#### Answer

$$
p(x)=2 x.
$$

#### Work Step by Step

Suppose that
$$
p(x)=a_{0}+a_{1} x+a_{2} x^{2}.
$$
Now, we have
$$
\begin{array}{l}
{p(2)=a_{0}+a_{1}(2)+a_{2}(2)^{2}=a_{0}+2a_{1}+4a_{2}=4} \\
{p(3)=a_{0}+a_{1}(3)+a_{2}(3)^{2}=a_{0}+3 a_{1}+9 a_{2}=6} \\
{p(5)=a_{0}+a_{1}(5)+a_{2}(5)^{2}=a_{0}+5 a_{1}+25 a_{2}=10}
\end{array}
$$
The above system gas the solution
$$a_{0}=0, \quad a_{1}=2, \quad a_{2}=0.$$
Hence, $$
p(x)=2 x.
$$