<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] Working $\,\bmod (x^2-1)\,$ we have that $\,x^2 \equiv 1\,$, then:
$$
f(x) = ax^3+bx^2+cx+d\equiv ax+b+cx+d=(a+c)x+(b+d)
$$
Comparing to the known remainder $\,2x-5\,$:
$$
\begin{cases}
a+c=2
\\ b+d = -5 \tag{1}
\end{cases}
$$
Repeating the steps $\,\bmod (x^2-4)\,$ and remainder $\,-3x + 4\,$:
$$
f(x) \equiv 4ax+4b+cx+d=(4a+c)x+(4b+d) \;\;\implies\;\;
\begin{cases}
4a+c=-3
\\ 4b+d = 4 \tag{2}
\end{cases}
$$
Solving the four equations $\,(1), (2)\,$ for $\,a,b,c,d\,$ gives the polynomial $\,f(x)\,$.
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