<s>[INST] 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 $x^2$.
 [/INST] Let $F(x)=P(x-1)$.  Of course, the constant term of $F(x)$ is $F(0)=P(-1)$
We write $$P(x)=g(x)\times (x-3)+18$$  Then, of course, $P(3)=18$
Using the equation $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$  We set $$x=-2\implies P(-1)=-6a+6$$
and $$x=2\implies P(3)=6a+6$$  Since we already know that $P(3)=18$ we deduce that $a=2$ hence $$P(-1)=-12+6=-6$$
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