Question 134723
Given: {{{((a^2+2a-24)/(a+1)) / ((a^2-a-12)/(a+1)^2) }}}

Multiply both top and bottom by (a+1)   {{{(a+1)/(a+1)}}}
Given: {{{((a^2+2a-24)/1) / ((a^2-a-12)/(a+1)) }}}

Factor 
{{{ (((a+6)(a-4))/1) / (((a-4)(a-3))/(a+1)) }}}

The term {{{(a-4)}}}{ occurs on both the numerator and the denominator, so it cancels to 1
{{{ (((a+6))/1) / (((a-3))/(a+1)) }}}

Now, as we did in the earlier problem, take the inverse of the denominator and multiply that times the numerator
{{{ (((a+6))/1) * (((a+1))/(a-3)) }}}
Simplify
{{{ ((a+6)(a+1))/(a-3) }}}