Question 89360
Start with the given expression


{{{(5x+9)(5x+9)}}}


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 {{{5x}}} and {{{5x}}})
O-Outer   (i.e. you multiply the outer terms in each parenthesis which in this case are {{{5x}}} and {{{9}}})
I-Inner   (i.e. you multiply the inner terms in each parenthesis which in this case are {{{9}}} and {{{5x}}})
L-Last    (i.e. you multiply the last terms in each parenthesis which in this case are {{{9}}} and {{{9}}})



So lets multiply the first terms: 

{{{5x*5x=25x^2}}}   multiply {{{5x}}} and {{{5x}}} to get {{{25x^2}}}




So lets multiply the outer terms: 

{{{5x*9=45x}}}   multiply {{{5x}}} and {{{9}}} to get {{{45x}}}




So lets multiply the inner terms: 

{{{9*5x=45x}}}   multiply {{{9}}} and {{{5x}}} to get {{{45x}}}




So lets multiply the last terms: 

{{{9*9=81}}}   multiply {{{9}}} and {{{9}}} to get {{{81}}}


Now lets put everything together

 {{{(5x+9)(5x+9)=25x^2+45x+45x+81=25x^2+90x+81}}}


 So the expression


 {{{(5x+9)(5x+9)}}}


 FOILs to:


 {{{25x^2+90x+81}}}


 -------------------------------------------------------------------------------