# If a polynomial function with rational coefficients has the zeros 2, -2+sqrt10, what are the additional zeros?

##### 1 Answer
Jan 15, 2017

$- 2 - \sqrt{10}$ must also be a zero and the function will be a multiple (scalar or polynomial) of:

$f \left(x\right) = {x}^{3} + 2 {x}^{2} - 14 x + 12$

#### Explanation:

Given that $2$ and $- 2 + \sqrt{10}$ are zeros, the rational conjugate $- 2 - \sqrt{10}$ must also be a zero.

The polynomial function will be a multiple (scalar or polynomial of this $f \left(x\right)$:

$f \left(x\right) = \left(x - 2\right) \left(x + 2 - \sqrt{10}\right) \left(x + 2 + \sqrt{10}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left(\left(x + 2\right) - \sqrt{10}\right) \left(\left(x + 2\right) + \sqrt{10}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({\left(x + 2\right)}^{2} - {\left(\sqrt{10}\right)}^{2}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({x}^{2} + 4 x + 4 - 10\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({x}^{2} + 4 x - 6\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} + 2 {x}^{2} - 14 x + 12$