Question 675962
the area of a rectangle is {{{A=(2x^2-11x+15)ft^2}}} 

the length of the rectangle is {{{L=(2x-5)ft}}} 

the width {{{W}}} will be


{{{A=L*W}}}


{{{W=A/L}}}


{{{W=(2x^2-11x+15)ft^2/(2x-5)ft}}}.....replace {{{-11x}}} with {{{-6x-5x}}}


{{{W=(2x^2-6x-5x+15)ft/(2x-5)}}}....group


{{{W=((2x^2-6x)-(5x-15))ft/(2x-5)}}}


{{{W=(2x(x-3)-5(x-3))ft/(2x-5)}}}


{{{W=((2x-5)(x-3))ft/(2x-5)}}}


{{{W=(cross((2x-5))(x-3))ft/cross((2x-5))}}}


{{{highlight(W=(x-3)ft)}}}