Answer
$$
p(x)=-6-3x+x^2- x^3+x^4.
$$
Work Step by Step
Suppose that
$$
p(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_3x^3+a_4x^4.
$$
Now, we have
$$
\begin{array}{l}
{p(-2)=a_{0}+a_{1}(-2)+a_{2}(-2)^{2}+a_3(-2)^3+a_4(-2)^4=a_{0}-2a_1+4a_2-8a_3+16a_4=28} \\
{p(-1)=a_{0}+a_{1}(-1)+a_{2}(-1)^{2}+a_3(-1)^3+a_4(-)^4=a_{0}-a_1+a_2-a_3 +a_4=0} \\
{p(0)=a_{0}+a_{1}(0)+a_{2}(0)^{2}+a_3(0)^3+a_4(0)^4=a_{0} =-6}\\
{p(1)=a_{0}+a_{1}(1)+a_{2}(1)^{2}+a_3(1)^3+a_4(1)^4=a_{0}+ a_{1}+ a_{2}+ a_3+a_4=-8}\\
{p(2)=a_{0}+a_{1}(2)+a_{2}(2)^{2}+a_3(2)^3+a_4(2)^4=a_{0}+2 a_{1}+4 a_{2}+8a_3+16a_4=0}\\
\end{array}
$$
The above system has the solution
$$a_{0}=-6, \quad a_{1}=-3, \quad a_{2}=1, \quad a_{3}=-1, \quad a_{4}=1.$$
Hence, $$
p(x)=-6-3x+x^2- x^3+x^4.
$$
