Daniel
I found in the Wolfram Documentation the following example
SolveAlways[(a - b) x + (a^2 - b + 1) x y + (a d - c + 2) x y z +
a b - c d == 0, {x, y, z}]
and the solution is
{{a -> (-1)^(1/3), b -> (-1)^(1/3), c -> 2 - (-1)^(1/3) + (-1)^(2/3),
d -> -1 + (-1)^(1/3)}, {a -> (-1)^(1/3), b -> (-1)^(1/3),
c -> 2 - (-1)^(1/3) + (-1)^(2/3),
d -> -1 + (-1)^(1/3)}, {a -> -(-1)^(2/3), b -> -(-1)^(2/3), c -> 1,
d -> -(-1)^(1/3)}, {a -> -(-1)^(2/3), b -> -(-1)^(2/3), c -> 1,
d -> -(-1)^(1/3)}}
So, for some examples, it seems to be possible to find the coefficients to make the polynomial == 0
That is what I want to do with my polynomial.
I guess that is going to be impossible.
Thanks anyway
Jesus