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Math Help - Need some help with the method

  1. #1
    Junior Member
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    Need some help with the method

    I've got an equation: 4x^3 - 24x^2 + 45 x - 23

    A tangent is drawn at point P. (1,2)

    dy/dx = 12x^2 - 48x +45

    At f´(1) = 12 -48 + 45 = 9

    2 = 9(1) +c

    y = 9x - 7

    Show that the x coordinate of point q is 4.

    9x-7 = 4x^3 - 24x^2 + 45 x - 23

    0 = 4x^3 - 24x^2 + 36 x - 16

    I found x = 4 using polyrootfinder on my gdc and was wondering if there is any other way of coming to that answer?
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  2. #2
    Junior Member Infophile's Avatar
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    I imagine q is the point where the tangent cross the curve, doesn't it ?

    Your method is good, but you can solve the equation if you notice that x=1 is solution, then put (x-1) in factor..

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  3. #3
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    That is correct. I see what you mean however I'm quite unfamiliar with factorisation with 3rd degree polynomials (I'm still a middle school student :S). Could you care to elaborate?
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  4. #4
    Junior Member Infophile's Avatar
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    Of course !

    P(x)=2x^3-12x^2+18x-8

    You have to test some values such as 1, 2, 3, -1, -2, -3.

    Indeed here : P(1)=0 so we can put (x-1) in factor :

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

    which we develop : P(x)=ax^3+(b-a)x^2+(c-b)x-c

    And we identify each coefficient \left\{\begin{array}{l}a=2\\b-a=-12\\c-b=18\\c=8\end{array}\right.

    After resolution we get : \left\{\begin{array}{l}a=2\\b=-10\\c=8\end{array}\right.

    Hence P(x)=(x-1)(2x^2-10x+8) and you know the factorisation 2nd degree polynomials.

    Is my english correct ?
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