# How do you find a polynomial function that has zeros 2, 4+sqrt5, 4-sqrt5?

Jul 22, 2017

$f \left(x\right) = {x}^{3} - 10 {x}^{2} + 27 x - 22$

#### Explanation:

If $x = a$ is a zero then $\left(x - a\right)$ is a factor.

So the simplest polynomial function of $x$ that has these zeros is:

$f \left(x\right) = \left(x - 2\right) \left(x - \left(4 + \sqrt{5}\right)\right) \left(x - \left(4 - \sqrt{5}\right)\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left(\left(x - 4\right) + \sqrt{5}\right) \left(\left(x - 4\right) - \sqrt{5}\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({\left(x - 4\right)}^{2} - 5\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({x}^{2} - 8 x + 16 - 5\right)$

$\textcolor{w h i t e}{f \left(x\right)} = \left(x - 2\right) \left({x}^{2} - 8 x + 11\right)$

$\textcolor{w h i t e}{f \left(x\right)} = {x}^{3} - 10 {x}^{2} + 27 x - 22$

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

In this example, we were asked for a function with zeros including both $4 + \sqrt{5}$ and its radical conjugate $4 - \sqrt{5}$. If only one had been specified then we would still have had to include the other if we wanted our function to have rational coefficients.