Question 90535
Start with the given expression


{{{(2x-3)(3x+2)}}}


When you FOIL, you multiply the terms in this order:

F-First   (i.e. you multiply the first terms in each parenthesis which in this case are {{{2x}}} and {{{2x}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{2x}}} and {{{-3}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{-3}}} and {{{2x}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{-3}}} and {{{-3}}})



So lets multiply the first terms: 

{{{2x*3x=6x^2}}}   multiply {{{2x}}} and {{{3x}}} to get {{{6x^2}}}




So lets multiply the outer terms: 

{{{2x*2=4x}}}   multiply {{{2x}}} and {{{2}}} to get {{{4x}}}




So lets multiply the inner terms: 

{{{-3*3x=-9x}}}   multiply {{{-3}}} and {{{3x}}} to get {{{-9x}}}




So lets multiply the last terms: 

{{{-3*2=-6}}}   multiply {{{-3}}} and {{{2}}} to get {{{-6}}}


Now lets put all the multiplied terms together

 {{{6x^2+4x-9x-6}}}


 Now combine like terms


 {{{6x^2-5x-6}}}


 So the expression


 {{{(2x-3)(3x+2)}}}


 FOILs to:


 {{{6x^2-5x-6}}}