Find a polynomial $p(x)$ that simultaneously has both the following properties.
$(i)$ When $p(x)$ is divided by $x^{100}$ the remainder is the constant polynomial $1$.
$(ii)$ When $p(x)$ is divided by $(x-2)^3$ the remainder is the constant polynomial $ 2$.
Though I could find the polynomial by taking the derivative (works better because the remainders are simple), I wanted to understand how the Chinese remainder theorem gives the polynomial.
The solution states the following:
$x^{100}$ and
$(x − 2)^3$ do not share a common factor, you know without any work that a polynomial with given properties must exist. The same Euclidean algorithm (but now with polynomials) gives a systematic way to find it. In the given problem we could use a different trick because the specified remainders here were rather simple (constants). But there is a conceptual way as well by implementing the Chinese remainder theorem.
I am new and poor with Latex. And thank you in advance.
Best Answer
We are given the system $$ \begin{cases} p(x) \equiv 1 \mod{x^{100}} \\ p(x) \equiv 2 \mod{(x-2)^3}. \end{cases}$$ The first line indicates that $$ p(x) = 1 + g(x) x^{100}$$ for some polynomial $g(x)$. Inserting into the second line gives that $$ 1 + g(x) x^{100} = 2 \mod (x-2)^3,$$ or rather that $$ g(x) x^{100} + h(x) (x-2)^3 = 1$$ for some polynomials $g(x)$ and $h(x)$. This equation has a solution since $x^{100}$ and $(x-2)^3$ are relatively prime. Note that this is very similar to questions looking like $$ 3x + 5y = 1,$$ or more generally $$ ax + by = 1,$$ which are traditionally solved using the Euclidean algorithm and back-substitution.
It happens to be that this Euclidean algorithm and back-substitution is gross, and I wouldn't recommend doing it by hand. But you can see the actual result using Wolfram|Alpha's implementation.