## Task A-2.1.

Let $x_{1}$ and $x_{2}$ be distinct solutions of the equation $x^{2}+5 x+3=0$. Calculate $\frac{x_{1}^{3} x_{2}+x_{1} x_{2}^{3}}{x_{1}+x_{2}}$.
## Solution.

According to Vieta's formulas, we have

$$
\begin{aligned}
x_{1}+x_{2} & =-5, & & 1 \text{ point} \\
x_{1} \cdot x_{2} & =3 . & & 1 \text{ point }
\end{aligned}
$$

From the formula for the square of a sum, it follows that

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=5^{2}-2 \cdot 3=19
$$

Therefore, the numerator of the desired fraction is

$$
x_{1}^{3} x_{2}+x_{1} x_{2}^{3}=x_{1} x_{2}\left(x_{1}^{2}+x_{2}^{2}\right)=3 \cdot 19=57
$$

and the desired value is

$$
\frac{x_{1}^{3} x_{2}+x_{1} x_{2}^{3}}{x_{1}+x_{2}}=\frac{57}{-5}=-\frac{57}{5}
$$

Note: The result can also be obtained by explicitly solving the quadratic equation, whose solutions are $x_{1,2}=\frac{-5 \pm \sqrt{25-12}}{2}=\frac{-5 \pm \sqrt{13}}{2}$. Scoring for such a solution follows the above point scheme: the correct calculation of the values of $x_{1}$ and $x_{2}$ earns 0 points, while determining the expressions involving $x_{1}$ and $x_{2}$ (as in the above solution) earns the corresponding points.