Question

($\sqrt{6} + \sqrt{32}$) × (√2-√24)

Answer 1

$(\sqrt6+\sqrt{32})\cdot(\sqrt2-\sqrt{24})=(\sqrt6\sqrt2-\sqrt6\sqrt{24})+(\sqrt{32}\sqrt2-\sqrt{32}\sqrt{24})\\=(\sqrt{12}+\sqrt{64}-\sqrt{144}-\sqrt{768})=\sqrt{12}+8-12-\sqrt{768}\\=2\sqrt3-4-\sqrt{2^{8}3}=2\sqrt3-2^4\sqrt3-4=\sqrt3(2-16)-4=\boxed{-14\sqrt3-4}$