# How do you find the polynomial function with roots 2 + 3sqrt3?

Aug 1, 2017

$f \left(x\right) = {x}^{2} - 4 x - 23$

#### Explanation:

If a function has the root $a + \sqrt{b}$, it must also have the root $a - \sqrt{b}$.

So, this function actually has the roots $2 + 3 \sqrt{3}$ and $2 - 3 \sqrt{3}$. Since these are roots, when $x$ takes on these values, $f \left(x\right) = 0$. Thus, each factor of the polynomial will be $x -$the root.

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

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

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

$f \left(x\right) = {x}^{2} - 4 x + 4 - 27$

$f \left(x\right) = {x}^{2} - 4 x - 23$