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Math Help - quadratic equation

  1. #1
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    quadratic equation

    Solve the following equation:x^3-2x^2-5x+6=0
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  2. #2
    Member Ranger SVO's Avatar
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    It factors to ( x - 1 )* ( x + 2 )* ( x - 3 ) = 0

    I bet you can get it from there
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  3. #3
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    Quote Originally Posted by gracy View Post
    Solve the following equation:x^3-2x^2-5x+6=0
    What is that property or rule or "definition"? Anyway, always test the sum of all the numerical coefficients first. If they add up to zero, then x = 1 is a root.

    x^3 -2x^2 -5x +6 = 0
    1 -2 -5 +6 = 0 -----------------so 1 is a root, and so (x-1) is a factor.

    Then divide the x^3 -2x^2 -5x +6 by (x-1) by long division or synthetic division. The quotient will be quadratic. Easy by then. Factor or use the Quadratic formula to find the other two factors or two roots.

    After division, you'd get x^2 -x -6 as the quotient.
    So,
    x^2 -x -6 = 0
    (x -3)(x+2) = 0
    Hence the other two facxtors are (x-3) and (x+2)
    Therefore, the original equation factors to
    (x-1)(x-3)(x+2) = 0
    Meaning,
    x = 1 or 3 or -2 ------------answer.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by gracy View Post
    Solve the following equation:x^3-2x^2-5x+6=0
    The "Rational Roots Theorem" states that if we have a polynomial equation:
    ax^n + bx^{n-1} + ... + cx + d = 0
    then any rational roots of the equation (if they exist, they many not!) will be of the form:
    x = (+/-)[factor of d]/[factor of a]

    In your case, the possible rational roots are going to be (+/-)1, 2, 3, 6. When we test each of these we find that x = -2, 1, 3 are solutions. Since this is a cubic equation, we know that 3 complex roots exist. Thus we have found all of them.

    -Dan
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