Question

How do you write a polynomial function in standard form with real coefficients whose zeros include `1`, `9i`, and `-9 i`?

Answer 1

The simplest such polynomial is:

`f(x) = x^3-x^2+81x-81`

Explanation:

Given zeros `1`, `9i`, `-9i`.

Any polynomial in `x` with these zeros will have linear factors including `(x-1)`, `(x-9i)` and `(x+9i)`.

So let:

`f(x) = (x-1)(x-9i)(x+9i)`

`color(white)(f(x)) = (x-1)(x^2-(9i)^2)`

`color(white)(f(x)) = (x-1)(x^2+81)`

`color(white)(f(x)) = x^3-x^2+81x-81`

So this `f(x)` is a suitable example and any polynomial in `x` with these zeros will be a multiple (scalar or polynomial) of this `f(x)`.

`color(white)()`
Footnote

If only the zeros `1` and `9i` had been requested, we would still require `-9i` to be a zero too in order to get real coefficients.

One way to think of this is to recognise that `i` and `-i` are essentially indistinguishable from the perspective of the real numbers. So if `a+ib` is a zero of a polynomial with real coefficients then so is `a-ib`.