Answer
$\color{blue}{P(x)=x^4-6x^3+10x^2-6x+9}$
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 zero of the given function is the conjugates of $-i$, which is ${\bf i}$.
The zeros of the polynomial function with real coefficients are:
$3$ (multiplicity 2)
$\\-i
\\i$
This means that the function is:
$P(x) = a(x+i) (x-i)(x-3)(x-3)
\\P(x) = a(x^2-i^2)(x^2-6x+9)
\\P(x) =ax^2-(-1)](x^2-6x+9)[
\\P(x) = a(x^2+1)(x^2-6x+9)
\\P(x) = a(x^4-6x^3+9x^2+x^2-6x+9)
\\P(x) = a(x^4-6x^3+10x^2-6x+9)$
Setting the leading coefficient $a=1$ gives:
$\color{blue}{P(x)=x^4-6x^3+10x^2-6x+9}$
