Question

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

Answer 1

`f(x)=C^{te} (x-3)(x+1-sqrt(3))((x-1)^2+4)`

Explanation:

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