How to divide polynomials $p(x) = 2$ and $q(x)=3$ using polynomial division?

Question

If we were to divide the polynomial $p(x) = 2$ by the polynomial $q(x)=3$ using polynomial division, will the resulting polynomial be $f(x) = 1$? I am asking this to understand whether in polynomial division, we always continue the process of dividing even when the divisor is greater than what we have of the dividend (i.e. the remainder thus far).

Edit: what if I were working in the system of some modulus doing modular arithmetic and the coefficients are only integers (can't have $2 \over 3$)? I am interested in the result both in general and in modular arithmetic.

Answer 1

Let's suppose for a start that we are dealing with polynomials with real our complex coefficients. Let's look at the definition. Given two polynomials $f(x)$ and $g(x)$, the Theorem of the Euclidean Division of polynomials say that there is only on pair of polynomials $(q(x), r(x))$ such that:

$$f(x)=g(x)q(x)+r(x)$$

and $\deg(r)\leq \deg(g)$ or $r=0$.

In the case given, we will have $f(x)=2$, $g(x)=3$ then $q(x)=2/3$ and $r(x)=0$ satisfy all the conditions required. Thanks to the uniqueness, yes, that is your answer.

You can do that with any ring of polynomials $K[x]$ where $K$ is a field. More generally, there is the concept of euclidean domain that essentially means a ring with a working concept of euclidean division. For example, $K=\mathbb{Z}_p$ is a field, then you have euclidean (polynomial) division in $K[x]$. However you can't do that in $\mathbb{Z}[x]$, that is, the ring of polynomials with integer coefficients is not an euclidean domain.

A quick reference for further reading: https://en.wikipedia.org/wiki/Euclidean_domain

Answer 2

In the context of "polynomial division", if you are dividing the polynomial $p(x) = 2$ by $q(x) = 3$, you continue the division process until the degree of the remaining polynomial (the dividend) is lower than the degree of the divisor. In other words, you divide until you can no longer get a term with a degree equal to or higher than the divisor's degree.

In your example, if you are dividing $p(x) = 2$ by $q(x) = 3$, you can indeed perform the division, but the result is not $f(x) = 1$. Instead, it's $f(x) = \frac{2}{3}$. In this case, $f(x)$ is a constant polynomial because $q(x)$ is a higher-degree polynomial.

In modular arithmetic with integer coefficients, the division would work similarly. If you are working in a modulus, such as modulo 5 (denoted as $\text{mod } 5$), and you want to perform the division of $p(x)$ by $q(x)$, you would still follow the same rule. However, you'll perform the division with integers, considering the modulus.

For example, if you are in modulo 5 and want to divide $p(x) = 2$ by $q(x) = 3$, you can perform the division:

$ \begin{align*} 2 &\equiv 2 \pmod{5} \quad \text{(because $2 \text{ mod } 5 = 2$)} \\ \end{align*} $

So, in modular arithmetic, $f(x)$ would be $2 \pmod{5}$, which is still $2$, not $1$.