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

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In summary, polynomial division involves finding the coefficients of a polynomial equation by dividing it by another polynomial. This is done by setting up an equation and finding the values of the coefficients through a step-by-step process. The divisor is then distributed over the dividend and the resulting equation is solved for the coefficients. This process is similar to dividing numbers in decimal form.
  • #1
roger
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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
 
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  • #2
What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?

Daniel.
 
  • #3
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..
 
  • #4
You're dividing two polyomials,you can't be doing anything else.

Daniel.
 
  • #5
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]
 
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Related to (2x^2 - 3x+6 ) / (2x+2) division

1. What is the quotient when dividing (2x^2 - 3x+6 ) by (2x+2)?

The quotient when dividing (2x^2 - 3x+6 ) by (2x+2) is x-2. This can be determined by using long division or synthetic division.

2. Can the division of (2x^2 - 3x+6 ) by (2x+2) be simplified further?

Yes, the division of (2x^2 - 3x+6 ) by (2x+2) can be simplified to x-2. This is the simplest form of the quotient and cannot be further simplified.

3. How does the division of (2x^2 - 3x+6 ) by (2x+2) relate to the concept of remainder?

The division of (2x^2 - 3x+6 ) by (2x+2) does not have a remainder because the polynomial division algorithm is used. However, if the division was done using long division, the remainder would be zero.

4. Is it possible to divide (2x^2 - 3x+6 ) by (2x+2) when x = -1?

Yes, it is possible to divide (2x^2 - 3x+6 ) by (2x+2) when x = -1. This would result in a quotient of -1 and a remainder of 0.

5. How can the division of (2x^2 - 3x+6 ) by (2x+2) be used to solve real-life problems?

The division of (2x^2 - 3x+6 ) by (2x+2) can be used to solve real-life problems involving polynomial equations, such as finding the roots of a quadratic equation. It can also be used to simplify complex algebraic expressions and solve for unknown variables.

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