How do you simplify #-4sqrt12 times 3 sqrt8 #?

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Answer 1 · 2016-06-19T12:59:46

#-48sqrt(6)#

you can write the equivalent expression:

#-4sqrt(2^2*3)*3sqrt(2^2*2)#

#-4*2sqrt(3)*3*2sqrt(2)#

#-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*3)=sqrt(4)*sqrt(3)=2sqrt(3)#

in the same way #sqrt(8)=2sqrt(2)# (try it).

then we have:

#-4sqrt(12)\times 3sqrt(8)#

#=-4*2*sqrt(3)\times3*2*sqrt(2)#

#=-8sqrt(3)\times6sqrt(2)#

#=-48sqrt(3)sqrt(2)#

#=-48sqrt(6)#.