## College Algebra 7th Edition

$\dfrac{3x+9}{x+4}$
Factor each polynomial completely to obtain: $=\dfrac{(x+3)(x-1)}{(x+4)(x+4)} \cdot \dfrac{3(x+4)}{x-1}$ Cancel common factors to obtain: $\require{cancel} \\=\dfrac{(x+3)\cancel{(x-1)}}{(x+4)\cancel{(x+4)}} \cdot \dfrac{3\cancel{(x+4)}}{\cancel{x-1}} \\=\dfrac{3(x+3)}{x+4} \\=\dfrac{3(x) + 3(3)}{x+4} \\=\dfrac{3x+9}{x+4}$