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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?

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How about

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

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