How do you simplify $- 4 \sqrt{12} \times 3 \sqrt{8}$ $-4sqrt12 times 3 sqrt8$ ?

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Answer 1 · 2016-06-19T12:59:46

$- 48 \sqrt{6}$ $-48sqrt(6)$

you can write the equivalent expression:

$- 4 \sqrt{2^{2} \cdot 3} \cdot 3 \sqrt{2^{2} \cdot 2}$ $-4sqrt(2^2*3)*3sqrt(2^2*2)$

$- 4 \cdot 2 \sqrt{3} \cdot 3 \cdot 2 \sqrt{2}$ $-4*2sqrt(3)*3*2sqrt(2)$

$- 48 \sqrt{6}$ $-48sqrt(6)$

Answer 2 · 2016-06-19T13:01:37

To solve this family of problems you have to see if the numbers in the root can be factorized.

$\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}$ $sqrt(12)=sqrt(4*3)=sqrt(4)*sqrt(3)=2sqrt(3)$

in the same way $\sqrt{8} = 2 \sqrt{2}$ $sqrt(8)=2sqrt(2)$ (try it).

then we have:

$- 4 \sqrt{12} \times 3 \sqrt{8}$ $-4sqrt(12)\times 3sqrt(8)$

$= - 4 \cdot 2 \cdot \sqrt{3} \times 3 \cdot 2 \cdot \sqrt{2}$ $=-4*2*sqrt(3)\times3*2*sqrt(2)$

$= - 8 \sqrt{3} \times 6 \sqrt{2}$ $=-8sqrt(3)\times6sqrt(2)$

$= - 48 \sqrt{3} \sqrt{2}$ $=-48sqrt(3)sqrt(2)$

$= - 48 \sqrt{6}$ $=-48sqrt(6)$ .

How do you simplify −4√12×3√8-4sqrt12 times 3 sqrt8 ?