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Division of polynomials (comparing coefficient method) how to find out remainder watch

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    I prefer using the "comparing coefficients" method (I think that's what it's called. It's not actually in my textbook.) because I'm completely hopeless at the other methods.

    This is the method I mean. It's just an example. (x^3 + 4x^2 + x -6)/(x - 2) = Ax^2 + Bx + C

    And multiply both sides by the divisor and just go through the letters A, B and C to find out the value.

    But what do I do when there's a remainder? How would I finish this question?

    (x^3 + 3x^2 + 6x + 1)/(x + 1)


    I make (x^3 + 3x^2 + 6x + 1) = (Ax^2 + Bx + C)(x + 1)

    coefficient of x^3 : 1 = A therefore A = 1
    coefficient of x^2 : 3 = A + B therefore B = 2
    coefficient of x^1 : 6 = B + C therefore C = 4
    coefficient of x^0 : 1 = C therefore C = 1

    Obviously my last value of C is wrong. How do I work out the remainder please? Rep on offer. Thanks.
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    Introduce an extra constant, the remainder R, by writing (x^3 + 3x^2 + 6x + 1) \equiv (Ax^2 + Bx + C)(x + 1) + R
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    (Original post by Kolya)
    Introduce an extra constant, the remainder R, by writing (x^3 + 3x^2 + 6x + 1) \equiv (Ax^2 + Bx + C)(x + 1) + R
    Righto! So once I've worked out my A, B and C values what do I do to work out R?
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    (Original post by AnythingButChardonnay)
    Righto! So once I've worked out my A, B and C values what do I do to work out R?

    What value does the polynomial take when x = -1?
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    (Original post by ghostwalker)
    What value does the polynomial take when x = -1?
    -3, which is the answer!

    So I just have to plug x = -1 into the polynomial every time to find out what the remainder is? If so that's great. Why does it work?
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    (Original post by AnythingButChardonnay)
    -3, which is the answer!

    So I just have to plug x = -1 into the polynomial every time to find out what the remainder is? If so that's great. Why does it work?

    NOOOOOO. You plug in x=-1 because (x+1) is a factor on the RHS, and hence that part will be zero, and you are left with R.


    If the factor was (x+a), you'd use x=-a.
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    When you have:

    (x+1)(x^2+2x+4), and you multiply them out, by inspection you can see the x^0 coefficient will be 4

    hence -3 will be needed to make the x^0 coefficient 1.

    (when using your method and working out the value of a,b and c, the higher coefficients always take priority, so c will be 4 rather than 1)

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    (Original post by ghostwalker)
    NOOOOOO. You plug in x=-1 because (x+1) is a factor on the RHS, and hence that part will be zero, and you are left with R.


    If the factor was (x+a), you'd use x=-a.
    :o:

    So if I was dividing by (2x + 3) I would plug in x = -3/2 ?

    (Original post by Robbie10538)
    When you have:

    (x+1)(x^2+2x+4), and you multiply them out, by inspection you can see the x^0 coefficient will be 4

    hence -3 will be needed to make the x^0 coefficient 1.
    Ah! That's good too. I would probably use that to check if I had time. It depends which would take longer, and which I'd be more likely to hash up.
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    (Original post by AnythingButChardonnay)
    :o:

    So if I was dividing by (2x + 3) I would plug in x = -3/2 ?
    Yes that is correct. The method he showed is the remainder theorem and so works as well. I'd use this to check you have the correct remainder.
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    Brilliant. Thanks so much. Rep to all 3 of you.
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    So if I was dividing by (2x + 3) I would plug in x = -3/2 ?
    Yep.
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    This is the method I use and I think it is so much easier to understand than long division.
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    Depends which way is fastest for you.
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    I like this method too, just very easy to see where you've gone wrong, if you have.
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    how about when you are dividing by an x squared, an x and a number
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    (Original post by bruce11)
    how about when you are dividing by an x squared, an x and a number
    Easiest to have a specific example you're working on, and how far you've got.
 
 
 
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