Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!

Hint:
Let $$\dfrac{f(x)}{(x-1)(x+1)(x-2)(x+2)}=\dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{x-2}+\dfrac D{x+2}$$
$$\implies f(x)=?$$
where $A,B,C,D$ which are arbitrary constants which can be found by putting  $x=1,-1,2,-2$ one by one.
For example, put $x=1,$ $$2\cdot1-5=A(1+1)(1-2)(1+2)$$
