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Math Help - Polynomial

  1. #1
    Member SengNee's Avatar
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    Polynomial

    Find the quotient and remainder when P(x)=2x^4+3x^3-x^2+5x-6 is divided by x^2+x-2.


    Answer:

    Quotient= 2x^2+x+2
    Remainder= 5x-2


    Is the answers right?
    If yes, then...

    When x=-1,

    P(-1)=-13
    D(-1)=-2
    Q(-1)=3
    R(-1)=-7

    (-13)/(-2)
    =6 remain 1
    ≠3 remain -7


    When x=0,

    P(0)=-6
    D(0)=-2
    Q(0)=2
    R(0)=-2

    (-6)/(-2)
    =3 remain 0
    ≠2 remain -2


    When x= 1,

    P(1)=3
    D(1)=0
    Q(1)=5
    R(1)=3

    (3)/(0)
    =undefined
    ≠5 remain 3


    When x=2,

    P(2)=56
    D(2)=4
    Q(2)=12
    R(2)=8

    (56)/(4)
    =14 remain 0
    ≠12 remain 8



    Why???
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  2. #2
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    Find the quotient and remainder when P(x)=2x^4+3x^3-x^2+5x-6 is divided by x^2+x-2.


    Answer:

    Quotient=2x^2+x+2
    Remainder=5x-2


    Is the answers right?
    yes

    When x=0,

    P(0)=-6
    D(0)=-2
    Q(0)=2
    R(0)=-2

    (-6)/(-2)
    =3 remain 0
    ≠2 remain -2
    But these are equal. Think about what you mean when you say "2 remain -2". You mean 2+ \frac {-2}{-2}. This is the same as 3.

    =6 remain 1
    ≠3 remain -7
    6 remain 1 is wrong.  6 * (-2) + 1 = -11 \not = -13.
    It should be 6 remain -1 or 7 remain 1, both of which are the same as 3 remain -7.

    When x= 1,

    P(1)=3
    D(1)=0
    Q(1)=5
    R(1)=3

    (3)/(0)
    =undefined
    ≠5 remain 3
    You have here \frac {3}{0} = \frac {3}{0} +5. It is entirely understandable to find this weird, but when you deal with infinities you get this kind of stuff.

    I think the main issue here is understanding what the quotient and remainder actually are. What you discovered at the very start was that
    <br />
\frac {2x^4+3x^3-x^2+5x-6}{x^2+x-2} = 2x^2+x+2+\frac {5x-2}{x^2+x-2}
    This is not contradicted by any substitution for x although things get weird if the divisor is 0.
    Last edited by badgerigar; January 8th 2008 at 12:31 AM. Reason: forgot math tags
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  3. #3
    Member SengNee's Avatar
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    Find the quotient and remainder when P(x)=2x^4+3x^3-x^2+5x-6 is divided by x^2+x-2.


    Answer:

    Quotient= 2x^2+x+2
    Remainder= 5x-2


    Is the answers right?
    If yes, then...


    Let,
    Quotient=Q(x)= 2x^2+x+2
    Remainder=R(x)= 5x-2
    Divisor=D(x)= x^2+x-2

    When x=-1,

    P(-1)=-13
    D(-1)=-2
    Q(-1)=3
    R(-1)=-7

    \frac {P(-1)}{D(-1)}
    = \frac {-13}{-2}
    =6 \frac {1}{2}
    =6 remain 1
    ≠Q(-1) remain R(-1)
    ≠3 remain -7


    When x=0,

    P(0)=-6
    D(0)=-2
    Q(0)=2
    R(0)=-2

    \frac {P(0)}{D(0)}
    = \frac {-6}{-2}
    =3
    =3 remain 0
    ≠Q(0) remain R(0)
    ≠2 remain -2


    When x= 1,

    P(1)=3
    D(1)=0
    Q(1)=5
    R(1)=3

    \frac {P(1)}{D(1)}
    = \frac {3}{0}
    =undefined
    ≠Q(1) remain R(1)
    ≠5 remain 3


    When x=2,

    P(2)=56
    D(2)=4
    Q(2)=12
    R(2)=8

    \frac {P(2)}{D(2)}
    = \frac {56}{4}
    =14
    =14 remain 0
    ≠Q(2) remain R(2)
    ≠12 remain 8



    Why???


    I still don't understand what you talking about...
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    do you know polynomial long division or synthetic division? if this remainder theorem stuff is confusing you, maybe you should try that approach
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  5. #5
    Member SengNee's Avatar
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    I know this, but the long division like not correct sometimes just like the example I gave upside.
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by SengNee View Post
    Find the quotient and remainder when P(x)=2x^4+3x^3-x^2+5x-6 is divided by x^2+x-2.


    Answer:

    Quotient= 2x^2+x+2
    Remainder= 5x-2


    Is the answers right?
    If yes, then...


    Let,
    Quotient=Q(x)= 2x^2+x+2
    Remainder=R(x)= 5x-2
    Divisor=D(x)= x^2+x-2

    When x=-1,

    P(-1)=-13
    D(-1)=-2
    Q(-1)=3
    R(-1)=-7

    \frac {P(-1)}{D(-1)}
    = \frac {-13}{-2}
    =6 \frac {1}{2}
    =6 remain 1
    ≠Q(-1) remain R(-1)
    ≠3 remain -7
    I think you are missing the main idea. What does the phrase "Q(-1) remain R(-1)" mean?

    We need to know that
    P(x) = Q(x)D(x) + R(x)
    From this, as long as D(x) \neq  0
    \frac{P(-1)}{D(-1)} = Q(-1) + \frac{R(-1)}{D(-1)}
    Now you have already calculated \frac{P(-1)}{D(-1)} = \frac{13}2
    so let us do the remaining,
    Q(-1) + \frac{R(-1)}{D(-1)} = 3 + \frac{-7}{-2} = \frac{6+7}{2} = \frac{13}{2}

    I still don't understand what you talking about...
    I think none of us here are understanding what you are saying!

    I know this, but the long division like not correct sometimes just like the example I gave upside.
    This generally does not happen in maths. If a method is proven, then it will work ALWAYS!
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  7. #7
    Member SengNee's Avatar
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    I confuse about it.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post



    I confuse about it.
    In the example you posted, the one that is incorrect is incorrect because |-7| > |-2|. Recall that the remainder must be less than the divisor. To make this clearer, I would always rearrange the negative signs so that the divisor is positive.

    When dealing with polynomials the idea is that the remainder polynomial be of a smaller degree than the divisor polynomial.

    -Dan
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