34. The number $2+\sqrt{3}$ is a root of the polynomial

$$
f(x)=x^{3}+4 x^{2}+a x+b(a \in Q, b \in Q)
$$

Find the other roots.
$\triangle$ First, let's find the values of the parameters $a$ and $b$.

Based on the statement of the reference problem 32, if $x_{1}=2+\sqrt{3}$ is a root of the given polynomial, then $x_{2}=2-\sqrt{3}$ is also its root. Therefore, the polynomial $f(x)$ is divisible by the expression

$\left(x-x_{1}\right)\left(x-x_{2}\right)=(x-2-\sqrt{3})(x-2+\sqrt{3})=(x-2)^{2}-3=x^{2}-4 x+1$.

Let's divide $f(x)$ by $x^{2}-4 x+1$ with a remainder:

$$
x^{3}+4 x^{2}+a x+b=\left(x^{2}-4 x+1\right)(x+8)+(a+31) x+(b-8)
$$

The remainder from this division must be identically zero:

$$
(a+31) x+(b-8) \equiv 0
$$

from which $a=-31, b=8$.

Then $x+8=0$, i.e., $x_{3}=-8$.

Answer: $2-\sqrt{3},-8$.