Algebraic numbers

Computation. $\alpha^2=2+3+2\sqrt6=5+2\sqrt6$. So $\alpha^2-5=2\sqrt6$. Squaring: $(\alpha^2-5)^2=24$.

Minimal polynomial. $\alpha$ is a root of $(X^2-5)^2-24=X^4-10X^2+1$.

Irreducibility. $X^4-10X^2+1$ factors over $\mathbb{Q}$? It has no rational root (candidates $\pm1$ give $-8$). If it factors into quadratics, $X^4-10X^2+1=(X^2+aX+b)(X^2-aX+c)$. Expanding and comparing coefficients gives $b+c-a^2=-10$, $bc=1$, $a(b-c)=0$. If $a=0$: $b+c=-10$, $bc=1$, giving $b,c$ roots of $Y^2+10Y+1=0$, discriminant $96$, not a square. Contradiction. If $b=c$: $2b=-10$, $b=-5$, $b^2=1$, impossible. So irreducible.

Conclusion. The minimal polynomial is $X^4-10X^2+1$, of degree $4$, and $\alpha$ is algebraic over $\mathbb{Q}$.