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How to factor x^3 - 3x^2 + 4 =0

Use D'Alembert's Rational Roots Theorem. Any rational roots of the form of p/q (p and q assumed to be relatively prime) must be such that p divides the constant term (in this case 4) and q divides the coefficient of the largest power of x (in this case 1).

So, the candidate roots are ±1, ±2, and ±4. If you try these, you see that
x = -1 and x = 2 are roots. There are no other roots, but there must be three roots when we count by multiplicity. So, one of the two roots we found must have a multiplicity of 2.

It isn't difficult to see that this must be the root x = 2, because then you get the correct factorization:

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

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