Question 93387
Start with the given expression


{{{(2x-5y)(x+y)}}}


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 {{{-5y}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{-5y}}} and {{{2x}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{-5y}}} and {{{-5y}}})



So lets multiply the first terms: 

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




So lets multiply the outer terms: 

{{{2x*y=2yx}}}   multiply {{{2x}}} and {{{y}}} to get {{{2yx}}}




So lets multiply the inner terms: 

{{{-5y*x=-5yx}}}   multiply {{{-5y}}} and {{{x}}} to get {{{-5yx}}}




So lets multiply the last terms: 

{{{-5y*y=-5y^2}}}   multiply {{{-5y}}} and {{{y}}} to get {{{-5y^2}}}


Now lets put all the multiplied terms together

 {{{2x^2+2yx-5yx-5y^2}}}


 Now combine like terms


 {{{2x^2-3yx-5y^2}}}


 So the expression


 {{{(2x-5y)(x+y)}}}


 FOILs to:


 {{{2x^2-3yx-5y^2}}}