Question 154557
<font size = 6 color = "red"><b>Stanbon's solution is wrong.
Correct solution by Edwin:

<pre><font size = 4 color = "indigo"><b>

I can't tell whether the divisor is {{{x-6}}} or {{{-x-6}}}
so I'll do it both ways.  With divisor {{{x-6}}}

{{{(x - (1/3)x^4 + 2x^3 + 2x^2)}}}÷{{{(x -6)}}}

Rearrange the numerator in descending order:

{{{(-1/3)x^4 + 2x^3 + 2x^2 + x)}}}÷{{{(x -6)}}}

There is a missing constant term on the right end
so we put 0

{{{(-1/3)x^4 + 2x^3 + 2x^2 + x+0)}}}÷{{{(x -6)}}}


             (-1/3)x³ + 0x² + 2x + 13 
     --------------------------------
x - 6) (-1/3)x<sup>4</sup> + 2x³ + 2x² +  x +  0
       (-1/3)x<sup>4</sup> + 2x³
       --------------
                  0x³ + 2x²
                  0x³ + 0x²
                  ---------
                        2x² +  x  
                        2x² -12x
                        --------
                             13x +  0
                             13x - 78
                             --------
                                   78

Answer:   {{{((-1/3)x^3 + 2x + 13) + 78/(x-6)}}}

--------------------------------

With divisor {{{-x-6)}}}
<pre><font size = 4 color = "indigo"><b>
{{{(x - (1/3)x^4 + 2x^3 + 2x^2)}}}÷{{{(x -6)}}}

Rearrange the numerator in descending order:

{{{(-1/3)x^4 + 2x^3 + 2x^2 + x)}}}÷{{{(x -6)}}}

There is a missing constant term on the right end
so we put 0

{{{(-1/3)x^4 + 2x^3 + 2x^2 + x+0)}}}÷{{{(-x -6)}}}


               (1/3)x³ -  4x² +  22x - 133 
      ------------------------------------
-x - 6) (-1/3)x4 + 2x³ +  2x² +    x +   0
        (-1/3)x4 - 2x³
       --------------
                   4x³ +  2x²
                   4x³ + 24x²
                  ---------
                        -22x² +    x  
                        -22x² - 132x
                        ------------
                                133x +   0
                                133x + 798
                             -------------
                                      -798 

Answer:   {{{((-1/3)x^3 - 4x^2 + 22x - 133) - 798/(x-6)}}}





Edwin</pre></font></b>