Question 151462


{{{(2x+4y)^2}}} Start with the given expression.



{{{(2x+4y)(2x+4y)}}} Expand. Remember something like {{{x^2=x*x}}}.



Now let's FOIL the expression.



Remember, when you FOIL an expression, you follow this procedure:



{{{(highlight(2x)+4y)(highlight(2x)+4y)}}} Multiply the <font color="red">F</font>irst terms:{{{(2*x)*(2*x)=4*x^2}}}.



{{{(highlight(2x)+4y)(2x+highlight(4y))}}} Multiply the <font color="red">O</font>uter terms:{{{(2*x)*(4*y)=8*x*y}}}.



{{{(2x+highlight(4y))(highlight(2x)+4y)}}} Multiply the <font color="red">I</font>nner terms:{{{(4*y)*(2*x)=8*x*y}}}.



{{{(2x+highlight(4y))(2x+highlight(4y))}}} Multiply the <font color="red">L</font>ast terms:{{{(4*y)*(4*y)=16*y^2}}}.



{{{4*x^2+8*x*y+8*x*y+16*y^2}}} Now collect every term to make a single expression.



{{{4*x^2+16*x*y+16*y^2}}} Now combine like terms.



So {{{(2x+4y)^2}}} FOILS to {{{4*x^2+16*x*y+16*y^2}}}.



In other words, {{{(2x+4y)^2=4*x^2+16*x*y+16*y^2}}}.