Question 194088
{{{(m^2-n^2-4m-4n)/(2m-2n-8)}}} Group the terms in the numerator as follows:
{{{((m^2-n^2)-highlight_green((4m+4n)))/(2m-2n-8)}}} Notice the change of sign in the second group.
Let's do the numerator first:
{{{(m^2-n^2)-(4m+4n)}}} Factor the first group as the difference of squares and factor a 4 from the second group to get:
{{{(m-n)*highlight((m+n))-4*highlight((m+n))}}} Now factor out the common factor of ({{{highlight(m+n)}}})
{{{(m+n)(m-n-4)}}} This is the numerator.
Now the denominator:
{{{2m-2n-8}}} Factor out a 2.
{{{2(m-n-4)}}} Putting it all together, we have:
{{{(m+n)highlight_green((m-n-4))/(2*highlight_green((m-n-4)))}}} Cancel the indicated factors to leave you with:
{{{highlight((m+n)/2)}}}