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Math Help - [SOLVED] Algebraic Long Division

  1. #1
    Super Member Quacky's Avatar
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    [SOLVED] Algebraic Long Division

    I have a few questions about algebraic division:

    a)As an example, how would I work out 2x^3-4x^2+6x-3 divided by x+1?
    I've never understood algebraic long division so a step by step guide (or link) would be greatly appreciated.

    b) If I have something like 2x^3-4x^2+6x-3 divided by 2x+1, how would I use the remainder theorem to work out the remainder? would I substiture in x=-0.5?

    Thanks in advance.
    Last edited by Quacky; January 9th 2010 at 08:04 AM. Reason: Clarified something.
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  2. #2
    Super Member
    earboth's Avatar
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    Quote Originally Posted by Quacky View Post
    I have a few questions about algebraic division:

    a)As an example, how would I work out 2x^3-4x^2+6x-3 divided by x+1?
    I've never understood algebraic long division so a step by step guide (or link) would be greatly appreciated.

    b) If I have something like 2x^3-4x^2+6x-3 divided by 2x+1, how would I use the remainder theorem to work out the remainder? would I substiture in x=-0.5?

    Thanks in advance.
    To a) You use the first summand of the divisor as divisor. To find the term which has to be subtracted you use the complete divisor. (Please read this sentence three times before you decide that you didn't understand it )
    Code:
      (2x^2 - 4x^2 + 6x - 3)  (x + 1) = 2x^2 - 6x + 12, R = -15
     -(2x^3 + 2x^2)
     ---------------
              -6x^2 + 6x
            -(-6x^2 - 6x)
            -------------
                      12x - 3
                    -(12x + 12)
            -----------------
                           -15
    to b) Yes.
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  3. #3
    Super Member Quacky's Avatar
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    You use the first summand of the divisor as divisor. To find the term which has to be subtracted you use the complete divisor.
    Wow, that's certainly a phrase to get my head around , but thanks for the explanations, I think I can follow your working when I try on paper now, I'll try some more examples by myself to see if I've understood.

    Thanks again.
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  4. #4
    Senior Member Dinkydoe's Avatar
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    (a)
    In general: Let f(x) = c_nx^n +c_{n-1}x^{n-1}+\cdots c_1x + c_0 be a polynomial of degree n. Say we want to divide this by a linear factor (ax+b). We will seek to try to find a polynomial q(x) of degree n-1 such that f(x) = q(x)(ax+b)+ r where r is the remainder of degree 0.(a constant)

    We reduce the polynomial step by step:
    Let g_1(x) = f(x) - \frac{c_n}{a}x^{n-1}\cdot (ax+b)= (c_{n-1}-\frac{c_nb}{a})x^{n-1}+c_{n-2}x^{n-2}+\cdots c_1x+c_0
    Let q_1(x) = \frac{c_n}{a}x^{n-1}

    We now reduced the polynomial with one degree and we have f(x) = q_1(x)(ax+b)+g_1(x)

    Continue in the same fashion with the new polynomial g_1(x) by repeatedly removing the head-coefficient.

    Your example: f(x) = 2x^3-4x^2+6x-3. Divide by x+1 :

    So we let q_1(x) = 2x^2
    and g_1(x) = f(x) - 2x^2(x+1) = -6x^2+6x-3
    q_2(x) = -6x
    g_2(x) = g_1(x)-(-6x(x+1)) =  12x - 3
    q_3(x) = 12
    g_3(x) = g_2(x) - 12(x+1) = -15

    Now q(x) = q_1(x)+q_2(x)+q_3(x) = 2x^2-6x+12 and f(x) = (2x^2-6x+12)(x+1) - 15.

    See now we got f into the form f(x) = q(x)(ax+b)+r with r=-15 as our remainder.

    (b) yes. The idea is that you got your polynomial into the form: f(x)= q(x)(ax+b)+r so if you plug in a zero of the linear factor we get f(\frac{-b}{a}) = r
    Last edited by Dinkydoe; January 9th 2010 at 09:51 AM.
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  5. #5
    Super Member Quacky's Avatar
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    Thanks for the explanation, I think between the two of you, everything I needed to know has been covered.
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