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 $x^2$.


the remainder of the division of polynomial $P(x)$ by $x−3$ is $18$ 

implies:
$$P(x)=(x-3)R(x)+18 \Rightarrow \\
P(x+1)=(x-2)R(x+1)+18=(x^2 -4)Q(x)+3ax+6\\
P(2+1)=18=6a+6 \Rightarrow a=2.$$
Hence:
$$P(\color{red}x-1)=P(x-2+1)=((x-2)^2-4)Q(x-2)+6(x-2)+6 =\\
(x^2-4x)Q(x-2)+6x-6 \Rightarrow \\
P(\color{red}0-1)=-6.$$
