Which is the highest fourth-degree $p(x)$ polynomial, which has $x_{1}=-3$ and $x_{2}=5$ as its zeros, and they are also points of local minimum? We also know that the polynomial $q(x)=p(x+1)$ is even, and the value of its local maximum is 256.
Solution. The function $p(x)$ cannot be constant. Suppose indirectly that $p(x)$ is constant. Since it has a zero, it can only be $p(x) \equiv 0$, in which case $q(x)=p(x+1) \equiv 0$ is also even, but it has no local maximum, contradicting the condition. Thus, we have proven that $p(x)$ cannot be constant.

The function $p(x)$ has zeros and minimum points at $x_{1}=-3$ and $x_{2}=5$. Knowing these roots, we can write its factorized form as:

$$
p(x)=(x+3)(x-5)\left(a x^{2}+b x+c\right)
$$

Since the function $p(x)$ has two minimum points, it must be at least a fourth-degree polynomial. It cannot be quadratic or cubic, as these can have at most one minimum point. This implies that $a \neq 0$.

Let's write the polynomial $q(x)=p(x+1)$:

$$
\begin{aligned}
q(x)=p(x+1) & =\left[a(x+1)^{2}+b(x+1)+c\right](x+1+3)(x+1-5)= \\
& =\left(a x^{2}+(2 a+b) x+a+b+c\right)(x+4)(x-4)= \\
& =\left[a x^{2}+(2 a+b) x+a+b+c\right]\left(x^{2}-16\right)
\end{aligned}
$$

We know that the polynomial $p(x+1)$ is even, which means that if $x_{0}$ is a root, then $-x_{0}$ is also a root, so

$$
a x^{2}+(2 a+b) x+a+b+c=a\left(x^{2}-x_{0}^{2}\right)
$$

and from this, $2 a+b=0$, and $q(x)=a\left(x^{2}-x_{0}^{2}\right)\left(x^{2}-16\right)$. According to the condition, $q(x)$ has minimum points at $\pm 4$; this can only happen if $x_{0}= \pm 4$, so $q(x)=a\left(x^{2}-16\right)^{2}$. This has a local maximum at $x=0$, so the value of the local maximum is $256=q(0)=a \cdot 16^{2}$, hence $a=1$.

The sought polynomial $p(x)$ is then:

$$
\begin{aligned}
p(x) & =q(x-1)=[(x+3)(x-5)]^{2}=\left(x^{2}-2 x-15\right)^{2}= \\
& =x^{4}-4 x^{3}-26 x^{2}+60 x+225
\end{aligned}
$$