Can you find all quartic polynomials with real, rational coefficients having 2-isqrt(3) and sqrt2 +1 as two of the zeros?

Apr 2, 2017

$f \left(x\right) = k \left({x}^{4} - 6 {x}^{3} + 14 {x}^{2} - 10 x - 7\right)$

for any rational $k \ne 0$

Explanation:

Note that:

$\left(x - \alpha\right) \left(x - \beta\right) = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$

So given $\alpha$ of the form $a + b \sqrt{c}$ with $a , b , c$ non-zero rational, we need $\beta$ to be a $a - b \sqrt{c}$ in order to make both $\alpha + \beta$ and $\alpha \beta$ rational.

In our example, that means that in addition to the given zeros:

$2 - i \sqrt{3} \text{ }$ and $\text{ } \sqrt{2} + 1$

we must have the conjugate zeros:

$2 + i \sqrt{3} \text{ }$ and $\text{ } - \sqrt{2} + 1$

So the simplest polynomial with rational coefficients and these zeros is:

$\left(x - 2 - i \sqrt{3}\right) \left(x - 2 + i \sqrt{3}\right) \left(x - 1 - \sqrt{2}\right) \left(x - 1 + \sqrt{2}\right)$

$= \left({\left(x - 2\right)}^{2} - {\left(i \sqrt{3}\right)}^{2}\right) \left({\left(x - 1\right)}^{2} - {\left(\sqrt{2}\right)}^{2}\right)$

$= \left(\left({x}^{2} - 4 x + 4\right) + 3\right) \left(\left({x}^{2} - 2 x + 1\right) - 2\right)$

$= \left({x}^{2} - 4 x + 7\right) \left({x}^{2} - 2 x - 1\right)$

$= {x}^{4} - 6 {x}^{3} + 14 {x}^{2} - 10 x - 7$

Any quartic with rational coefficients and these zeros will be a rational multiple of this one.

So the possible quartics are:

$f \left(x\right) = k \left({x}^{4} - 6 {x}^{3} + 14 {x}^{2} - 10 x - 7\right)$

for any rational $k \ne 0$