Question 46976
  
  {{{(z^2+8z+12)/(z^2+9z+14)}}} divided by {{{(z^2+6z)/(z^2+11z+28)}}}
  
= {{{((z^2+8z+12)/(z^2+9z+14))((z^2+11z+28)/(z^2+6z))}}}  .....  ...dividend to be multiplied by the multiplicative inverse of the divisor. Now both the denominators & numerators are to be factorised first.
  
= {{{(highlight((z^2+8z+12))/highlight((z^2+9z+14)))(highlight((z^2+11z+28))/highlight((z^2+6z)))}}} = {{{((z+6)(z+2)/(z+7)(z+2))((z+7)(z+4)/z(z+6))}}}
  
= {{{(cross((z+6))cross((z+2))/cross((z+7))cross((z+2)))(cross((z+7))(z+4)/cross((z+6))z)}}}
  
=  {{{(z+4)/z}}}   Answer.     
 
I hope this is clear to you.  
 
gsm.