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Math Help - Cubic

  1. #1
    Junior Member phgao's Avatar
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    Cubic

    Another that has stumpted me... (sorri about the typos ticbol)

    I have

    P(x) = x^3 - 3x + 2
    So i find that P(2) is a value as it = zero.

    So I have:

    P(x) = (x-2)(ax^2 + bx + 1) (typo)

    Then I divide x-2 / x^3 - 3x + 2

    But I get x^2 +2x + 1 Remainder 4!!

    So what went wrong, how do you factorise the equation?! Need help!
    Last edited by phgao; February 12th 2006 at 03:05 AM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by phgao
    Another that has stumpted me...

    I have

    P(x) = x^2 - 3x + 2
    So i find that P(2) is a value as it = zero.

    So I have:

    P(x) = (x-2)(ax^2 + bx + 1) (typo)

    Then I divide x-2 / x^2 - 3x + 2

    But I get x^2 +2x + 1 Remainder 4!!

    So what went wrong, how do you factorise the equation?! Need help!
    You were stumped by that?

    If there is somebody who will not be stumped by that as well, he must be a genius.

    I thought I knew Math.

    You may want to rewrite or correct the data as given.
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  3. #3
    Junior Member phgao's Avatar
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    Very very sorry!!!

    Yeah its all fixed now.
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  4. #4
    Junior Member phgao's Avatar
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    Fixed... i think so. Hope you understand it now.

    Another that has stumpted me... (sorri about the typos ticbol)

    Polynumeral(x) = x^3 - 3x + 2
    One value of P = 2 as when you sub it in you get 0.


    So I have:

    P(x) = (x-2)(ax^2 + bx + 1)

    Then I divide x-2 / x^3 - 3x + 2 (by long division)

    But I get x^2 +2x + 1 Remainder 4!!


    Another Q I had went like this:

    P(x) = x^3 + 7x^2 + 14x + 8
    P(x) = (x+a)(x+b)(x+c)
    abc = 8

    So, I found that P(-1) is a possible answer.

    Then Eventually after long division I got:

    P(x) = (x+1)(x^2+6x+8)
    = (x+1)(x+2)(x+4)
    Last edited by phgao; February 12th 2006 at 03:14 AM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by phgao
    Fixed... i think so. Hope you understand it now.

    Another that has stumpted me... (sorri about the typos ticbol)

    Polynumeral(x) = x^3 - 3x + 2
    One value of P = 2 as when you sub it in you get 0.
    If a is a root of a polynomial P(x) (that is P(a)=0) then (x-a)
    is a factor of P(x). Now in your case P(x)=x^3-3x+2, and so
    P(2)=4 so (x-2) is not a factor of P(x)=x^3-3x+2.

    Now P(1)=0, so (x-1) is a factor of P(x).

    Factorising gives:

    P(x)=x^3-3x+2=(x-1)(x^2+x-2) =(x-1)(x-1)(x+2).

    RonL
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  6. #6
    Super Member
    earboth's Avatar
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    Quote Originally Posted by phgao
    Fixed... i think so. Hope you understand it now.

    Another that has stumpted me... (sorri about the typos ticbol)

    Polynumeral(x) = x^3 - 3x + 2
    One value of P = 2 as when you sub it in you get 0.
    Hello,

    I'm little puzzled, 'cause when I put 2 instaed of x I don't get zero:
    (2)^3-3\cdot2+2=8-6+2

    Take P(-2):

    (-2)^3-3\cdot(-)2+2=(-8)+6+2

    Bye
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  7. #7
    Junior Member phgao's Avatar
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    Thanks!
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