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
You have
$$f(x)=(x^2-1)(ax+b)+(2x-5)$$
and
$$f(x)=(x^2-4)(cx+d)+(-3x+4)$$
From the second you get $f(1)=-3(c+d)+1$ and from the first you get $f(1)=-3$. Thus
$$\color{red}{c+d=\frac{4}{3}}$$
From the first you get $f(2)=3(2a+b)-1$ and from the second you get $f(2)=-2$. Thus
$$\color{red}{2a+b=\frac{-1}{3}}$$
You also get $$f(0)=-b-5=-4d+4 \implies\color{red}{4d-b=9}$$
NOTE You can make a simple observation about the coefficients  of $x^3$ and $x^2$ in both the expressions for $f(x)$ to conclude that $a=c$ and $b=d$. With that the first two equations can suffice.
