Question 146198
I'm assuming that the expression for the area is {{{4x^2-9y^2}}}



{{{4x^2-9y^2}}} Start with the given expression


{{{(2x)^2-9y^2}}} Rewrite {{{4x^2}}} as {{{(2x)^2}}}


{{{(2x)^2-(3y)^2}}} Rewrite {{{9y^2}}} as {{{(3y)^2}}}



Now use the difference of squares to factor the expression. Remember, the difference of squares formula is {{{A^2-B^2=(A+B)(A-B)}}} where in this case {{{A=2x}}} and {{{B=3y}}}


{{{4x^2-9y^2=(2x+3y)(2x-3y)}}} Plug in {{{A=2x}}} and {{{B=3y}}}


So the expression


{{{4x^2-9y^2}}}


factors to


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


Notice that if you foil the factored expression, you get the original expression. This verifies our answer.




So this means that the width is {{{2x+3y}}} and the length is {{{2x-3y}}} (or vice versa since the order does not matter).