Answer
remaining zeros: $-i$ and $1-i$
The equation of the function is:
$f(x)= x^4-2x^3+3x^2-2x+2$
Work Step by Step
The Conjugate Pairs Theorem states that when a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. This means that, when $(p +i \ q)$ is a zero of a polynomial function with real number coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
We notice that the function has a degree of $4$, so it has $4$ complex or real zeros. We are given $2$ complex zeros, so the remaining zeros are also complex (because they come in pairs). Thus, the remaining $2$ zero are: $-i$ and $1-i$, which are the conjugates of $i$ and $1+i$, by the Conjugate Pairs Theorem.
We can use the zeros to form factors, which we multiply to find the polynomial:
$f(x)=(x-i) [x-(1+i)][x-(-i)][x-(1-i)] \\=(x^2-i^2) [(x-1)^2 -i^2] \\=(x^2+1) (x^2-2x+1+1) \\ = x^4-2x^3+3x^2-2x+2$
