# How do you find a fourth-degree polynomial equation, with integer coefficients, for which -sqrt11 and 2i are solutions?

Aug 21, 2017

$P \left(x\right) = A \left({x}^{2} - 11\right) \left({x}^{2} + 4\right)$

Where $A$ is an arbitrary integer.

#### Explanation:

By the fundamental Theorem of Algebra, any polynomial of degree $4$ can be written in the form:

$P \left(x\right) = A \left(x - \alpha\right) \left(x - \beta\right) \left(x - \gamma\right) \left(x - \delta\right)$

Where, $\alpha , \beta , \gamma , \delta$ are the roots (or zeros) of the equation $P \left(x\right) = 0$

We are given that $- \sqrt{11}$ and $2 i$ are solutions (presumably, although not explicitly stated, of $P \left(x\right) = 0$, thus, wlog, we can write:

$\alpha = - \sqrt{11}$
$\beta = 2 i$

We also know that complex roots must appear as complex conjugate pairs, so $- 2 i$ must be another root, so in addition:

$\gamma = - 2 i$

And so we have:

$P \left(x\right) = A \left(x + \sqrt{11}\right) \left(x - 2 i\right) \left(x + 2 i\right) \left(x - \delta\right)$

We can multiply the complex products to get real coefficients quadratic:

$\left(x - 2 i\right) \left(x + 2 i\right) = {x}^{2} - {\left(2 i\right)}^{2}$
$\text{ } = {x}^{2} - 4 {i}^{2}$
$\text{ } = {x}^{2} + 4$

And using this we get:

$P \left(x\right) = A \left(x + \sqrt{11}\right) \left({x}^{2} + 4\right) \left(x - \delta\right)$

We require that $P \left(x\right)$ has integer coefficients, rather than irrational coefficients, and this would require that the product $\left(x + \sqrt{11}\right)$ and $\left(x - \delta\right)$ has integer coefficients, Now:

$\left(x + \sqrt{11}\right) \left(x - \delta\right) = {x}^{2} + \left(\sqrt{11} - \delta\right) x - \sqrt{11} \delta$

And if we set $\delta = - \sqrt{11}$ this condition is met, and the product of the irrational coefficients are:

$\left(x + \sqrt{11}\right) \left(x + \sqrt{11}\right) = {x}^{2} - {\left(\sqrt{11}\right)}^{2}$
$\text{ } = {x}^{2} - 11$

Hence, we have

$P \left(x\right) = A \left({x}^{2} - 11\right) \left({x}^{2} + 4\right)$

Where $A$ is an arbitrary integer.