<s>[INST] 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 !!
 [/INST] $f(x)$ leaves remainder $2x-5$ when divided by $x^2-1.$
Then,
$$f(1)=2(1)-5=-3$$
$$f(-1)=2(-1)-5=-7$$
$f(x)$ leaves remainder $-3x+4$ when divided by $x^2-4.$
So we have,
$$f(x)=(x^2-4)\,q(x)-3x+4$$
Since $f(x)$ is of degree $3$, the degree of $q(x)$ should be $1.$
Let $q(x)=ax+b.$ Then,
$$f(x)=(x^2-4)(ax+b)-3x+4$$
$$f(x)=ax^3+bx^2-(4a+3)x-(4b-4)$$
Let $x=1$ then,
$$f(1)=a(1)^3+b(1)^2-(4a+3)(1)-(4b-4)=-3$$
$$a+b=\dfrac{4}{3}\quad\quad\dots(1)$$
Let $x=-1$ then,
$$f(-1)=a(-1)^3+b(-1)^2-(4a+3)(-1)-(4b-4)=-7$$
$$a-b=-\dfrac{14}{3}\quad\quad\dots(2)$$
Solving (1) and (2) we get,
$$a=-\dfrac{5}{3}$$
$$b=3$$
Therefore,
$$f(x)=-\dfrac{5}{3}x^3+3x^2+\dfrac{11}{3}x-8$$
Now,
$$\underline{\underline{f(-3)=53}}$$
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