It's difficult to see how you get this code without seeing the reference. I altered it based on a Wikipedia article.

bairs[pol_, p0_, m_] := 
 Module[{n, a, r, s, j, b, c, d, f, , g, h, tol},
  tol = 1/10000; n = Length[pol] - 1;
  Table[a[i] = pol[[n - i + 1]], {i, 0, n}];
  r = p0[[1]]; s = p0[[2]];
  Do[
   b[n] = 0; b[n - 1] = 0;
   Table[b[i] = a[i + 2] + r b[i + 1] + s b[i + 2], {i, n - 2, 0, -1}];
   c = a[1] + r*b[0] + s*b[1];
   d = a[0] + s*b[0];
   f[n] = 0; f[n - 1] = 0;
   Table[f[i] = b[i + 2] + r f[i + 1] + s f[i + 2], {i, n - 2, 0, -1}];
   g = b[1] + r*f[0] + s*f[1];
   h = b[0] + s*f[0];
   {r, s} = {r, s} + 
     1/(-s*g^2 + h*(h + r*g))*{{-h, g}, {g*s, -g*r - h}}.{c, d};
   Print["{r,s} ", {r, s}], {m}];
  Print["El factor cuadratico es: x^2", If[-r > 0, "+", ""], N[-r, 4],
    "x", If[-s > 0, "+", ""], N[-s, 4]]]

This version seems to work reasonably well. It was taken, with very minor adjustment, from:

http://en.wikipedia.org/wiki/Bairstow%27s_method

I'm not convinced that that's a viable reference, at least not for the method in the Wikipedia article. There was loop nesting to three levels in that code. Only two should be necessary. One to compute the new coefficients, and one to iterate some number of times (or until convergence).

here is the definition of the function bair

bair[pol_, p0_, m_, p_] := Module[{n, a, r, i, s, j, b, c, tol},
  tol = 0.0001;
  n = Length[pol] - 1;
  (* primer for*)For[i = 0, i <= n, i++,
   a[i] = SetPrecision[pol[[n - i + 1]], p]];
  r = SetPrecision[p0[[1]], p]; s = SetPrecision[p0[[2]], p];
  b[0] = SetPrecision[2 tol, p];
  b[1] = SetPrecision[2 tol, p];
  (* segundo for*)For[j = 1, j <= m, j++,
   b[n] = SetPrecision[a[n], p];
   b[n - 1] = SetPrecision[a[n - 1] + r b[n], p];
   (* tercer for*)For[i = n - 2, i >= 0, i--,
    b[i] = SetPrecision[a[i] + r b[i + 1] + s b[i + 2], p]];
   c[n] = SetPrecision[b[n], p];
   c[n - 1] = SetPrecision[b[n - 1] + r c[n], p];
   (* ultimo for*)For[i = n - 2, i >= 1, i--,
    c[i] = SetPrecision[b[i] + r c[i + 1] + s c[i + 2], p]];
   r = SetPrecision[r + (-b[1] c[2] + b[0] c[3])/(c[2]^2 - c[1] c[3]),
      p];
   s = SetPrecision[s + (-b[0] c[2] + b[1] c[1])/(c[2]^2 - c[1] c[3]),
      p]];
  Print["El factor cuadratico es: x^2", If[-r > 0, "+", ""], -r, "x", 
   If[-s > 0, "+", ""], -s]]

now make comparisons runtime between the two functions with the same arguments

(* Using the bair function*)
Timing[bair[{1, -7, 14, 2, -20}, {5, 3}, 7, 10]]

(* Using the bairs function*)
Timing[bairs[{1, -7, 14, 2, -20}, {5, 3}, 7]]

(* Using the bair function*)
Timing[bair[{1, -3, -9, -27, -162}, {2, 7}, 7, 10]]

(* Using the bairs function*)
Timing[bairs[{1, -3, -9, -27, -162}, {2, 7}, 7]]

if anyone has any idea of the situation happens for this time difference, please comment on this, because I have not seen what the conflict to have this result, thanks in advance

attached my file so they can make comparisons

Attachments: