# How do you write a polynomial in standard form given the zeros x= -3i and sqrt3?

Jun 24, 2016

According to what you want, the simplest polynomials are:

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

${x}^{3} - \sqrt{3} {x}^{2} + 9 x - 9 \sqrt{3}$

${x}^{4} + 6 {x}^{2} - 27$

#### Explanation:

A polynomial of lowest degree with these zeros is:

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

Typically we would be interested in the polynomial having Real coefficients. If so, then any non-Real Complex zeros occur in Complex conjugate pairs. Hence $x = 3 i$ would also be a zero and the simplest polynomial is:

$\left(x - 3 i\right) \left(x + 3 i\right) \left(x - \sqrt{3}\right)$

$= \left({x}^{2} + 9\right) \left(x - \sqrt{3}\right)$

$= {x}^{3} - \sqrt{3} {x}^{2} + 9 x - 9 \sqrt{3}$

If we also want the polynomial to have rational coefficients, then any irrational zeros of the form $a + \sqrt{b}$ occur in radical conjugate pairs. That is, if $a + \sqrt{b}$ is a zero then so is $a - \sqrt{b}$.

Hence the simplest polynomial with the given zeros and rational coefficients is the quartic:

$\left(x - 3 i\right) \left(x + 3 i\right) \left(x - \sqrt{3}\right) \left(x + \sqrt{3}\right)$

$= \left({x}^{2} + 9\right) \left({x}^{2} - 3\right)$

$= {x}^{4} + 6 {x}^{2} - 27$