Algebra 1

$3\sqrt{2}-4\sqrt{3}$
Using the distribution property, $\sqrt{3}(\sqrt{6}-4)= \sqrt{3} \ast \sqrt{6}-\sqrt{3}\ast 4$. Since $6=3\ast2$, $\sqrt{3}(\sqrt{6}-4)=\sqrt{3}\ast \sqrt{3\ast 2}-\sqrt{3}\ast 4$. By simplifying this, we get $\sqrt{3}(\sqrt{6}-4)=3\sqrt{2}-4\sqrt{3}$