Answer
$(\sqrt{6}-4\sqrt{2})(3\sqrt{6}+1)=18+\sqrt{6}-24\sqrt{3}-4\sqrt{2}$
Work Step by Step
$(\sqrt{6}-4\sqrt{2})(3\sqrt{6}+1)$
Evaluate the product:
$(\sqrt{6}-4\sqrt{2})(3\sqrt{6}+1)=3\sqrt{6^{2}}+\sqrt{6}-12\sqrt{12}-4\sqrt{2}=...$
$...=3(6)+\sqrt{6}-12\sqrt{12}-4\sqrt{2}=...$
$...=18+\sqrt{6}-12\sqrt{12}-4\sqrt{2}=...$
Rewrite this expression as $18+\sqrt{6}-12\sqrt{4\cdot3}-4\sqrt{2}$ and simplify:
$...=18+\sqrt{6}-12\sqrt{4\cdot3}-4\sqrt{2}=...$
$...=18+\sqrt{6}-12(2)\sqrt{3}-4\sqrt{2}=$
$...=18+\sqrt{6}-24\sqrt{3}-4\sqrt{2}$
