Answer
$$
p(x)=4.
$$
Work Step by Step
Suppose that
$$
p(x)=a_{0}+a_{1} x+a_{2} x^{2}.
$$
Now, we have
$$
\begin{array}{l}
{p(2)=a_{0}+a_{1}(2)+a_{2}(2)^{2}=a_{0}+2a_{1}+4a_{2}=4} \\
{p(3)=a_{0}+a_{1}(3)+a_{2}(3)^{2}=a_{0}+3 a_{1}+9 a_{2}=4} \\
{p(4)=a_{0}+a_{1}(4)+a_{2}(4)^{2}=a_{0}+4 a_{1}+16 a_{2}=4}
\end{array}
$$
The above system gas the solution
$$a_{0}=4, \quad a_{1}=0, \quad a_{2}=0.$$
Hence, $$
p(x)=4.
$$
