All right, roger:
Suppose you are to perform the divison 46/5.
What that means, is that you want to find a number "a", given in decimal form, so that 46=5*a
Let us set a=a_{1}*10+a_{0}*1+a_{-1}*\frac{1}{10}+a_{-2}*\frac{1}{100}+++
We must therefore determine coeficients a_{i}, i=1,0,-1,-2.. between 0 and 9, so that the equation holds:
46=a_{1}*50+a_{0}*5+a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++)
First, we see that the only valid choice for a_{1} between 0 and 9 is a_{1}=0
Thus, we have gained:
46=a_{0}*5+a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++)
Now, it follows that we must choose a_{0}=9 that is, we get, by subtracting 9*5 from both sides:
1=a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++),a_{0}=9
We now see that by setting a_{-1}=2 we get, by subtracting 2*\frac{5}{10} from both sides:
0=a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++),a_{0}=9, a_{-1}=2
Thus, the equation is fulfilled by setting the rest of the coefficients equal to zero, i.e.
46/5=9.2
You should think in a similar manner about polynomial division:
We want to find to find a function F(x)=P(x)+R(x), so that
(2x^2 - 3x+6 ) = (2x+2)*F(x), and
where P(x) is a polynomial (R(x) is then the "rest")
Let us set P(x)=a_{1}x+a_{0}, where a_{1},a_{0} are real numbers we want to find.
We have:
2x^{2}-3x+6=a_{1}2x^{2}+(2a_{1}+2a_{0})x+2a_{0}+(2x+2)*R(x)
Thus, we set a_{1}=1, and subtract 2x^{2} from both sides:
-3x+6=(2*1+2a_{0})x+2a_{0}+(2x+2)*R(x), a_{1}=1
Now, we set 2+2a_{0}=-3, that is, a_{0}=-\frac{5}{2} and, subtract -3x from both sides:
6=-\frac{5}{2}*2+(2x+2)*R(x), a_{1}=1,a_{0}=-\frac{5}{2}
from which we can determine R(x):
R(x)=\frac{11}{2x+2}
Thus, we have established:
\frac{2x^{2}-3x+6}{2x+2}=x-\frac{5}{2}+\frac{11}{2x+2}
