Question 274656
<pre>
       3x^3+ 4x^2+ 2x  + 4
      ________________________
2x+3 / 6x^4+17x^3+16x^2+14x+12
       6x^4+ 9x^3
       ----------
             8x^3+16x^2
             8x^3+12x^2
             ----------
                   4x^2+14x
                   4x^2+ 6x
                   --------
                         8x+12
                         8x+12
                         -----
                             0
</pre>
So {{{(6x^4+17x^3+16x^2+14x+12)/(2x+3) = 3x^3+ 4x^2+ 2x  + 4}}} (No remainder)