4. The quadratic trinomial $p(x)=a x^{2}+b x+c, a>0$ when divided by ( $x-1$ ) gives a remainder of 4, and when divided by ( $x-2$ ) - a remainder of 15. Find the maximum possible value of the ordinate of the vertex of the parabola $y=p(x)$ under these conditions. For what value of $x$ is it achieved?
4. Solution. By the Remainder Theorem, the remainder of the division of a polynomial $p(x)$ by $x-a$ is $p(a)$.

Therefore, the conditions of the problem are equivalent to the system:

$\left\{\begin{array}{l}4 a+2 b+c=15 \\ a+b+c=4\end{array} \Rightarrow\left\{\begin{array}{l}b=11-3 a \\ c=2 a-7\end{array}\right.\right.$.

Thus, $p(x)=a x^{2}+(11-3 a) x+2 a-7$. The minimum value of the quadratic trinomial $p(x)$ (the ordinate of the vertex of the parabola) is achieved at its vertex $x_{s}=\frac{3 a-11}{2 a}$. It is equal to $p_{\text {min }}(x)=p\left(x_{k}\right)=p\left(\frac{3 a-11}{2 a}\right)=-\frac{a^{2}-38 a+121}{4 a}=\frac{19}{2}-\frac{1}{4}\left(a+\frac{121}{a}\right)=4-\left(\sqrt{a}-\frac{11}{\sqrt{a}}\right)^{2}$.

This value is maximal if and only if $\sqrt{a}-\frac{11}{\sqrt{a}}=0$, that is, $a=11$. When $a=11$, $x_{s}=1$,
and $p\left(x_{s}\right)=4$.