Question 1024937
You can divide,
{{{(x^3-11x^2+55x-125)/(x-5)=x^2-6x+25}}}
This quadratic only has complex roots since {{{b^2-4ac=(-6)^2-4(1)(25)<0}}}
So then,
{{{x^2-6x+25=0}}}
{{{x^2-6x+9+25=9}}}
{{{(x-3)^2+25=9}}}
{{{(x-3)^2=-16}}}
{{{x-3=0 +- 4i}}}
{{{x=3 +- 4i}}}