11. Construct a quadratic trinomial $p(x)$ such that $p(a)=a^{4}$, $p(b)=b^{4}$ and $p(c)=c^{4}$ (here $a, b$ and $c$ are distinct given numbers).
Solution 1. Let $q(x)=x^{4}-x(x-a)(x-b)(x-c)$. It is clear that this is a cubic polynomial such that $q(a)=a^{4}, q(b)=b^{4}$ and $q(c)=c^{4}$. Its leading coefficient is the number $a+b+c$. Therefore, the desired quadratic trinomial is the trinomial

$$
\begin{gathered}
p(x)=x^{4}-x(x-a)(x-b)(x-c)-(a+b+c)(x-a)(x-b)(x-c)= \\
=\left(a^{2}+b^{2}+c^{2}+a b+b c+a c\right) x^{2}-(a+b)(b+c)(c+a) x+a b c(a+b+c)
\end{gathered}
$$

Topic: "Roots of Polynomials"