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

  • Context: High School 
  • Thread starter Thread starter roger
  • Start date Start date
  • Tags Tags
    Division
Click For Summary

Discussion Overview

The discussion revolves around the division of the polynomial (2x^2 - 3x + 6) by (2x + 2). Participants explore the concept of polynomial division, seeking clarification on the process and underlying principles.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about polynomial division and requests a step-by-step explanation, questioning why terms cannot be treated separately in the division.
  • Another participant challenges the first poster's understanding of polynomial division, suggesting a lack of foundational knowledge.
  • A further reply emphasizes the distinction between performing a task and understanding the underlying concepts of polynomial division.
  • One participant attempts to draw an analogy between polynomial division and simple numerical division, outlining a method to find coefficients in a polynomial context.
  • This participant proposes a specific approach to polynomial division, detailing the steps to express the polynomial as a product of the divisor and a quotient plus a remainder.

Areas of Agreement / Disagreement

There is no consensus among participants. Some express confusion and seek clarification, while others assert differing levels of understanding and approach to the topic.

Contextual Notes

Participants' contributions reflect varying degrees of familiarity with polynomial division, and there are unresolved assumptions regarding foundational knowledge and the application of division principles.

roger
Messages
318
Reaction score
0
I don't understand how this works :


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

I don't 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 isn't the 2x and +2 separately distributive over the numerator ?


Roger
 
Mathematics news on Phys.org
What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?

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

Daniel.
 
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:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
3K