## Precalculus: Mathematics for Calculus, 7th Edition

$\dfrac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}}=\dfrac{(x+2)^{2}(x-13)}{(x-3)^{3}}$
$\dfrac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}}$ Take out common factor $(x+2)^{2}(x-3)$ from the numerator: $\dfrac{3(x+2)^{2}(x-3)^{2}-(x+2)^{3}(2)(x-3)}{(x-3)^{4}}=...$ $...=\dfrac{(x+2)^{2}(x-3)[3(x-3)-2(x+2)]}{(x-3)^{4}}=...$ Simplify the expression inside brackets in the numerator: $...=\dfrac{(x+2)^{2}(x-3)[3x-9-2x-4]}{(x-3)^{4}}=...$ $...=\dfrac{(x+2)^{2}(x-3)(x-13)}{(x-3)^{4}}=...$ Simplify the rational expression: $...=\dfrac{(x+2)^{2}(x-13)}{(x-3)^{3}}$