find the value of $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is a root of the polynomial? Find the value of  $a\in\Bbb{Z}$ such that $2+\sqrt{3}$ is  a root of the   polynomial 
$$x^3-5x^2 +ax -1$$ 
I got   the answer    and value of $a = -10 -5\sqrt{3} -1$.
Is  my answer  is correct or not?

If $a \in \mathbb Z$, then if $2+\sqrt3$ is a root, then so is $2-\sqrt3$.
Therefore, $x^3-5x^2 +ax -1$ is divisible by $x^2 - 4 x + 1$, which has $2\pm\sqrt3$ as roots.
Now, $x^3-5x^2 +ax -1 = (x^2 - 4 x + 1)(x-1)+(a-5)x$ and so $a=5$.
