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Thread: Sythetic division

  1. #1 Sythetic division 
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    Is synthetic devision possible only with a linear diviser?

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  3. #2  
    Forum Professor river_rat's Avatar
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    You can extend the method to work with higher order divisors but standard synthetic division only works with linear divisors.


    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  4. #3 Thanks 
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    hi,
    Thanks alot. Do you know any site that can explain the extended method.

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  5. #4  
    Forum Professor river_rat's Avatar
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    I could do one better and walk you through the method

    Before i start, do you know why normal synthetic division works?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  6. #5 hi 
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    Woow thanks alot. Well umm a bit shamed to say, but I actually don't know why the normal synthetic devision works. Its only that I know the mechanism. I know this is wrong. But can you really help me?

    Thanks alot
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  7. #6  
    Forum Professor river_rat's Avatar
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    Hi Dilan

    The power of synthetic division comes from the polynomial remainder theorem (which you should have met at some point at school) which says that if you divide a polynomial P(x) by a (x-a) then the remainder is P(a). You use this fact quite a bit when you solve polynomial equations, as (x-a) is a factor iff its remainder when dividing P(x) is 0 i.e. P(a) = 0. An example might help clear this up, consider P(x) = x<sup>2</sup> - 1 and say we wish to divide it by (x+1) and see what the remainder is. Well in this case our a = -1 and a quick calculation gives that P(-1) = 0 i.e. (x+1) is a factor of P(x) which infact is P(x) = (x+1)(x-1).

    Now synthetic division uses this fact to maximum effect by utilising the following little trick. We can write the term b x<sup>n+1</sup> as
    b x<sup>n+1</sup> = b x<sup>n</sup> (x-a) + a b x<sup>n</sup>

    This is in fact the synthetic division method writen out term for term, perhaps an example may help here again. Consider the same polynomial P(x) = x<sup>2</sup> - 1 = x<sup>2</sup> +0 x - 1 and lets try divide it by (x+1) using the above recipe for guidance (it may help if you do the normal synthetic division steps on paper as you read this).

    1. Starting with the x<sup>2</sup> term, we know we can write that as x (x+1) - x. So our polynomial now looks like P(x) = x(x+1) - x - 1
    2. Lets look at the -x term, we can write that according to our recipe as -(x+1) + 1, so our polynomial now looks like P(x) = x(x+1) - (x+1) + 0
    3. Now here comes the remainder theorem, remember that the remainder when you divide P(x) by (x+1) is given by P(-1). But twe have writen our polynomial in such a way that you can see the remainder immediately since all the terms involving (x+1) equate to 0. So our remainder is 0 (so once again (x+1) is a factor of P(x) )


    So thats how normal synthetic division works - if thats all clear then i can show you the extended method.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  8. #7 ok 
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    Hi,

    Thanks alot. Everything clear as crystal. Ya it was easy to understand when I was writing down on a paper. Thanks alot.
    ok now for the extended method . Hope I will understand

    Thanks alot
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  9. #8  
    Forum Professor river_rat's Avatar
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    Hi Dilan

    Now here is a form of synthetic division for higher order divisors. Its easier to explain the method using an example and as it does translate well into the standard block format on a forum i wont try use that.

    So lets try divide x<sup>3</sup> + 3 x<sup>2</sup> + 2 x + 1 by x<sup>2</sup> + 1. We use the same trick as we did for normal synthetic division (i.e. we try write each term as something times the divisor) and arrive at the following steps.

    x<sup>3</sup> + 3 x<sup>2</sup> + 2 x + 1 (First we handle the x<sup>3</sup> term)
    = x(x<sup>2</sup> + 1) + 3 x<sup>2</sup> + x + 1 (Then the x<sup>2</sup>)
    = x(x<sup>2</sup> + 1) + 3(x<sup>2</sup> + 1) + x - 2 (We cant take out any more x<sup>2</sup> so we are finished)

    Which shows us that x<sup>3</sup> + 3 x<sup>2</sup> + 2 x + 1 = (x<sup>2</sup> + 1)(x + 3) + (x - 2)

    That long method translates quite nicely to the block method of normal synthetic division (though you do end up using a few lines extra for the intermediate steps). So there we have division of two polynomials with out ever dividing anything
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  10. #9 OMG 
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    OMG, I just can't believe this. All these days I divided using the long division. But this is soo easy. Woow, thanks alot. This reduces the errors also. Long division is sort of a little difficult when your dealing with polynomials.

    Thanks alot river rat. You really helped me alot.

    Thanks again :-D
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