Question 621327
{{{x^2-6x-216=(x+12)(x-18)}}} so
a rectangle whose area is {{{x^2-6x-216}}} square feet
could be {{{x+12}}} feet long and {{{x-18}}} feet wide.
Its dimensions could also be {{{x^2-6x-216}}} feet and 1 foot.
It could be {{{2sqrt(x^2-6x-216)}}} feet long by {{{sqrt(x^2-6x-216)/2}}} feet wide.
Maybe the dimensions could be {{{root(3,(x^2-6x-216)^2)}}} by {{{root(3,x^2-6x-216)}}}.
That rectangle has infinite options for its dimensions.