We have roots at x = 1/2 and x = i
For any polynomial with a nonreal, complex, root at a+bi
the complex conjugate a-bi is also a root.
If x=i is a root, then so is x=-i
Your polynomial is thus:
= (x-1/2)(x2 +1)
= x3 - (1/2)x2 + x - 1/2
This is, of course, the same answer as Josue H gave you...
But with no need to find that 2nd root in the manner in
which Josue did it.
If p(x) has a root at 3+4i then it must have a root at 3-4i
and so forth..
i = 0 + 1i
Conjugate: 0 - 1i = -i
So the roots were immediately known to be x = 1/2, i , -i