## Elementary Linear Algebra 7th Edition

$$p(x)=-6-3x+x^2- x^3+x^4.$$
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, $$p(x)=-6-3x+x^2- x^3+x^4.$$