how can you factor
x^3-3x^2-2 = 0 ?
x^3-3x^2-2 = 0 ?
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The only possible rational roots are +/-1, +/-2
since none of these give zero, then it does not factor.
since none of these give zero, then it does not factor.
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The equation has a real root between 3 and 4 (an irrational one) and two complex roots. It cannot be factored in the rationals, factors into linear and quadratic parts in the reals, and factors fully only when one enters the complex domain. The way one can see this easily is by finding the zeros of the derivative. The expression on the left, which is what you presumably are asking for factors of (an equation not being called for), has a local maximum at x=0 and a local minimum at x=2 (with an inflection at x=1). Beyond x=2 it is strictly increasing and changes sign in the interval mentioned. It is not difficult to show this zero cannot be at a rational value. It's possible (though pointless here) to give explicit representation to all three roots, and this is true for all cubics and quartics but requires special functions beyond radicals (squareroots, cuberoots, and so on) for general equations of fifth and higher degree.
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Add 2 take the derivative of both sides. So x^3-3x^2=2 then 3x^2-6x voila!