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I would like to solve this polynomial of third order for its roots:

Reduce[{1 > a > 0, 1 > b > 0, 1 > c > 0, 1 > d > 0, 1 > f > 0, 1 > g > 0,
(a (x + c + d) (x + f) + (b (x + d) + x (x + c +  d)) (x + g + f)) == 0}, x]

Is it possible and if so is there a better way than reduce with the above assumptions?

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FindInstance[{1 > a > 0, 1 > b > 0, 1 > c > 0, 1 > d > 0, 1 > f > 0, 1 > g > 0, 
(a (x + c + d) (x + f) + (b (x + d) + x (x + c + d)) (x + g + f)) == 0}, {a, b, c, d, f, g, x}]

(* {{a -> 1/2, b -> 1/2, c -> 1/2, d -> 1/2, f -> 1/2, g -> 1/2, x -> 1/2 (-2 - Sqrt[2])}} *)

a = b = c = d = f = g = 1/2;
Reduce[(a (x + c + d) (x + f) + (b (x + d) + x (x + c + d)) (x + g + f)) == 0, x]

enter image description here

These are the only real roots, I think so.

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  • $\begingroup$ Thank you very much. I am interested in complex roots though, because I want to inverse transform this using the residue theorem. $\endgroup$ Apr 1, 2016 at 8:08
  • $\begingroup$ @ Asking Questions You have an expression in x^3 and you have 3 roots (real). What do you want more? $\endgroup$
    – user36273
    Apr 1, 2016 at 9:05

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