# (2x^2 - 3x+6 ) / (2x+2) division

1. May 15, 2005

### roger

I dont understand how this works :

(2x^2 - 3x+6 ) / (2x+2)

I dont really understand the case using numbers either but we never divide by a+b+c etc , (ie we just divide by a single number rather than 3+5 for example)

Please explain step by step how it works. Also, why isnt the 2x and +2 separately distributive over the numerator ?

Roger

2. May 15, 2005

### dextercioby

Daniel.

3. May 15, 2005

### roger

But there is a difference between understanding and simply performing a task according to a given set of rules.

Can somebody explain how it works ?

As I said, I can get the answer, but I'm not quite sure what, I'm doing.. :grumpy:

4. May 15, 2005

### dextercioby

You're dividing two polyomials,you can't be doing anything else.

Daniel.

5. May 15, 2005

### arildno

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:
$$\frac{2x^{2}-3x+6}{2x+2}=x-\frac{5}{2}+\frac{11}{2x+2}$$

Last edited: May 15, 2005