Answer
$\color{blue}{P(x)=x^5-4x^4+7x^3-8x^2+6x-4}$
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 $1+i$, which are
${\bf i, 1-i}$.
The zeros of the polynomial function with real coefficients are:
$2
\\-i
\\i
\\1+i
\\1-i$
This means that the function is:
$P(x) = a(x-2)(x-i)(x+i) \cdot [x-(1+i)][x-(1-i)]
\\P(x) = a(x-2)(x^2-i^2)(x-1-i)(x-1+i)
\\P(x) =a(x-2)(x^2-(-1)] \cdot [(x-1)-i][(x-1)+i)]]
\\P(x) = a(x-2)(x^2+1)[(x-1)^2-i^2]
\\P(x) = a(x-2)(x^2+1)[x^2-2x+1-(-1)]
\\P(x) = a(x-2)(x^2+1)(x^2-2x+1+1)
\\P(x)=a(x-2)(x^2+1)(x^2-2x+2)
\\P(x) = a(x-2)(x^4-2x^3+2x^2+x^2-2x+2)
\\P(x)=a(x-2)(x^4-2x^3+3x^2-2x+2)
\\P(x)=a(x^5-2x^4+3x^3-2x^2+2x-2x^4+4x^3-6x^2+4x-4)
\\P(x)=a(x^5-4x^4+7x^3-8x^2+6x-4)$
Setting the leading coefficient $a=1$ gives:
$\color{blue}{P(x)=x^5-4x^4+7x^3-8x^2+6x-4}$
