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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?

user961447
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2 Answers2

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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.

dxiv
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  • My answer here shows how the same approach works for the other question linked by the OP. – dxiv Jan 29 '22 at 04:44
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    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 $x^3 \equiv \beta x^2+\gamma x + \delta$ – user961447 Jan 29 '22 at 05:35
  • @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 '22 at 06:01
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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.