Polynomial function $P(x)$ satisfies that
$P(x) = (x+1)^3 Q_1 (x) + 2x^2 +1$
$P(x) = (x-1)^3 Q_2 (x) + 24x^2 -32x +19$
$P(1) = 11, P(-1)=3$
$P(x) = (x+1)^3 (x-1)^3 Q_3 (x) + R(x) $
Find $R(x)$ .
I thought of comparing the next two expressions $R(x) = (x+1)^3(ax^2 +bx+c ) +2x^2 +1 = (x-1) ^3 (a x^2 + dx+e ) + 24x^2 -32x +19$
but hard to find each coefficient.
The question can be rewritten like this:
Given a polynomial $P(x)$ that satisfies:
$$ \begin{cases} P(x) \equiv 2x^2 + 1 \mod{(x+1)^3} \\ P(x) \equiv 24x^2 - 32x + 19 \mod{(x-1)^3} \end{cases} $$
Find $R(x)$ such that: $$ P(x) \equiv R(x) \mod{((x+1)^3(x-1)^3)} $$
Chinese remainder theorem can solve this system of congruence just like in integer modulo arithmetic (See Chinese remainder theorem). This only works if the moduli are pairwise coprime, in this case we can verify that $(x+1)^3$ and $(x-1)^3$ are coprime using Euclidean algorithm. If we set up variables like this ($A^{-1} \mod{B}$ denotes the modulo multiplicative inverse of A with respect to modulus B): $$ \begin{cases} a_1 = 2x^2 + 1 \\ a_2 = 24x^2 - 32x + 19 \\ y_1 = (x-1)^3 \\ y_2 = (x+1)^3 \\ z_1 \equiv [(x-1)^3]^{-1} \mod{(x+1)^3} \\ z_2 \equiv [(x+1)^3]^{-1} \mod{(x-1)^3} \end{cases} $$ then the solution will be: $$ R(x) \equiv a_1y_1z_1 + a_2y_2z_2 \mod{((x+1)^3(x-1)^3)} $$ By using extended Euclidean's algorithm, we can find one solution of $z_1$ and $z_2$: $$ z_1 = \frac{-3}{16}x^2 - \frac{9}{16}x - \frac{1}{2} $$ $$ z_2 = \frac{3}{16}x^2 - \frac{9}{16}x + \frac{1}{2} $$ Distributing everything, we have: $$ R(x) \equiv \frac{33}{8}x^7 - 6x^6 - \frac{83}{8}x^5 + 20x^4 + \frac{75}{8}x^3 - 17x^2 + \frac{7}{8}x + 10 \mod{((x+1)^3(x-1)^3)} $$ or equivalently: $$ R(x) \equiv 2x^5 + 2x^4 - 3x^3 + x^2 + 5x + 4 \mod{((x+1)^3(x-1)^3)} $$
