Question 636887
Your error is in going from
{{{(x(2x^3-3x+5)-2)/(x+2)}}}
to
{{{(2x^3-3x+5-1)/1}}}
When reducing fractions, <b><i>only</i></b> factors cancel!! The "x" in the numerator is only a factor of part of the numerator, not the whole numerator. And the "x" in the denominator is not a factor at all. And the 2's are not factors either. So neither is x nor the 2 cancels. In fact, the numerator is prime. It will not factor. So there no common factors to cancel. It is not possible to reduce this fraction.<br>
But it is possible to divide the numerator by the denominator. Long division can be used:
<pre>
      2x^3 - 4x^2 + 5x    -  5
     _________________________________
x+2 / 2x^4 + 0x^3 - 3x^2  +  5x  -2
      2x^4 + 4x^3
      -----------
            -4x^3 -3x^2
            -4x^3 -8x^2
           ------------
                   5x^2 +  5x
                   5x^2 + 10x
                   ---------
                          -5x -  2
                          -5x - 10
                          --------
                                 8
</pre>
The remainder is 8. And as you may recall from when you first learned long division, you make a fraction out of the remainder and the divisor. So
{{{(2x^4 - 3x^2  +  5x  -2)/(x+2) = 2x^3 - 4x^2 + 5x    -  5 + 8/(x+2)}}}<br>
With a divisor of x+2, synthetic division could also be used:
<pre>
-2 |  2   0   -3    5   -2
         -4    8  -10   10
      --------------------
      2  -4    5   -5    8
</pre>
As we should have expected, we get a remainder of 8 (in the lower right corner). And the quotient is represented in the rest of the bottom row. The "2  -4    5   -5" translates into {{{2x^3 - 4x^2 + 5x    -  5}}}. So with synthetic division we get the same answer:
{{{(2x^4 - 3x^2  +  5x  -2)/(x+2) = 2x^3 - 4x^2 + 5x    -  5 + 8/(x+2)}}}<br>
P.S. Note that in both methods of division, it was important to use "0x^3" for the missing x^3 term. <i>Any</i> missing terms should be handled this way when dividing.