Question 36459
Hey, hey!!  You wrote this one so I can see it!!  Good job.  Here is the equation in an algebra.com equation box:

{{{ ((m^2-16)/(m^2-6m+9))}}} divided by {{{ ((8-2m)/(m^2-4m-12)) }}}

Of course, your first step is to FACTOR everything that can be factored, and invert the second number and multiply:

{{{ (((m-4)*(m+4))/((m-3)*(m-3)))}}} times {{{ (((m-6)*(m+2))/(2*(4-m))) }}}


It is indeed strange that not much divides out here.  In fact, NOTHING matches up exactly, but there is a factor of (m-4) in the first numerator, and a factor of (4-m) in the second denominator that are NEGATIVES of one another.  When you divide a number by its negative, the result is always a factor of -1.  So these two factors divide out, leaving a factor of -1.  It is preferred to place this -1 factor in the numerator, so the final answer is this:


{{{ (-1*(m+4)*(m-6)*(m+2))/(2* (m-3)^2)}}}


R^2 at SCC