Question 661194
Note: I'm going to use 'x' in place of 'y' in the long division


Use polynomial long division to get


<img src="http://i150.photobucket.com/albums/s91/jim_thompson5910/Algebra%20dot%20com/10-3-20121-12-39PM.png">


So {{{(y^4+2y-3)/(y^2+8)=y^2-8}}} remainder {{{2y+61}}}


Note: I'm switching from x back to y.


You can also say that


{{{(y^4+2y-3)/(y^2+8)=y^2-8+(2y+61)/(y^2+8)}}}