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

  1. May 15, 2005 #1
    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 ?

  2. jcsd
  3. May 15, 2005 #2


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    What do you know about polynomial division...?My guess is:nothing.So how about do some reading ?Okay?

  4. May 15, 2005 #3

    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:
  5. May 15, 2005 #4


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

  6. May 15, 2005 #5


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    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:
    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:
    Now, it follows that we must choose [tex]a_{0}=9[/tex] that is, we get, by subtracting 9*5 from both sides:
    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.

    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:
    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):
    Thus, we have established:
    Last edited: May 15, 2005
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