<PRE><B>Question 16772</B>
(6x^5 - 5x^4 - 6x^3 - 10x^2 + 21x - 9)/(2x - 3)

Straight out let me tell you I have never liked long division. And looking at that,I suddenly remembered why. But anyhow, heres how it goes:&lt;br&gt;

&lt;pre&gt;
2x - 3 &lt;b&gt;|&lt;/b&gt; 6x^5 - 5x^4 - 6x^3 - 10x^2 + 21x - 9 &lt;b&gt;|&lt;/b&gt; 3x^4 + 2x^2 - 5x + 3
       &lt;b&gt;|&lt;/b&gt; 6x^5 - 9x^4
               -     +
        &lt;b&gt;---------------&lt;/b&gt;
                       4x^4 - 6x^3
                       4x^4 - 6x^3
                      -     +
                      &lt;b&gt;------------&lt;/b&gt;
                                    -10x^2 + 21x
                                    -10x^2 + 15x
                                    +      -
                                   &lt;b&gt;------------&lt;/b&gt;
                                             6x - 9
                                             6x - 9
                                            -   +
                                           &lt;b&gt;----------&lt;/b&gt;
                                             [remainder = 0]

&lt;/pre&gt;

&lt;P&gt;
As you can see,the quotient is (3x^4+2x^3-5x+3) and there is no remainder. Hope this helps.
 - Prabhat</PRE>