\(\frac{x^2 + 8x+16}{x+3} \cdot \frac{x^2-9}{2x+8} = \frac{(x+4)^2}{x+3} \cdot \frac{(x+3)(x-3)}{2(x+4)}= \frac{(x+4)(x-3)}{2} = \frac{x^2+x-12}{2}\)