Answer
$\dfrac{3x+9}{x+4}$
Work Step by Step
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:
$\require{cancel}
\\=\dfrac{(x+3)\cancel{(x-1)}}{(x+4)\cancel{(x+4)}} \cdot \dfrac{3\cancel{(x+4)}}{\cancel{x-1}}
\\=\dfrac{3(x+3)}{x+4}
\\=\dfrac{3(x) + 3(3)}{x+4}
\\=\dfrac{3x+9}{x+4}$
