Question 581628
The idea here is to factor this expression


{{{x^2+3x+5x+15}}} Start with the given expression



{{{(x^2+3x)+(5x+15)}}} Group the terms in two pairs



{{{x(x+3)+5(x+3)}}} Factor out the GCF {{{x}}} out of the first group. Factor out the GCF {{{5}}} out of the second group



{{{(x+5)(x+3)}}} Since we have the common term {{{x+3}}}, we can combine like terms



So {{{x^2+3x+5x+15}}} factors to {{{(x+5)(x+3)}}}



In other words, {{{x^2+3x+5x+15=(x+5)(x+3)}}}



So because {{{x^2+3x+5x+15}}} factors to {{{(x+5)(x+3)}}}, this means that the length and width are x+5 and x+3 units