Elementary Linear Algebra 7th Edition

$$p(x)=-\frac{3}{2}x+2x^2+\frac{1}{2}x^3.$$
Suppose that $$p(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_3x^3.$$ Now, we have $$\begin{array}{l} {p(-1)=a_{0}+a_{1}(-1)+a_{2}(-1)^{2}+a_3(-1)^3=a_{0}-a_{1}+a_{2}-a_3=3} \\ {p(0)=a_{0}+a_{1}(0)+a_{2}(0)^{2}+a_3(0)^3=a_{0} =0} \\ {p(1)=a_{0}+a_{1}(1)+a_{2}(1)^{2}+a_3(1)^3=a_{0}+ a_{1}+ a_{2}+a_3=1}\\ {p(4)=a_{0}+a_{1}(4)+a_{2}(4)^{2}+a_3(4)^3=a_{0}+4 a_{1}+16 a_{2}+64a_3=58}\\ \end{array}$$ The above system gas the solution $$a_{0}=0, \quad a_{1}=-\frac{3}{2}, \quad a_{2}=2, \quad a_{3}=\frac{1}{2}.$$ Hence, $$p(x)=-\frac{3}{2}x+2x^2+\frac{1}{2}x^3.$$