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

Apr 19, 2017

The simplest such polynomial is:

$f \left(x\right) = {x}^{3} - {x}^{2} + 81 x - 81$

#### Explanation:

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

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

So let:

$f \left(x\right) = \left(x - 1\right) \left(x - 9 i\right) \left(x + 9 i\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 1\right) \left({x}^{2} - {\left(9 i\right)}^{2}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 1\right) \left({x}^{2} + 81\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} - {x}^{2} + 81 x - 81$

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

$\textcolor{w h i t e}{}$
Footnote

If only the zeros $1$ and $9 i$ had been requested, we would still require $- 9 i$ 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 $a + i b$ is a zero of a polynomial with real coefficients then so is $a - i b$.