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Math Help - Serious Polynomial Fun

  1. #1
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    Cool Serious Polynomial Fun

    This is for anyone who wants a challenge with polynomials because they are too difficult for me thank you for any help

    i have found the answers

    a)-2
    b)-11
    c)12
    d)-11/12
    e)-1/6
    f)0
    g)-132
    h)26
    i)13/72

    would someone be able to check these
    Attached Thumbnails Attached Thumbnails Serious Polynomial Fun-polynomial.jpg  
    Last edited by nath_quam; June 18th 2006 at 01:10 AM.
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  2. #2
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    Quote Originally Posted by nath_quam
    i have found the answers for
    a)-2
    b)-11
    c)12
    d)-11/12
    e)-1/6
    f)0
    g)?
    h)26
    i)?
    would someone be able to check these and solve/tell me how to do the unknowns
    Hello,

    all your results are correct. (Congratulations!)

    with g): Plug in the values you know and you'll have:
    (-12)^2\cdot (-1)+(-3)^2\cdot (-4)+(4)^2\cdot 3 = -132

    with i): You may know that n^{-2}=\frac{1}{n^2}. With your solutions you get:

    \left( {1\over(-12)}\right)^2+\left( {1\over(-3)}\right)^2+\left( {1\over(4)}\right)^2=\left( {13\over 72}\right)

    Greetings

    EB
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  3. #3
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    Hello, nath_quam!

    These are great problems!
    They give me a chance to show-off . . . LOL!

    The first three are easy . . . if you know this theorem:
    If a,b,c are roots of the cubic x^3 + px^2 + qx + r\,=\,0
    . . then: . a + b + c \,= \,-p,\;\;ab + bc + ac \,= \,q,\;\;pqr \,= \,-r


    If a,b,c are roots of x^3 + 2x^2 - 11x - 12\:=\:0, find:

    a)\;\;a + b + c \;=\;-2
    b)\;\;ab + ac + bc\;=\;-11
    c)\;\;abc\;=\;12

    Then it gets interesting!
    These problems lend themselves to some clever manipulations.


    d)\;\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\;=\;\frac{ab + ac + bc}{abc} \;= \;\frac{-11}{12}\;=\;-\frac{11}{12}


    e)\;\;\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}\;=\;\frac{a + b + c}{abc}\;=\;\frac{-2}{12}\;=\;-\frac{1}{6}


    f)\;\;(a + 1)(b + 1)(c + 1)\;=\;abc + (ab + ac + bc) +  (a + b + c) + 1
    . . . . . =\;12 + (-11) + (-2) + 1 \;= \;0


    g)\;\;(ab)^2c + (ac)^2b + (bc)^2a \;=\;a^2b^2c + a^2bc^2 + ab^2c^2
    . . . . . = \;abc(ab + ac + bc) \;= \;(12)(-11) \;= \; -132


    h)\;\;a^2 + b^2 + c^2 . . . This one requires Olympic-level gymnastics.

    We know that: . a + b + c = -2
    . . . . Hence: . (a + b + c)^2 \;=\;4

    Expand: . a^2 + 2ab + 2ac + b^2 + 2bc + c^2 \;=\;4

    We have: . (a^2 + b^2 + c^2) + 2(ab + bc + ac) \;=\;4

    Hence: . (a^2 + b^2 + c^2) + 2(-11)\;=\;4\quad\Rightarrow\quad a^2+b^2+c^2\:=\:26


    i)\;\;(ab)^{-2} + (ac)^{-2} + (bc)^{-2}\;=\;\frac{1}{a^2b^2} + \frac{1}{a^2c^2} +  \frac{1}{b^2c^2} \;= \;\frac{a^2 + b^2 + c^2}{a^2b^2c^2}

    . . . . . =\;\frac{a^2 + b^2 + c^2}{(abc)^2}\;=\;\frac{26}{12^2}\;=\;\frac{13}{72  }

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