Answer
$$ p(x)=42-41x-2x^2+ x^3. $$
Work Step by Step
Suppose that $$ p(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_3x^3. $$ Now, we have $$ \begin{array}{l} {p(0)=a_{0}+a_{1}(0)+a_{2}(0)^{2}+a_3(0)^3=a_{0}=42} \\ {p(1)=a_{0}+a_{1}(1)+a_{2}(1)^{2}+a_3(01)^3=a_{0}+a_1+a_2+a_3 =0} \\ {p(2)=a_{0}+a_{1}(2)+a_{2}(2)^{2}+a_3(2)^3=a_{0}+ 2a_{1}+ 4a_{2}+8a_3=-40}\\ {p(3)=a_{0}+a_{1}(3)+a_{2}(3)^{2}+a_3(3)^3=a_{0}+3 a_{1}+9 a_{2}+27a_3=-72}\\ \end{array} $$ The above system gas the solution $$a_{0}=42, \quad a_{1}=-41, \quad a_{2}=-2, \quad a_{3}=1.$$ Hence, $$ p(x)=42-41x-2x^2+ x^3. $$
