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 !!
 A: 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.
A: 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 $x=1,$ $$2\cdot1-5=A(1+1)(1-2)(1+2)$$
A: You have (for instance) that $f(x) = Q(x)(x+1)(x-1) + R(x)$, where $R(x) = 2x-5$.
Putting $x=1$ then $x=-1$ gives you $f(1) = - 3$ and $f(-1)=-7$.
Doing the same with $x=2, x=- 2$ allows you to work out two more points on the cubic.
Four distinct points allow you to completely characterise a cubic, you have a system of four linear equations to solve for the coefficients.
A: $f(x)$ 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,
$$\underline{\underline{f(-3)=53}}$$
A: 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 $\,f(x)\,$.
