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

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.

• No. When it comes to division of a constant polynomial by another constant polynomial, the quotient is just the quotient of the constants. In this case, $p(x)/q(x) = 2/3$ Sep 6 at 22:53
• The common polynomial division algorithm works only when the lead coef of the divisor is a unit (invertible), e.g. when the coef ring is a field, e.g. $\Bbb Z_p,\,$ for $p$ prime. When zero-divisors are present (e.g. $\Bbb Z_n$ for $n$ composite) divisibility matters are much more complex, e.g. $\bmod 6\!:\ x\equiv (2x-3)(3x+2)\ \$ Sep 6 at 23:12
• Polynomials with coefficients where? The question is not well formed until you specify where the coefficients come from. Sep 7 at 15:18

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

• More generally the quotient is the quotient of the lead coef's when $\,\deg f = \deg g.\ \$ Sep 6 at 23:35

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 $$1$$.

• No, $\!\bmod 5\,$ the quotient is still $\,2/3\equiv 2/(-2)\equiv -1\ \$ Sep 15 at 1:33