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Math Help - Solve for X values

  1. #1
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    Solve for X values

    x^3-3x^2+3
    Im not sure how to factor this
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  2. #2
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    Okay

    So, first would probably be grouping.

    You would take the x^3 and group it with the 3x^2.

    Factor that into x^2 (x-3) + 3

    Then it would just be (x^2+3) (x-3)
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  3. #3
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    so factor for this eqn will be

    x= 3, and 2 imaginary fators will be there with "i"(iota)
    Last edited by mr fantastic; October 20th 2009 at 06:22 PM. Reason: Removed link advertising commercial website.
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  4. #4
    Senior Member pacman's Avatar
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    you have 3 real roots, see the graph. By Descartes law of signs, (+) 2 sign changes, (-) only 1. So we have two positive roots and one negative root.
    h



    This cubic equation is hard to factor by grouping . . . .
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  5. #5
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    Quote Originally Posted by Mycc82 View Post
    So, first would probably be grouping.

    You would take the x^3 and group it with the 3x^2.

    Factor that into x^2 (x-3) + 3

    Then it would just be (x^2+3) (x-3)
    This is wrong!

    (x^2+3)(x-3) = x^2(x-3) + 3(x-3) By the law of distributivity

    Obviously, x^2(x-3) + 3 \neq x^2(x-3) + 3(x-3) !!!
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  6. #6
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    Quote Originally Posted by jeremyparker10 View Post
    so factor for this eqn will be

    x= 3, and 2 imaginary fators will be there with "i"(iota)
    This is wrong. Please check the original question before making bold statements that rely on someone elses work (work which, as has been pointed out, is wrong).
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  7. #7
    Senior Member pacman's Avatar
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    x^3-3x^2+3 = 0

    let x = 1/y,

    (1/y)^3 - 3(1/y)^2 + 3 = 0, MBS by y^3,

    1 - 3y + 3y^3 = 0,

    y^3 - y + 1/3 = 0. Very hard to factor by grouping.

    y^3 = y - 1/3 = Ay + B, where A = 1 and B = -1/3

    Let p = A/3 = 1/3
    q = B/2 = (-1/3)/2 = -1/6

    q^2 - p^3 = (-1/6)^2 - (1/3)^3 = 1/36 - 1/27 = -1/108(negative) < 0.

    CASE III. q^2 - p^3 < 0, 3 roots are real and distinct. Determine an angle u between 0 and 180 degrees, such that cos u = q/(p^3/2) = (-1/6)/(1/3)^3/2 = -(1/2)(3)^(1/2)

    u = 150 degrees or 210 will do but we choose 150 because of the restriction.

    y1 = (2p^1/2) cos (u/3) = 2(1/3)^(1/2) cos (150/3) = 0.74228

    y2 = (2p^1/2) cos (u/3 + 120) = 2(1/3)^(1/2) cos (150/3 + 120) = 2(1/3)^(1/2) cos (170) = 1.08310

    y3 = (2p^1/2) cos (u/3 + 240) = 2(1/3)^(1/2) cos (150/3 + 240) = 0.39493

    x1 = 1/y1 = 1/0.74228 = 1.3472

    x2 = 1/y2 = 1/1.08310 = -0.8794

    x3 = 1/y3 = 1/0.39493 = 2.532

    See graph below.
    Attached Thumbnails Attached Thumbnails Solve for X values-ok.gif  
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  8. #8
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    Quote Originally Posted by Nightasylum View Post
    x^3-3x^2+3
    Im not sure how to factor this
    By the "rational roots theorem", the only possible rational roots will be 1, -1, 3, or -3. (1)^2- 2(1)^2+ 3= 1- 2+ 3= 2, (-1)^3-2(-1)+ 3= -1- 2+ 3= 0. (3^3- 2(3^2)+ 3= 27- 18+ 3= 12, and (-3)^3- 3(-3)^2+ 3= -27- 3+ 3= 27.

    So x= -1 is the only rational root. You can divide x^3- 3x^2+ 3 by x+ 1 to see that x^2 - 3x+ 3 is the other factor. Since its discriminant is negative, there is no further factorization (with integer coefficients).

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