Question 90554
Rewrite {{{(2x-5)^2}}} as {{{(2x-5)(2x-5)}}}


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



So lets multiply the first terms: 

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




So lets multiply the outer terms: 

{{{2x*-5=-10x}}}   multiply {{{2x}}} and {{{-5}}} to get {{{-10x}}}




So lets multiply the inner terms: 

{{{-5*2x=-10x}}}   multiply {{{-5}}} and {{{2x}}} to get {{{-10x}}}




So lets multiply the last terms: 

{{{-5*-5=25}}}   multiply {{{-5}}} and {{{-5}}} to get {{{25}}}


Now lets put all the multiplied terms together

 {{{4x^2-10x-10x+25}}}


 Now combine like terms


 {{{4x^2-20x+25}}}


 So the expression


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


 FOILs to:


 {{{4x^2-20x+25}}}