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Math Help - Remainder with polynomial division

  1. #1
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    Remainder with polynomial division

    The question: The remainder obtained when 3x^4 + 7x^3 + 8x^2 -2x -3 is divided by x + 1 is?

    The answer is 3. My book says to plug in -1 for x (which results with 3). Why should I do this?
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    MHF Contributor Siron's Avatar
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    Re: Remainder with polynomial division

    Because that's the way you find a remainder.
    For example if you want to solve the equation f(x)=3x^4+7x^3+8x^2-2x-3=0, in first instance you're going to search for a number x_1 wherefore f(x_1)=0, that means if you divide f(x) by (x-x_1) you'll obtain remainder 0, in this case if you enter -1 you see f(x)=3 so the remainder is 3. To have a closer look you can make the Horner scheme.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Remainder with polynomial division

    Quote Originally Posted by benny92000 View Post
    The question: The remainder obtained when 3x^4 + 7x^3 + 8x^2 -2x -3 is divided by x + 1 is?

    The answer is 3. My book says to plug in -1 for x (which results with 3). Why should I do this?
    Read here:

    Polynomial long division - Wikipedia, the free encyclopedia
    Attached Thumbnails Attached Thumbnails Remainder with polynomial division-33333.gif  
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    Re: Remainder with polynomial division

    Hmm, thanks. Makes a little more sense. I guess I still need to work a few hands-on examples to make sense of why we do that. Let me ask a few questions to get a more thorough understanding:
    1.) Why do we set the first equation equal to 0?
    2.) Why does dividing by the the x-x1 give us remainder 0?
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    MHF Contributor Siron's Avatar
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    Re: Remainder with polynomial division

    That was just an example (another case), and because x_1 is a zero therefore if you divive f(x) by (x-x_1) you'll obtain remainder 0 (this is a theoretical look to it).
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    Re: Remainder with polynomial division

    Oh... Ok. That seems to make sense now. Thanks!
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    MHF Contributor Siron's Avatar
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    Re: Remainder with polynomial division

    You're welcome!
    Beside the 'polynomial long division' (see Also sprach Zarathrusta's post) you can also use Horner scheme to evaluate the remainder.
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