3. It is known that the quadratic trinomial $x^{2}+b x+c$ has two distinct roots. If we add the coefficients $b$ and $c$ and the two roots (four numbers), we get the number -3, and if we multiply these same four numbers, we get the number 36. Find all such quadratic trinomials.
Answer: $x^{2}+4 x-3$, the only trinomial.

Solution. Consider the roots of the quadratic trinomial $x^{2}+b x+c$. Let's denote them as $x_{1}$ and $x_{2}$. Then, according to the problem, we have the following pair of relations:

$$
\left\{\begin{array}{l}
b+c+x_{1}+x_{2}=-3 \\
b \cdot c \cdot x_{1} \cdot x_{2}=36
\end{array}\right.
$$

where, applying Vieta's theorem for the reduced quadratic equation ( $x_{1}+x_{2}=-b, x_{1}$. $\left.x_{2}=c\right)$, we get

$$
\left\{\begin{array}{l}
b+c-b=-3 \\
b \cdot c \cdot c=36
\end{array}\right.
$$

from which

$$
\left\{\begin{array}{l}
c=-3 \\
b=4
\end{array}\right.
$$

Thus, the only solution to the problem is the quadratic trinomial $x^{2}+4 x-3$. Note that the roots of such a quadratic trinomial are the numbers $x_{1,2}=-2 \pm \sqrt{7}$. Solutions are possible where the roots of the quadratic trinomial are expressed through its coefficients $b$ and $c$ (or vice versa).

Remarks on evaluation. An answer without justification - 0 points. If the relations for the 4 numbers use Vieta's theorem or the expression of the roots of the quadratic trinomial - 2 points. A fully justified answer - full points.