# How do you form a polynomial function whose zeros are 3, 1-2√3, and 1+2i?

$f \left(x\right) = {C}^{t e} \left(x - 3\right) \left(x + 1 - \sqrt{3}\right) \left({\left(x - 1\right)}^{2} + 4\right)$
Let $f \left(x\right)$ be such a function, a real valued function. If $f \left(3\right) = 0$ then certainly $f \left(x\right) = \left(x - 3\right) g \left(x\right)$. Now regarding the next root, appliying the same procedure, $f \left(x\right) = \left(x - 3\right) \left(x + 1 - \sqrt{3}\right) h \left(x\right)$. Now resuming, if $1 + 2 i$ is a root, $1 - 2 i$ also must be also a root because only this way the function $f \left(x\right)$ will be real valued. So $f \left(x\right) = \left(x - 3\right) \left(x + 1 - \sqrt{3}\right) \left(x - 1 + 2 i\right) \left(x - 1 - 2 i\right) r \left(x\right)$ arriving at
$f \left(x\right) = \left(x - 3\right) \left(x + 1 - \sqrt{3}\right) \left({\left(x - 1\right)}^{2} + 4\right) r \left(x\right)$. Now if $f \left(x\right)$ is a polynomial and has no more roots,$r \left(x\right) = {C}^{t e}$ which is a constant
$f \left(x\right) = {C}^{t e} \left(x - 3\right) \left(x + 1 - \sqrt{3}\right) \left({\left(x - 1\right)}^{2} + 4\right)$