How do you simplify #(sqrt [6] + 2sqrt [2])(4sqrt[6] - 3sqrt2)#?

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Answer 1 · 2018-06-28T13:22:53

#12+5sqrt12#

Given: #(sqrt6+2sqrt2)(4sqrt6-3sqrt2)#.

Use the #"FOIL"# theorem, which states that #(a+b)(c+d)=ac+ad+bc+bd#.

So, we get:

#=sqrt6*4sqrt6-3sqrt2*sqrt6+2sqrt2*4sqrt6-2sqrt2*3sqrt2#

#=4*6-3sqrt12+8sqrt12-6*2#

#=24-12+5sqrt12#

#=12+5sqrt12#

Answer 2 · 2018-06-28T13:26:36

#color(crimson)(=> 2 (6 - 5 sqrt 3)#

#(sqrt 6 + 2 sqrt 2) (4 sqrt 6 - 3 sqrt 2)#

#=> sqrt 6 * 4 sqrt 6 + 2 sqrt 2 * 4 sqrt 6 - sqrt 6 * 3 sqrt 2 - 2 sqrt 2 * 3 sqrt 2#

#=> 4 * 6 + 8 sqrt 12 - 3 sqrt 12 - 6 * 2#

#=> 24 - 12 + 8 sqrt 12 - 3 sqrt 12#

#=> 12 + 5 sqrt (4 * 3)#

#=> 12 - 10 sqrt 3#

#color(crimson)(=> 2 (6 - 5 sqrt 3)#