Author 
Comment/Response 
yehuda

12/29/12 00:17am
In Response To 'Re: Re: Re: Polynomial factorization by roots?'  Well, this is a Mathematica forum rather than a forum glorying HP50G and alike
Another solution which is general but shorter than my previous answer is
p = x^4 + 2 x^2 + 1;
Factor[p, Extension > (x /. Solve[p == 0, x])]
Another example
p1 = x^3  3*x^2  2*x + 6;
Factor[p1, Extension > (x /. Solve[p1 == 0, x])]
so this is soluble in one short line of Mathematica
yehdua
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