Answer
$x^2+6x+8$
Work Step by Step
Given: $\dfrac{x^2-4}{x+3} \times \dfrac{x^2+7x+12}{x-2}$
$\dfrac{x^2-4}{x+3} \times \dfrac{x^2+7x+12}{x-2}=\dfrac{(x^2-2^2)(x+3)(x+4)}{(x-2)(x+3)}$
$=\dfrac{(x-2)(x+2)(x+3)(x+4)}{(x-2)(x+3)}$
$=(x+2)(x+4)$
$=x^2+6x+8$
