Question 107625
{{{(8sqrt(6) + 3sqrt(2))}}}{{{(4sqrt(6) - 5sqrt(2))}}}

The {{{FOIL}}} technique:

{{{F}}} stands for first terms, 
{{{O}}} stands for outer terms, 
{{{I}}} stands for inner terms, and 
{{{L}}} stands for last terms.


{{{F}}} is the product of the first terms: {{{(8sqrt(6))(4sqrt(6)) = 32*6 = 192}}}

{{{O }}}is the product of the outer terms: {{{(3sqrt(2))(-5sqrt(2)) = -15*2 = -30}}} 

{{{I}}} is the product of the inner two terms: {{{(3sqrt(2))(4sqrt(6)) = 12sqrt(2)*sqrt(6) }}}

{{{L}}} is the product of the last two terms: {{{(8sqrt(6))(-5sqrt(2)) = - 40sqrt(6)sqrt(2)}}}
 

then we have: 

{{{(8sqrt(6) + 3sqrt(2))}}}{{{(4sqrt(6) - 5sqrt(2))}}}

= {{{192}}} – {{{30}}} + {{{12sqrt(2)sqrt(6)}}} – {{{40 sqrt(6)sqrt(2)}}} 

= {{{162 + (sqrt(6)sqrt(2))(12-40 )}}} 

= {{{162}}} – {{{28 (sqrt(6)sqrt(2))}}} 

= {{{162}}} – {{{28(2.45)(1.41)}}}
 
= {{{162 - 96.726}}} 

= {{{65.274}}}