Question 46781
  
  {{{(x^2+2x-15)/4x^2}}} divided by {{{(x^2-25)/(2x-10)}}}
 
= {{{(x^2+2x-15)/4x^2}}} multiplied by the inverse of {{{(x^2-25)/(2x-10)}}}
 
= {{{((x^2+2x-15)/((2x)^2))((2x-10)/(x^2-25))}}} = {{{(highlight((x^2+2x-15))/(highlight((2x)^2)))(highlight((2x-10))/highlight((x^2-25)))}}} 
 
= {{{((x+5)(x-3)/(2(2x^2)))(2(x-5)/(x+5)(x-5))}}} = {{{(cross((x+5))(x-3)/(cross(2)(2x^2)))(cross(2)cross((x-5))/cross((x+5))cross((x-5)))}}} = {{{(x-3)/2x^2}}} Answer. 

 
gsm