# How do you find a fourth degree polynomial given roots sqrt2 and 2-sqrt3?

Sep 11, 2016

${x}^{4} - 4 {x}^{3} - {x}^{2} + 8 x - 2$

#### Explanation:

If a polynomial with all rational coefficients, then if it has one irrational root, its conjugate will also be a root.

So here, since we have a root of $\sqrt{2}$, we also know that $- \sqrt{2}$ is a root. This goes the same for $2 - \sqrt{3}$. We know that $2 + \sqrt{3}$ also must be a root. (Note these these four roots will give us a fourth degree polynomial.)

Recall that if $x = a$ is a root of a polynomial, one of the polynomial's factors will be $\left(x - a\right)$. Taking our four roots in mind, we see that the polynomial function is:

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

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

Note that $\left(x - \sqrt{2}\right) \left(x + \sqrt{2}\right) = {x}^{2} - {\left(\sqrt{2}\right)}^{2} = {x}^{2} - 2$, as a difference of squares:

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

Note that $\left(\left(x - 2\right) + \sqrt{3}\right) \left(\left(x - 2\right) - \sqrt{3}\right)$ is also a difference of squares, with the terms being $\left(x - 2\right)$ and $\sqrt{3}$, that is:

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

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

So:

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

Expanding:

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

$p \left(x\right) = {x}^{4} - 4 {x}^{3} + {x}^{2} - 2 {x}^{2} + 8 x - 2$

$p \left(x\right) = {x}^{4} - 4 {x}^{3} - {x}^{2} + 8 x - 2$