# 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?

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.

• My answer here shows how the same approach works for the other question linked by the OP.
– dxiv
Jan 29 at 4:44
• Please explain how you got $x^4 \equiv \beta(\beta x^2+\gamma x + \delta)+\gamma x + \delta = \beta'x^2 + \gamma' x + \delta'$ from $x^3 \equiv \beta x^2+\gamma x + \delta$ Jan 29 at 5:35
• @user961447 There was a typo, now fixed in the edit. It's just $\,x^4 = x \cdot x^3$ $= x \cdot \left(\beta x^2 + \gamma x + \delta\right)$ $=\beta \color{red}{x^3} + \gamma x^2 + \delta x \,$ then replace $\,\color{red}{x^3}=\beta x^2 + \gamma x + \delta\,$ again.
– dxiv
Jan 29 at 6:01

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.