f(x)=(x+5)(x-4-3i)(x+a+ib) represents the function with an added unknown complex factor. We need to find a and b to identify the third zero. We know that f(x)=91 when x=2, so we can write 91=7(-2-3i)(2+a+ib). That is:

-13=(2+3i)(2+a+ib)=4+2a+2ib+6i+3ia-3b.

There is no complex component in the number 91, so we can write 4+2a-3b=-13 and 2b+6+3a=0. The first of these equations gives b in terms of a: b=(4+2a+13)/3, and this can be substituted into the second equation:

(8+4a+26)/3+6+3a=0, or 4a+34+18+9a=0. So 13a=-52, making a=-4 and b=3 by substituting for a in either equation. f(x) then becomes the cubic (n, the order, is 3): f(x)=(x+5)(x-4-3i)(x-4+3i)=(x+5)((x-4)^2+9)=(x+5)(x^2-8x+25). So,

f(x)=x^3-3x^2-15x+125.