Prove its an algebraic integer

Relevant wiki: Algebraic Number Theory

Let $\alpha = 3+2\sqrt 5$ be a root of a monic polynomial. We can assume another root $\beta = 3-2\sqrt 5$ ; so that by Vieta's formula , we have $\alpha + \beta = 3+2\sqrt 5 + 3-2\sqrt 5 = 6$ and $\alpha \beta = \left( 3+2\sqrt 5 \right) \left( 3-2\sqrt 5 \right) = 3^2 - \left(2\sqrt 5 \right)^2 = 9 - 20 = - 11$ to give integer coefficients. Therefore, the monic polynomial is $\boxed{x^2-6x-11}$ .

In order for $3+2\sqrt{5}$ to be an algebraic integer, it must be a root of a monic polynomial with integer coefficients. The number in question is a root of every given polynomial other than $x^2+6x+11.$ Among those polynomials, the only one that has integer coefficients is $\boxed{x^2-6x-11}.$