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

  • Thread starter roger
  • Start date
319
0
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
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,925
501
What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?

Daniel.
 
319
0
dextercioby said:
What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?

Daniel.

Your guess is wrong.

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:
 

dextercioby

Science Advisor
Homework Helper
Insights Author
12,925
501
You're dividing two polyomials,you can't be doing anything else.

Daniel.
 

arildno

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
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 [tex]a=a_{1}*10+a_{0}*1+a_{-1}*\frac{1}{10}+a_{-2}*\frac{1}{100}+++[/tex]
We must therefore determine coeficients [tex]a_{i}, i=1,0,-1,-2..[/tex] between 0 and 9, so that the equation holds:
[tex]46=a_{1}*50+a_{0}*5+a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++)[/tex]
First, we see that the only valid choice for [tex]a_{1}[/tex] between 0 and 9 is [tex]a_{1}=0[/tex]
Thus, we have gained:
[tex]46=a_{0}*5+a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++)[/tex]
Now, it follows that we must choose [tex]a_{0}=9[/tex] that is, we get, by subtracting 9*5 from both sides:
[tex]1=a_{-1}*\frac{5}{10}+a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++),a_{0}=9[/tex]
We now see that by setting [tex]a_{-1}=2[/tex] we get, by subtracting [tex]2*\frac{5}{10}[/tex] from both sides:
[tex]0=a_{-2}*\frac{5}{100}+5*(a_{-3}*\frac{1}{1000}++),a_{0}=9, a_{-1}=2[/tex]
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 [tex]P(x)=a_{1}x+a_{0}[/tex], where [tex]a_{1},a_{0}[/tex] are real numbers we want to find.
We have:
[tex]2x^{2}-3x+6=a_{1}2x^{2}+(2a_{1}+2a_{0})x+2a_{0}+(2x+2)*R(x)[/tex]
Thus, we set [tex]a_{1}=1[/tex], and subtract [tex]2x^{2}[/tex] from both sides:
[tex]-3x+6=(2*1+2a_{0})x+2a_{0}+(2x+2)*R(x), a_{1}=1[/tex]
Now, we set [tex]2+2a_{0}=-3[/tex], that is, [tex]a_{0}=-\frac{5}{2}[/tex] and, subtract -3x from both sides:
[tex]6=-\frac{5}{2}*2+(2x+2)*R(x), a_{1}=1,a_{0}=-\frac{5}{2}[/tex]
from which we can determine R(x):
[tex]R(x)=\frac{11}{2x+2}[/tex]
Thus, we have established:
[tex]\frac{2x^{2}-3x+6}{2x+2}=x-\frac{5}{2}+\frac{11}{2x+2}[/tex]
 
Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top