Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$ 
Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$
  and that the remainder of the division of polynomial $P(x)$ by $x-3$ is  $18$. Evaluate the constant term of polynomial $P(x-1)$.

All I could see so far is that the polynomial $P(x)$ should be quadratic because $Q(x)$ is multiplied by quadratic term, which is $x^2$.

We are given that $P(x+1) = (x^2 -4)Q(x)+3ax+6$ $(\ast)$. Plugging $x=2$ into $(\ast)$ yields $P(3)=6a+6$. On the other hand, for every $b$, $P(b)$ is the remainder of the division of polynomial $P(x)$ by $x-b$ hence $P(3)=18$, which shows that $a=2$. 
Finally, plugging $x=-2$ into $(\ast)$ yields $P(-1)=-6a+6=-6$ hence the constant term of $P(x-1)$ is $P(-1)=-6$.
