| 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
URL: , |
|
