Question 1209677
<pre>
Let the dividend polynomial be g(x). Let the quotient polynomial be q(x).


      <u>  q(x)    </u>
x<sup>2</sup>-x-6) g(x)
       ---------
            2x+7

The smallest degree g(x) could be is 2.  So let's see if we can find a
2nd degree polynomial that g(x) could be:

Let {{{g(x) = ax^2+bx+c}}}

          <u>                     a</u>
x<sup>2</sup> - x - 6)ax<sup>2</sup> +     bx +      c
           <u>ax<sup>2</sup> -     ax -     6a</u>
                 (b+a)x + (c+6a)    <--- remainder must equal 2x + 7

So we must have b+a = 2 and c+6a = 7.

We see immediately that a = b = c = 1 satisfies that. So

          <u>          1</u>
x<sup>2</sup> - x - 6)x<sup>2</sup> + x + 1
           <u>x<sup>2</sup> - x - 6</u>
               2x + 7

So one possibility would be {{{g(x)=x^2+x+1}}}

Then {{{g(8)=8^2+8+1=64+9=73}}} for this particular polynomial.

However g(8) is probably not unique, because there are many other
possibilities for a, b, and c, such that
b+a = 2 and c+6a = 7.  

Another such case is a = 3, b = -1, and c = -11
Those values also satisfy b+a = 2 and c+6a = 7
In that case,

          <u>             3</u>
x<sup>2</sup> - x - 6)3x<sup>2</sup> -  x - 11
           <u>3x2 - 3x - 18</u>
                 2x +  7


So another possibility would be {{{g(x)=3x^2-x-11}}}

Then {{{g(8)=3*8^2-8-11=3*64-19=173}}} for this particular polynomial.

Note that we only found solutions when g(x) has degree 2.  g(x) could
could have any degree 2 or greater.

This problem has infinitely many solutions. g(8) is not unique.

Edwin</pre>