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x^3 - 4x + 1 = 0 Solve for x..?

How would I do this please explain

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  • Anonymous
    1 decade ago
    Favorite Answer

    1A. Solve by Factorization

    x^3 -4*x +1 = 0

    Separate : ( x^3 +2.1149*x^2 ) + ( -2.1149*x^2 -4.4728*x ) + ( 0.4728*x +1 ) = 0

    Commutative Law : ( x^3 -2.1149*x^2 +0.4728*x ) + ( 2.1149*x^2 -4.4728*x +1 ) = 0

    Distributive Law : x*( x^2 -2.1149*x +0.4728 ) + 2.1149*( x^2 -2.1149*x +0.4728 ) = 0

    Factor : ( x +2.1149 )*( x^2 -2.1149*x +0.4728 ) = 0

    Separate : ( x +2.1149 )*( ( x^2 -0.2541*x ) + ( -1.8608*x +0.4728 ) ) = 0

    Commutative Law : ( x +2.1149 )*( ( x^2 -1.8608*x ) + ( -0.2541*x +0.4728 ) ) = 0

    Distributive Law : ( x +2.1149 )*( x*( x -1.8608 ) + -0.2541*( x -1.8608 ) ) = 0

    Factor : ( x +2.1149 )*( x -0.2541 )*( x -1.8608 ) = 0

    So the Polynomial have 3 roots :

    x1 = -2.1149

    x2 = 0.2541

    x3 = 1.8608

  • 1 decade ago

    x1 = -2,1149075414767555

    x2 = 0,2541016883650524

    x3 = 1,8608058531117033

    Source(s): It's explained here: http://en.wikipedia.org/wiki/Cubic_equation
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