# Finding monic polynomial of lowest degree

I came across this following question.

Find the monic polynomial $P(x)$ of lowest degree with rational coefficients such that $\sqrt{2} + 3i$ is a root of $P(x) = 0$.

What I did was write $P(x)=[x-(\sqrt{2}+3i)][x-(\sqrt{2}-3i)]$ by the Complex Conjugates Theorem. However, what I got was $x^2-2\sqrt{2}x+11$. This probably means that I need to have a fourth degree polynomial. But if I try squaring it, I get more radicals.

So what should I multiply the equation by and why?

$$P(x)=(x^{2}+11-2\sqrt{2}x)(x^{2}+11+2\sqrt{2}x)=(x^{2}+11)^{2}-8x^{2}=x^{4}+14x^{2}+121$$