Question 381711
If {{{x=-i}}} is a zero, then so is {{{x=i}}}
{{{f(x)=a(x-1)(x-2)(x-i)(x+i)}}}
{{{f(0)=a(-1)(-2)(-i)(i)=8}}}
{{{a(-2)(-i^2)=8}}}
{{{a=-4}}}
{{{f(x)=-4(x-1)(x-2)(x-i)(x+i)}}}
However the leading coefficient is {{{-4}}}. 
There is no solution as the problem is currently set up.
Please check the problem setup and re-post.