How do you simplify $(2 \sqrt{2} + 2 \sqrt{24}) \cdot \sqrt{3}$ $(2sqrt2 + 2sqrt24) * sqrt3$ ?

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Answer 1 · 2016-06-09T12:04:48

$(2 \sqrt{2} + 2 \sqrt{24}) \cdot \sqrt{3} = 2 \sqrt{6} + 12 \sqrt{2}$ $(2sqrt2+2sqrt24)*sqrt3=2sqrt6+12sqrt2$

$(2 \sqrt{2} + 2 \sqrt{24}) \cdot \sqrt{3}$ $(2sqrt2+2sqrt24)*sqrt3$

= $2 \sqrt{2} \cdot \sqrt{3} + 2 \sqrt{24} \cdot \sqrt{3}$ $2sqrt2*sqrt3+2sqrt24*sqrt3$

and as $\sqrt{a} \times \sqrt{b} = \sqrt{a} b$ $sqrtaxxsqrtb=sqrtab$ , this is equal to

$2 \sqrt{6} + 2 \sqrt{72}$ $2sqrt6+2sqrt72$

= $2sqrt(2xx3)+2sqrt(ul(2xx2)xx2xxul(3xx3))$

= $2sqrt6+12sqrt2$

How do you simplify (2sqrt2 + 2sqrt24) * sqrt3(2√2+2√24)⋅√3(2sqrt2 + 2sqrt24) * sqrt3?