Problem 11.3. A polynomial $G(x)$ with real coefficients takes the value 2022 at exactly five different points $x_{1}<x_{2}<x_{3}<x_{4}<x_{5}$. It is known that the graph of the function $y=G(x)$ is symmetric with respect to the line $x=-8$.

(a) (2 points) Find $x_{1}+x_{3}+x_{5}$.

(b) (2 points) What is the smallest degree that $G(x)$ can have?
Answer: (a) -24. (b) 6.

Solution. (a) Since the graph of $G(x)$ is symmetric with respect to the line $x=-8$, the points at which it takes the same value must be divided into pairs of symmetric points, except possibly one point that lies on the axis of symmetry. Therefore, the middle of the five given points should lie on the axis of symmetry, i.e., $x_{3}=-8$. The first and fifth points should be symmetric with respect to the line $x=-8$, so $x_{1}+x_{5}=2 \cdot(-8)=-16$. Therefore, $x_{1}+x_{3}+x_{5}=-24$.

(b) Consider the auxiliary polynomial $F(x)=G(x-8)-2022$. Clearly, the degree of $F(x)$ coincides with the degree of $G(x)$. The graph of $F(x)$ is obtained by a parallel shift of the graph of $G(x)$ 8 units to the right and 2022 units down. Then this graph is symmetric with respect to the line $x=0$, and at five points $F(x)$ takes the value 0. It is clear that $F(x)$ has exactly 5 distinct roots, so its degree is at least 5. On the other hand, since the graph of $F(x)$ is symmetric with respect to the line $x=0$, the function $F(x)$ is even, so the degree of $F(x)$ is also even. Therefore, the degree of $F(x)$, and hence the degree of $G(x)$, is at least 6.

As an example of a polynomial of degree 6, we can take $F(x)=(x+2)(x+1)x^{2}(x-1)(x-2)$. It corresponds to $G(x)=(x+10)(x+9)(x+8)^{2}(x+7)(x+6)+2022$.