Answer
$\color{blue}{P(x)=x^4-2x^3+6x^2-2x+5}$
Work Step by Step
RECALL:
If $a+bi$ is a zero of a polynomial function with real number coefficients, then its conjugate $a-bi$ is also a zero of the function.
Thus, the missing zeros of the given function are the conjugates of $i$ and $1+2i$, which are
${\bf-1, 1-2i}$.
The zeros of the polynomial function with real coefficients are:
$i
\\1+2i
\\-i
\\1-2i$
This means that the function is:
$P(x) = a(x-i)(x+i) \cdot [x-(1+2i)][x-(1+2i)]
\\P(x) = a(x^2-i^2)(x-1-2i)(x-1+2i)
\\P(x) =a(x^2-(-1)] \cdot [(x-1)-2i][(x-1)+2i)]]
\\P(x) = a(x^2+1)[(x-1)^2-4i^2]
\\P(x) = a(x^2+1)[x^2-2x+1-4(-1)]
\\P(x) = a(x^2+1)(x^2-2x+1+4)
\\P(x)=a(x^2+1)(x^2-2x+5)
\\P(x) = a(x^4-2x^3+5x^2+x^2-2x+5)
\\P(x)=a(x^4-2x^3+6x^2-2x+5))$
Setting the leading coefficient $a=1$ gives:
$\color{blue}{P(x)=x^4-2x^3+6x^2-2x+5}$