Answer
$x^5-4x^4+7x^3-8x^2+6x-4$
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 a real number of the coefficients, then its conjugate $(p –i q)$, is also a zero of the function.
We are given that the zeros of the function are: $2, 1\pm i, \pm i$ .
We write the factors of the function as $(x-zero)$ and multiply them to get the polynomial.
Therefore, we can write the equation of the function as:
$f(x)=[x-(1+i)][x-(1-i)](x-i)(x-(-i)) (x-2) \\=(x-1-2i)(x-1+i)(x-i)(x+i)(x-2)\\=x^5-4x^4+7x^3-8x^2+6x-4$
