# How do you write a polynomial function of least degree given the zeros sqrt6, -3+sqrt5?

Apr 8, 2017

$\left(x - \sqrt{6}\right) \left(x - \left(- 3 + \sqrt{5}\right)\right) = {x}^{2} + \left(3 - \sqrt{5} - \sqrt{6}\right) x + \sqrt{30}$

#### Explanation:

Since $\sqrt{6}$ is a zero, we must have $x - \sqrt{6}$ is a factor.

And since $- 3 + \sqrt{5}$ is also a zero, we must have $x - \left(- 3 + \sqrt{5}\right)$ is also a factor.

To get a polynomial of least degree use just these two factors (and a non-0 constant factor if you wish).

Bonus

If we wish the polynomial to have integer or rational coefficients, the the conjugates of the given zeros are also zeros.

That would make the zeros: $\sqrt{6}$, $- \sqrt{6}$, and $- 3 + \sqrt{5}$, and $- 3 - \sqrt{5}$

The factors in this case would be

$\left(x - \sqrt{6}\right) \left(x + \sqrt{6}\right) \left(x - \left(- 3 + \sqrt{5}\right)\right) \left(x - \left(- 3 - \sqrt{5}\right)\right)$.

When we expand, we get

$\left({x}^{2} - 6\right) \left({x}^{2} + 6 x + 4\right)$ which is

${x}^{4} + 6 {x}^{3} - 2 {x}^{2} - 36 x - 24$

We could also multiply by any non-0 constant.