Answer
$-4\sqrt{3} - 2\sqrt{6}i$
Work Step by Step
RECALL:
(1) $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab} \text{ where } a, b, \gt 0$
(2) $a(b+c) =ab+ac$
(3) $i^2=-1$
(4) For any real number $a \gt 0$, $\sqrt{-a} = i\sqrt{a}$
Use rule (4) above to obtain:
$=i\sqrt{12}(i\sqrt{4}-\sqrt{2})
\\=i\sqrt{4(3)}(i\sqrt{2^2}-\sqrt{2})
\\=i\sqrt{2^2(3)}(2i-\sqrt{2})
\\=i\cdot 2\sqrt{3}(2i-\sqrt{2})
\\=2i\sqrt{3}(2i-\sqrt{2})$
Use rules (1) and (2) to obtain:
$=4i^2\sqrt{3} - 2\sqrt{6}i$
Use rule (3) above to obtain:
$=4(-1)\sqrt{3} - 2\sqrt{6}i
\\=-4\sqrt{3} - 2\sqrt{6}i$
