Answer
$f(x)=x^4-6x^3+10x^2-6x+9$
Work Step by Step
The Conjugate Pairs Theorem says that if a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. That is, if $a + bi$ is a zero then so is $a – bi$ and vice-versa.
Therefore, the zeros of the function are:
$\pm i$ and 3 with multiplicity 2
Hence, the function could be:
$f(x) = (x-i)[x-(-i)](x-3)(x-3)\\
f(x) = (x-i)(x+i)(x^2-6x+9)\\
f(x) = (x^2-i^2)(x^2-6x+9)\\
f(x) = (x^2+1)(x^2-6x+9)\\
f(x) = x^2(x^2-6x+9)+1(x^2-6x+9)\\
f(x) = x^4-6x^3+9x^2+x^2-6x+9\\
f(x) = x^4-6x^3+10x^2-6x+9$
