Finding a polynomial from its remainders. 
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 the value of $f(-3)$.

This question is taken from this.
I found that the question can be solved very easily using Lagrange's Interpolation Formula, since we can compute the value of $f(x)$ at $4(=3+1)$ points. All other answers to the above question were based on utilising the "zeroes" of the divisors. But suppose this question is modified a bit by extending the degree of $f(x)$ (say to $5$) and the degree of divisors to $3$. It seems to be possible to calculate $f(x)$ since $3+3 = 5+1$ (relate this to original question : $2+2 = 3+1$, i.e., sum of degree of divisors $=$ degree of main polynomial $+ 1$).
Modified version: $f(x)$ is a polynomial of degree $5$ and its remainders are $a_1x^2 +b_1x+c_1$ and $a_2x^2 +b_2x+c_2$ when divided by $A_1x^3 +B_1x^2 +C_1x + D_1$ and $A_2x^3 + B_2x^2 + C_2x + D_2$ where each of the divisors have $3$ distinct real roots (not necessarily rational). Find $f(x)$
Realising that it is very difficult to find the roots of $3$ degree polynomial, is there any other way to approach this problem?
 A: Let $\,\beta=-B/A, \gamma=-C/A, \delta=-D/A\,$, then working $\,\bmod (A x^3 +B x^2 +C x + D)\,$ :
$$
\begin{align}
x^3 &\equiv \beta x^2+\gamma x + \delta
\\ x^4 &\equiv \beta\,(\beta x^2+\gamma x + \delta)+\gamma x^2 + \delta x = \beta'x^2 + \gamma' x + \delta' 
\\ x^5 &\equiv \beta'\,(\beta x^2+\gamma x + \delta)+\gamma' x^2 + \delta' x = \beta''x^2 + \gamma'' x + \delta'' 
\end{align}
$$
Then:
$$
\begin{align}
f(x) &= px^5+qx^4+rx^3+sx^2+tx+u
\\ &\equiv p(\beta''x^2 + \gamma'' x + \delta'') +q (\beta'x^2 + \gamma' x + \delta' )+r(\beta x^2+\gamma x + \delta) + sx^2+tx+u
\\ &= s' x^2 + t' x + u'
\end{align}
$$
Equating the latter to the known remainder gives three equations in the coefficients of $f$, then repeating it for the second pair of divisor and remainder gives three more equations.

[ EDIT ] $\;$ The above is essentially a shortcut to calculating the remainder of the division directly, without calculating the quotient (which is not needed here), and with less overhead than the full Euclidean (long) polynomial division.
A: One way to approach this problem is to find the remainder by using long division of polynomials and the comparing it's coefficients with the remainder given in the problem.
