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Math Help - Nonlinear factor polynomial divisions + remainder theorem question

  1. #1
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    Nonlinear factor polynomial divisions + remainder theorem question

    Hi,

    I'm abit confused with these two things. Firstly, I was taught this quick shortcut for dividing polynomials (although I struggle with using it for dividing by non-linear factors). I'll try explain it and see if anyone can help me with it.

    To do it, using this example: 5x^5 + 13x^4 - 2x^2 - 6 / x + 1

    I write out the (x + 1) five times (since highest power is 5) and to make it like this:

    (x + 1) (x + 1) (x + 1) (x + 1) (x + 1)

    Then I'd go: 5x^4 is the first number, since that's what needed to do the first power. That would leave 5x^4 left. Then I'd need + 8x^4 to get to 13x^4 so my equation would now look like:

    5x^4(x + 1) + 8x^3 (x + 1)   (x + 1)   (x + 1)   (x + 1)

    Then so on until i get to:

    5x^4(x + 1) + 8x^3 (x + 1) - 8x^2(x + 1) + 6x(x + 1) - 6(x + 1)

    and then I can factor to find solution is 5x^4 + 8x^3 - 8x^2 + 6x - 6

    Dunno if anyone else knows that method. But if so, how would I do it if there was a remainder? Also, how could I use this to divide by non linear factors (ie: (x^2 + 1) which has a remainder when it is x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1 is divided by it)

    Also, I was wondering. with that above question (find remainder when x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1 is divided by (x^2 + 1)), how could I use the remainder thereom with it? Is it possible?

    Thanks!
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  2. #2
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    Hi

    Iīm a bit confused. Partially because you speak of division and but the equation looks

    y = aX^5+ bX^4 + cX^3 +dX^2 + 1/x. or you mean:
    ( aX^5+ bX^4 + cX^3 +dX^2)/(x+1) ??

    If there is a ramainder that canīt be devided, you donīt devide it, for example say the reminder is the constant 7, gives you
    ax^3 + bx^2 + 7/(x+1) <--------the reminder cant be devided.

    If you wanīt some more explaining search youtube. Some videos on this topic are great.
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  3. #3
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    Quote Originally Posted by Rudey View Post
    Hi,

    I'm abit confused with these two things. Firstly, I was taught this quick shortcut for dividing polynomials (although I struggle with using it for dividing by non-linear factors). I'll try explain it and see if anyone can help me with it.

    To do it, using this example: 5x^5 + 13x^4 - 2x^2 - 6 / x + 1

    I write out the (x + 1) five times (since highest power is 5) and to make it like this:

    (x + 1) (x + 1) (x + 1) (x + 1) (x + 1)

    Then I'd go: 5x^4 is the first number, since that's what needed to do the first power. That would leave 5x^4 left. Then I'd need + 8x^4 to get to 13x^4 so my equation would now look like:

    5x^4(x + 1) + 8x^3 (x + 1)   (x + 1)   (x + 1)   (x + 1)

    Then so on until i get to:

    5x^4(x + 1) + 8x^3 (x + 1) - 8x^2(x + 1) + 6x(x + 1) - 6(x + 1)

    and then I can factor to find solution is 5x^4 + 8x^3 - 8x^2 + 6x - 6

    Dunno if anyone else knows that method.
    I've seen it but I find it awkward to use. Perhaps just because I learned 'regular division' first.

    But if so, how would I do it if there was a remainder?
    Nothing new. you would just arrive a constant, not 0 term in the last step.

    Also, how could I use this to divide by non linear factors (ie: (x^2 + 1) which has a remainder when it is x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1 is divided by it)
    You wouldn't. At least not in the form you give here which consists of writing powers of your divisor.

    Also, I was wondering. with that above question (find remainder when x^5 - 3x^4 + 2x^3 - 2x^2 + 3x + 1 is divided by (x^2 + 1)), how could I use the remainder thereom with it? Is it possible?
    The "remainder theorem" says that the remainder when polynomial p(x) is divided by x-a is just p(a). You can't use that here because x^2+ 1 is not of the form "x- a".

    Of course, the "Euclidean algorithm" tells you that you can divide and that the remainder will be either a constant or a first degree polyomial.

    Thanks!
    Last edited by HallsofIvy; January 10th 2010 at 04:40 AM.
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  4. #4
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    how could I use the remainder thereom with it? Is it possible?
    Well in fact you can, but you must understand polynomial division first:

    If you divide a polynomial p(x) by x^2+1 we can find a polynomial q(x) such that p(x) = q(x)(x^2+1)+r

    so filling in a zero of (x^2+1), that is x_0 = \pm i gives p(x_0) = r

    (observe that (x^2+1)= (x-i)(x+1) wich gives zero's \pm i. Do you know what complex numbers are?))
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  5. #5
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    So how could I solve it? Could someone please work through it, so I can see how it's done and apply it to similar questions. Here is the question:

    Let P(x)=x^5-3x^4+2x^3-2x^2+3x+1
    a) Show that neither (x-1) nor (x+1) is a factor of P(x)

    This I did by using the factor theorem and showing that P(1) and P(-1) do not equal 0


    b) Given that P(x) can be written in the form
    (x^2 - 1)Q(x) + ax + b where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when P(x) is divided by (x^2-1)
    The solution is: 6x - 4
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  6. #6
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    The first one.

    Given P(x) = x^5 - 3x^4 + 2x^3 - 2x^2 +3x +1

    Showing that P(1) is not 0
    P(1) = 1^5 - 3*1^4 +2*1^3 -2*1^2 +3*1 +1 = 2
    so its not zero. That shows you that when x = 1; y = 2 so the curve does not intersect X-axis when X is 1 correct: (x-1) can therefor not be a factor.

    If you wanīt to show that neither x-1; x+1 are factors just divide.

    (1^5 - 3*1^4 +2*1^3 -2*1^2 +3*1 +1)/(x+1)

    IF there are no remainders then it factors but if there is a remainder it can not factor. If you donīt know how to devide I can follow you through. A tip is to check out the youtube videos on polynomial division, there are som very good once. You will be solving this in notime yourself, when u get how to do it.
    Last edited by Henryt999; January 9th 2010 at 02:32 PM.
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  7. #7
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    Hi, thanks for reply. But I managed to get that, what I was having trouble with was how to solve the second part (part b)
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