Of course, I considered $p(x) = (x+1)(x-2)(x-3)$, but it is clear that $p(0) \neq 1$. A bit stumped by this I see that the book solution says that we MUST have $p(x) = c(x+1)(x-2)(x-3)$ for some $c$. I understand why this is an idea to get to a polynomial that satisfies the conditions. What I don't understand is how it is obvious $p(x)$ MUST be of this form. If I knew this I believe I would have considered this idea on my own.
Is there a rigorous argument to see why $p(x)$ MUST be able to be reduced in this way (preferably using only precalculus knowledge)?
Suppose that $-1$, $2$, and $3$ are roots of $p(x)$ and that $\deg p(x)=3$. Divide $p(x)$ by $(x+1)(x-2)(x-3)$; you will get$$p(x)=(x+1)(x-2)(x-3)q(x)+r(x)$$where $\deg r(x)<3$. But then since $\deg p(x)=3$, $\deg q(x)=0$; that is, $q(x)=c$, for some constant $c$. So, you have$$p(x)=c(x+1)(x-2)(x-3)+r(x)$$and, since $r(-1)=r(2)=r(3)=0$ and $\deg r(x)<3$, $r(x)=0$.