# How do you find a polynomial function that has zeros 1+sqrt3, 1-sqrt3?

Aug 14, 2017

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

#### Explanation:

Since these are the zeros, we can make the following equation:

$\left(x - \left(1 + \sqrt{3}\right)\right) \left(x - \left(1 - \sqrt{3}\right)\right) = 0$

Or

$\left(x - 1 - \sqrt{3}\right) \left(x - 1 + \sqrt{3}\right) = 0$

When we expand this, we get

${x}^{2} - x + x \sqrt{3} - x + 1 - \sqrt{3} - x \sqrt{3} + \sqrt{3} - 3 = 0$

Combining like terms:

color(blue)(ulbar(|stackrel(" ")(" "f(x) = x^2 - 2x - 2" ")|)

Aug 15, 2017

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

#### Explanation:

The simplest polynomial with distinct zeros $\alpha$ and $\beta$ is:

$\left(x - \alpha\right) \left(x - \beta\right) = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$

With $\alpha = 1 + \sqrt{3}$ and $\beta = 1 - \sqrt{3}$, we find:

$\left\{\begin{matrix}\alpha + \beta = \left(1 + \sqrt{3}\right) + \left(1 - \sqrt{3}\right) = \textcolor{red}{2} \\ \alpha \beta = \left(1 + \sqrt{3}\right) \left(1 - \sqrt{3}\right) = {1}^{2} - {\left(\sqrt{3}\right)}^{2} = 1 - 3 = \textcolor{b l u e}{- 2}\end{matrix}\right.$

So a suitable polynomial function would be:

$f \left(x\right) = {x}^{2} - \textcolor{red}{2} x \textcolor{b l u e}{- 2}$

Any polynomial function in $x$ with these two zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$.