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