Finding polynomials $p(-1)$

Question

When the third-order polynomial $p(x)$ is divided by $x-1$, $x-2$, and $x-3$, the remainders were all $1$ . If the constant term of $p(x)$ is $2$, what is the value of $p(-1)$?

Answer 1

Let $p(x)=ax^3+bx^2+cx+d$ $$p(0)=2=d$$ $$p(1)=1$$ $$p(2)=1$$ $$p(3)=1$$ Then $$ \left\{ \begin{array}{c} a+b+c+2=1 \\ 8a+4b+2c+2=1 \\ 27a+9b+3c+2=1 \end{array} \right. $$ Which gives us $$a=-\frac{1}{6}$$ $$b=1$$ $$c=-\frac{11}{6}$$ And $$p(-1)=-1*(-\frac{1}{6})^3+1*1^2+1*(-\frac{11}{6})+2=5$$