## Task B-4.7.

The roots of the polynomial $p(x)=x^{3}-3 x^{2}+a x+3$ are real numbers $x_{1}, x_{2}, x_{3}$ that form an increasing arithmetic sequence. Determine the coefficient $a$ and the roots of the given polynomial.
## First Solution.

The polynomial $p(x)$ is written in its factored form:

$$
\begin{aligned}
& x^{3}-3 x^{2}+a x+3=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)= \\
& =x^{3}-\left(x_{1}+x_{2}+x_{3}\right) x^{2}+\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right) x-x_{1} x_{2} x_{3} .
\end{aligned}
$$

By the theorem of equality of polynomials, we obtain the system of equations (Vieta's formulas)

$$
\left\{\begin{array}{l}
-\left(x_{1}+x_{2}+x_{3}\right)=-3 \\
x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=a \\
-x_{1} x_{2} x_{3}=3
\end{array}\right.
$$

According to the problem, $x_{1}, x_{2}, x_{3}$ are terms of an increasing arithmetic sequence, so we write

$$
x_{1}=t-d, x_{2}=t, x_{3}=t+d, t \in \mathbb{R}
$$

From the first equation of the system, we get

$$
t-d+t+t+d=3 \Rightarrow t=1
$$

From the second equation of the system, we obtain

$$
\begin{aligned}
& x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=a \\
& a=(1-d) \cdot 1+(1-d) \cdot(1+d)+1 \cdot(1+d) \\
& a=3-d^{2}
\end{aligned}
$$

From the third equation of the system, we get

$$
\begin{aligned}
& x_{1} x_{2} x_{3}=-3 \\
& (1-d) \cdot 1 \cdot(1+d)=-3 \\
& 1-d^{2}=-3 \\
& d^{2}=4
\end{aligned}
$$

Since the sequence is increasing, $d>0$, so $d=2$. 1 point

The coefficient $a$ is equal to $a=3-2^{2}=-1$, 1 point

and the roots of the given polynomial are $-1, 1, 3$. 1 point

Note: The student does not have to derive Vieta's formulas; they can use them without proof.