancel{(x-5)}\cancel{(x+3)}}{\cancel{(x-5)}\cancel{(x-1)}} \cdot \dfrac{\cancel{(x-1)}(x+1)}{(x-4)\cancel{(x+3)}} \\=\dfrac{x+1}{x-4}$