Generalizing the Classical Remainder Theorem: A Reflection-Based Methodological Strategy

Abstract

The framework of this paper is the presentation of a case study in which university students are required to extend a particular problem of division of polynomials in one variable over the field of real numbers (as generalizing action) clearly influenced by prior strategies (as reflection generalization). Specifically, the objective of this paper is to present a methodology for generalizing the classical Remainder Theorem to the case in which the divisor is a product of binomials ${(x-a_1)}^{n_1}{(x-a_2)}^{n_2}\cdots {(x-a_k)}^{n_k}$, where $a_1,a_2,\cdots ,a_k\in \mathbb{R}$ and $n_1,n_2,\cdots ,n_k\in \mathbb{N}$. A first approach to this issue is the Taylor expansion of the dividend $P\left(x\right)$ at a point a, which clearly shows the quotient and the remainder of the division of $P\left(x\right)$ by ${(x-a)}^k$, where the degree of $P\left(x\right)$, say n, must be greater than or equal to k. The methodology used in this paper is the proof by induction which allows to obtain recurrence relations different from those obtained by other scholars dealing with the generalization of the classical Remainder Theorem.

1. Introduction

In a classical paper, ref. [1] points out generalization as a powerful tool that permeates the whole science. Specifically, he states that generalization “has helped to extend the frontier of mathematical knowledge in every direction” and, moreover, to the creation of new branches. In the same line, ref. [2] states that “generalization is a critical component of mathematical activity and has garnered increased attention in school mathematics at all levels” and in all areas of mathematics. Moreover, she claims that researchers have highlighted several social and pedagogical influences on generalizing activities by students. According to [3], generalizing is increasingly recognized as the essence of thinking in algebra because it is an act of abstraction. Additionally, when generalizing in problem situations, “students learn to suppress detail and hide the irrelevant features of a situation while paying attention to ideas, qualities, and properties”.

Dumitrascu [4] states that activity theory framework provides basic principles which allows to consider generalization as “an activity that is socially and historically developed through tools and artifacts mediations, internalization of social knowledge, and that is transformed through learning and development”.

Ellis [5] presents several types of generalizations which students have constructed when reasoning in mathematics. Her taxonomy distinguishes between generalizing actions (relating, searching, and extending) and reflection generalizations (identifications or statements, definitions, and the influence of prior ideas or strategies). This paper presents a case study in which university students are required to extend a particular problem of division of polynomials in one variable over the field of real numbers (as generalizing action) clearly influenced by prior strategies (as reflection generalization); that is to say, the presentation of a particular problem which can be more effectively approached in a more general form [6,7,8]. This contribution is similar to [9], with the difference that our case study is presented with the aim that university students can see what is expected from them.

Specifically, the objective of this paper is to generalize the classical Remainder Theorem to the case in which the divisor is a product of binomials ${(x-a_1)}^{n_1}{(x-a_2)}^{n_2}\cdots {(x-a_k)}^{n_k}$, where $a_1,a_2,\cdots ,a_k\in \mathbb{R}$ and $n_1,n_2,\cdots ,n_k\in \mathbb{N}$. In this way, ref. [10] proposed a generalization for polynomials over commutative coefficient rings which allows calculating the remainder without using the long division method (see also [11]).

The simplest case of this topic is the classical Remainder Theorem, according to which the remainder of the division of a polynomial $P\left(x\right)$ by the binomial $x-a$ is the numerical value of $P\left(x\right)$ for $x=a$; that is to say, $P\left(a\right)$. A second approach to this issue is the Taylor expansion of the dividend $P\left(x\right)$ at a point a which clearly exhibits the quotient and the remainder of the division of $P\left(x\right)$ by ${(x-a)}^k$, where the degree of $P\left(x\right)$, say n, must be greater than or equal to k. In effect, the Taylor expansion of $P\left(x\right)$ at point $x=a$ is:

$$P\left(x\right)={\displaystyle \frac{P^{\left(n\right)}\left(a\right)}{n!}}{(x-a)}^n+\cdots +{\displaystyle \frac{P^{\left(k\right)}\left(a\right)}{k!}}{(x-a)}^k+\cdots +{\displaystyle \frac{P^{\prime }\left(a\right)}{1!}}(x-a)+P\left(a\right),$$

(1)

where $P^{\left(k\right)}\left(a\right)$ denotes the numerical value of the k-th derivative of $P\left(x\right)$ at point $x=a$. Obviously, Equation (1) can be written as

$$P\left(x\right)=\left[{\displaystyle \frac{P^{\left(n\right)}\left(a\right)}{n!}}{(x-a)}^{n-k}+\cdots +{\displaystyle \frac{P^{\left(k\right)}\left(a\right)}{k!}}\right]{(x-a)}^k+\cdots +{\displaystyle \frac{P^{\prime }\left(a\right)}{1!}}(x-a)+P\left(a\right),$$

(2)

which clearly exhibits that the quotient $Q\left(x\right)$ of this division is

$$Q\left(x\right):={\displaystyle \frac{P^{\left(n\right)}\left(a\right)}{n!}}{(x-a)}^{n-k}+\cdots +{\displaystyle \frac{P^{\left(k\right)}\left(a\right)}{k!}}$$

(3)

and the remainder $R\left(x\right)$ is

$$R\left(x\right):={\displaystyle \frac{P^{(k-1)}\left(a\right)}{(k-1)!}}{(x-a)}^{k-1}+\cdots +{\displaystyle \frac{P^{\prime }\left(a\right)}{1!}}(x-a)+P\left(a\right).$$

(4)

Wanicharpichat [12] analyzed the general case in which the divisor is a product of binomials of the form ${(x-a_1)}^{n_1}{(x-a_2)}^{n_2}\cdots {(x-a_k)}^{n_k}$, where $a_1,a_2,\cdots ,a_k\in \mathbb{R}$ and $n_1,n_2,\cdots ,n_k\in \mathbb{N}$, obtaining a solution for the remainder of the division which involves the successive derivatives of the dividend. The main contribution of the solution presented in this paper is that it does not use the derivatives but a solution derived by recurrence.

On the other hand, ref. [13] presented a generalization of the factor theorem for a given polynomial in the variable x by which, under certain conditions, there exists a factor of the form $x^n-b$, for a positive integer n and complex b. These scholars also show the way quadratic factors could be obtained. Finally, ref. [14] proposed a computational methodology which, using vector algebra, generalizes the remainder theorem to divisions for polynomials of any degree over any integral domain.

This paper is organized as follows. After this introduction, Section 2 solves successive generalizations of the classical Remainder Theorem until arriving to a general formula where the divisor is of the form ${(x-a_1)}^{n_1}{(x-a_2)}^{n_2}\cdots {(x-a_k)}^{n_k}$, where $a_1,a_2,\cdots ,a_k\in \mathbb{R}$ and $n_1,n_2,\cdots ,n_k\in \mathbb{N}$. Finally, Section 3 summarizes and concludes.

2. Generalizing the Classical Remainder Theorem

The content of this section will be developed progressively according to the following subsections.

2.1. Case of a Quadratic Divisor

Let $P\left(x\right)$ be a polynomial of degree $n\ge 2$. Assume that $P\left(x\right)$ has to be divided by the product of binomials $(x-a)(x-b)$. By first dividing $P\left(x\right)$ by $(x-a)$, one has

$$P\left(x\right)=C\left(x\right)(x-a)+R,$$

(5)

where $C\left(x\right)$ is the quotient and R is the remainder of the division. As obviously $C\left(x\right)$ is a polynomial of degree $n-1\ge 1$, it can be divided by $(x-b)$, leading to

$$C\left(x\right)=D\left(x\right)(x-b)+S,$$

(6)

where $D\left(x\right)$ is the quotient and S is the remainder of the new division. By substituting (6) in (5), one has

$$P\left(x\right)=D\left(x\right)(x-a)(x-b)+S(x-a)+R.$$

(7)

Therefore, the remainder of the division $P\left(x\right):\left[\right(x-a\left)\right(x-b\left)\right]$ is

$$\mathcal{R}:=S(x-a)+R.$$

(8)

By the classical Remainder Theorem,

$$R=P\left(a\right)$$

(9)

and

$$S=C\left(b\right).$$

(10)

As $P\left(b\right)=C\left(b\right)(b-a)+R$, taking into account (9) and (10), one has

$$P\left(b\right)=S(b-a)+P\left(a\right),$$

(11)

from where:

$$S={\displaystyle \frac{P\left(b\right)-P\left(a\right)}{b-a}}.$$

(12)

If $P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0$, a practical arrangement can be used as represented in Figure 1, where calculations have been performed according to Horner’s scheme.

In Section 2.3, we are going to generalize Equation (12) to the case of a cubic divisor. For this purpose, we will write Equation (12) in the following form:

$$S={\displaystyle \frac{P\left(a\right)}{a-b}}+{\displaystyle \frac{P\left(b\right)}{b-a}}.$$

(13)

Thus, the remainder of the division $P\left(x\right):\left[\right(x-a\left)\right(x-b\left)\right]$ remains as

$$\mathcal{R}=\left[{\displaystyle \frac{P\left(a\right)}{a-b}}+{\displaystyle \frac{P\left(b\right)}{b-a}}\right](x-a)+P\left(a\right).$$

(14)

Taking into account that the former remainder is the same as the remainder of the division $P\left(x\right):\left[\right(x-b\left)\right(x-a\left)\right]$, then Formula (14) can be also written as

$$\mathcal{R}=\left[{\displaystyle \frac{P\left(b\right)}{b-a}}+{\displaystyle \frac{P\left(a\right)}{a-b}}\right](x-b)+P\left(b\right).$$

(15)

Example 1.

To calculate the remainder of the division of the polynomial $P\left(x\right)=3x^3-7x^2+5x-1$ by $Q\left(x\right)=x^2-5x+6$, take into account that

• $x^2-5x+6=(x-2)(x-3)$.
• $P\left(2\right)=24-28+10-1=5$.
• $P\left(3\right)=81-63+15-1=32$.

Then,

1.

By considering Formula (14),

$$\mathcal{R}=\left({\displaystyle \frac{5}{-1}}+{\displaystyle \frac{32}{1}}\right)(x-2)+5=27(x-2)+5=27x-49.$$

(16)

2.

By considering Formula (15),

$$\mathcal{R}=\left({\displaystyle \frac{32}{1}}+{\displaystyle \frac{5}{-1}}\right)(x-3)+32=27(x-3)+32=27x-49.$$

(17)

2.2. Particular Case

In the case in which $a=b$, Formula (12) is indeterminate. So, we can take

$$S=\underset{b\to a}{lim}{\displaystyle \frac{P\left(b\right)-P\left(a\right)}{b-a}}=P^{\prime }\left(a\right).$$

(18)

Then,

$$\mathcal{R}=P^{\prime }\left(a\right)(x-a)+P\left(a\right).$$

(19)

Example 2.

To calculate the remainder of the division of the polynomial $P\left(x\right)=3x^3-7x^2+5x-1$ by $Q\left(x\right)=x^2-4x+4$, take into account that

• $x^2-4x+4={(x-2)}^2$.
• $P\left(2\right)=5$.
• $P^{\prime }\left(x\right)=9x^2-14x+5$.
• $P^{\prime }\left(2\right)=36-28+5=13$.

Then, by considering Formula (19),

$$\mathcal{R}=13(x-2)+5=13x-21.$$

(20)

2.3. Case of a Cubic Divisor

Let $P\left(x\right)$ be a polynomial of degree $n\ge 3$. Assume that $P\left(x\right)$ has to be divided by $(x-a)(x-b)(x-c)$. By first dividing $P\left(x\right)$ by $(x-a)$, one has

$$P\left(x\right)=C\left(x\right)(x-a)+R,$$

(21)

where $C\left(x\right)$ is the quotient and R is the remainder of the division. As $C\left(x\right)$ is obviously a polynomial of degree $n-1\ge 2$, it can be divided by $(x-b)$, leading to

$$C\left(x\right)=D\left(x\right)(x-b)+S,$$

(22)

where $D\left(x\right)$ is the quotient and S is the remainder of the new division. Once again, as obviously $D\left(x\right)$ is a polynomial of degree $n-2\ge 1$, it can be divided by $(x-c)$, leading to

$$D\left(x\right)=E\left(x\right)(x-c)+T,$$

(23)

where $E\left(x\right)$ is the quotient and T is the remainder of the third division. By substituting successively (22) and (23) in (21), one has

$$P\left(x\right)=E\left(x\right)(x-a)(x-b)(x-c)+T(x-a)(x-b)+S(x-a)+R.$$

(24)

Therefore, the remainder of the division $P\left(x\right):\left[\right(x-a\left)\right(x-b\left)\right(x-c\left)\right]$ is

$$\mathcal{R}:=T(x-a)(x-b)+S(x-a)+R.$$

(25)

By the classical Remainder Theorem,

$$R=P\left(a\right),$$

(26)

$$S=C\left(b\right)$$

(27)

and

$$T=D\left(c\right).$$

(28)

As $P\left(b\right)=C\left(b\right)(b-a)+R$, taking into account (26) and (27), one has

$$P\left(b\right)=S(b-a)+P\left(a\right),$$

from where

$$S={\displaystyle \frac{P\left(b\right)-P\left(a\right)}{b-a}}={\displaystyle \frac{P\left(a\right)}{a-b}}+{\displaystyle \frac{P\left(b\right)}{b-a}}.$$

(29)

On the other hand, by (24),

$$P\left(c\right)=T(c-a)(c-b)+S(c-a)+R$$

(30)

or, equivalently,

$$P\left(c\right)=T(c-a)(c-b)+{\displaystyle \frac{P\left(b\right)-P\left(a\right)}{b-a}}(c-a)+P\left(a\right),$$

(31)

from where

$$T={\displaystyle \frac{P\left(c\right)-P\left(a\right)}{(c-a)(c-b)}}-{\displaystyle \frac{P\left(b\right)-P\left(a\right)}{(b-a)(c-b)}}={\displaystyle \frac{P\left(c\right)-P\left(a\right)}{(c-a)(c-b)}}+{\displaystyle \frac{P\left(b\right)-P\left(a\right)}{(b-a)(b-c)}}.$$

(32)

If $P\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0$, a practical arrangement can be used as represented in Figure 2, where calculations have been performed according to Horner’s scheme.

Simple calculation shows that

$$T={\displaystyle \frac{P\left(a\right)}{(a-b)(a-c)}}+{\displaystyle \frac{P\left(b\right)}{(b-a)(b-c)}}+{\displaystyle \frac{P\left(c\right)}{(c-a)(c-b)}}.$$

(33)

Thus, the remainder of the division $P\left(x\right):\left[\right(x-a\left)\right(x-b\left)\right(x-c\left)\right]$ remains as

$$\begin{array}{ccc}\hfill \mathcal{R}& =& \left[{\displaystyle \frac{P\left(a\right)}{(a-b)(a-c)}}+{\displaystyle \frac{P\left(b\right)}{(b-a)(b-c)}}+{\displaystyle \frac{P\left(c\right)}{(c-a)(c-b)}}\right](x-a)(x-b)+\hfill \\ & & \left[{\displaystyle \frac{P\left(a\right)}{a-b}}+{\displaystyle \frac{P\left(b\right)}{b-a}}\right](x-a)+\hfill \\ & & P\left(a\right).\hfill \end{array}$$

(34)

Observe that the solution of $\mathcal{R}$ obtained in Equation (34) is the same as the solution obtained in Equation (15), but adds up a new term (specifically, the quadratic term).

Example 3.

To calculate the remainder of the division of the polynomial $P\left(x\right)=x^5$ by $Q\left(x\right)=x^3-x$, take into account that

• $x^3-x=x(x-1)(x+1)$.
• $P\left(0\right)=0$.
• $P\left(1\right)=1$.
• $P(-1)=-1$

Then, by considering Formula (34),

$$\mathcal{R}=\left({\displaystyle \frac{0}{-1}}+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{-1}{2}}\right)(x-0)(x-1)+\left({\displaystyle \frac{0}{-1}}+{\displaystyle \frac{1}{1}}\right)(x-0)+0=x.$$

(35)

2.4. Particular Case

In the case in which $a=b=c$, Formula (32) is indeterminate. So, we can take

$$T=\underset{(b,c)\to (a,a)}{lim}\left[{\displaystyle \frac{P\left(c\right)-P\left(a\right)}{(c-a)(c-b)}}-{\displaystyle \frac{P\left(b\right)-P\left(a\right)}{(b-a)(c-b)}}\right].$$

(36)

Let $c-a:=h$ and $c-b=b-a={\displaystyle \frac{h}{2}}$. Therefore, the former limit remains as

$$T=\underset{h\to 0}{lim}\left[{\displaystyle \frac{P(a+h)-P\left(a\right)}{h^2/2}}-{\displaystyle \frac{P(a+h/2)-P\left(a\right)}{h^2/4}}\right].$$

(37)

Making simple calculations,

$$\begin{array}{ccc}\hfill T& =& \underset{h\to 0}{lim}{\displaystyle \frac{P(a+h)-P\left(a\right)-1/2\left[P\right(a+h/2)-P(a\left)\right]}{h^2/2}}\hfill \\ & =& {\displaystyle \frac{P^{\prime }(a+h)-1/2P^{\prime }(a+h/2)}{h}}\hfill \\ & =& P^{\prime \prime }\left(a\right)-{\displaystyle \frac{1}{2}}{P}^{\prime \prime }\left(a\right)={\displaystyle \frac{1}{2}}{P}^{\prime \prime }\left(a\right).\hfill \end{array}$$

(38)

Then,

$$\mathcal{R}={\displaystyle \frac{1}{2}}{P}^{\prime \prime }\left(a\right)(x-a)(x-b)+P^{\prime }\left(a\right)(x-a)+P\left(a\right).$$

(39)

Example 4.

To calculate the remainder of the division of the polynomial $P\left(x\right)=5x^4-8x^3+6x^2-5x+2$ by $Q\left(x\right)=x^3+3x^2+3x+1$, take into account that

• $x^3+3x^2+3x+1={(x+1)}^3$.
• $P(-1)=26$.
• $P^{\prime }\left(x\right)=20x^3-24x^2+12x-5$
• $P^{\prime \prime }\left(x\right)=60x^2-48x+12$
• $P^{\prime }(-1)=-20-24-12-5=-61$.
• $P^{\prime \prime }(-1)=60+48+12=120$.

Then, by considering Formula (39),

$$\mathcal{R}={\displaystyle \frac{120}{2!}}{(x+1)}^2-61(x+1)+26=60x^2+59x+25.$$

(40)

2.5. General Case

Let $P\left(x\right)$ be a polynomial of degree n. Assume that $P\left(x\right)$ has to be divided by $(x-a_1)(x-a_2)\cdots (x-a_n)$. By recurrence, it can be shown that

$$\begin{array}{ccc}\hfill \mathcal{R}& :=& \left[\sum _{i=1}^n{\displaystyle \frac{P\left(a_i\right)}{{\prod }_{\stackrel{j=1}{j\ne i}}^n(a_i-a_j)}}\right]\prod _{i=1}^n(x-a_i)\hfill \\ & +& \left[\sum _{i=1}^{n-1}{\displaystyle \frac{P\left(a_i\right)}{{\prod }_{\stackrel{j=1}{j\ne i}}^{n-1}(a_i-a_j)}}\right]\prod _{i=1}^{n-1}(x-a_i)\hfill \\ & +& \cdots \hfill \\ & +& \left[{\displaystyle \frac{P\left(a\right)}{a-b}}+{\displaystyle \frac{P\left(b\right)}{b-a}}\right](x-a_1)\hfill \\ & +& P\left(a_1\right)\hfill \\ & =& \sum _{k=1}^n\left[\sum _{i=1}^k{\displaystyle \frac{P\left(a_i\right)}{{\prod }_{\stackrel{j=1}{j\ne i}}^k(a_i-a_j)}}\right]\prod _{i=1}^k(x-a_i).\hfill \end{array}$$

(41)

Remark 1.

Observe that if $\mathcal{R}$ is the remainder of dividing the polynomial P by $(x-a_1)\cdots (x-a_n)$, then $\mathcal{R}$ is the interpolation polynomial in Newton’s form for the set of points $(a_1,P\left(a_1\right)),\cdots$, $(a_n,P\left(a_n\right))$; hence, the natural way to find this $\mathcal{R}$.

In general, if the divisor is ${(x-a_1)}^{n_1}{(x-a_2)}^{n_2}\cdots {(x-a_k)}^{n_k}$, then the remainder of the division is

$$\mathcal{R}=R_1\left(x\right)+{(x-a_1)}^{n_1}\{R_2\left(x\right)+{(x-a_2)}^{n_2}[R_3\left(x\right){(x-a_3)}^{n_3}\cdots +R_k\left(x\right)]\},$$

(42)

where $R_1\left(x\right)$ is the remainder of the division $P\left(x\right):{(x-a_1)}^{n_1}$, whose quotient is $Q_1\left(x\right)$; $R_2\left(x\right)$ is the remainder of the division $Q_1\left(x\right):{(x-a_2)}^{n_2}$, whose quotient is $Q_2\left(x\right)$; …; and, finally, $R_k\left(x\right)$ is the remainder of the division $Q_{k-1}\left(x\right):{(x-a_k)}^{n_k}$, whose quotient is $Q_k\left(x\right)$, in such a way that the degree of $Q_k\left(x\right)$ is less than $n_{k+1}$.

Example 5.

To calculate the remainder of the division of the polynomial $P\left(x\right)=5x^4-6x^3-2x^2+7x+9$ by $Q\left(x\right)=x^3+2x^2$, take into account that $Q\left(x\right)=x^2(x+2)$. Obviously, one has

$$P\left(x\right)=(5x^2-6x-2)x^2+(7x+9),$$

from where

$$Q_1\left(x\right)=5x^2-6x-2$$

and

$$R_1\left(x\right)=7x+9.$$

Moreover,

$$Q_1\left(x\right)=(5x-16)(x+2),$$

from where

$$Q_2\left(x\right)=5x-16$$

and

$$R_2\left(x\right)=30.$$

Consequently,

$$\mathcal{R}\left(x\right)=R_1\left(x\right)+x^2\left[R_2\left(x\right)\right]=(7x+9)+30x^2=30x^2+7x+9.$$

3. Conclusions

In this paper, we have presented a case study where a generalization of a topic related to the division of polynomials is presented to university students as a model to follow the strategy of generalizing a mathematical result influenced by previous strategies. Specifically, we have solved the general problem of determining the remainder of the division of a polynomial $P\left(x\right)$ by a product of binomials $D\left(x\right)={(x-a_1)}^{n_1}{(x-a_2)}^{n_2}\cdots {(x-a_k)}^{n_k}$. The main advantage of the solution presented in the present paper is that it does not use the successive derivatives of $P\left(x\right)$, as in [12]. In effect, one contribution of [12] is Theorem 2.3, which provides a result concerning a special remainder theorem when the polynomial $P\left(x\right)$ is divided by another polynomial $D\left(x\right)$ of the form ${(x-b)}^m$. Additionally, Theorem 2.5 presents the expression of the quotient in the same case as Theorem 2.3. In these two results, the successive derivatives of $P\left(x\right)$ have been used. Finally, Theorem 4.1 states the so-called Generalized Quotient and Remainder Theorem, which uses the successive derivatives of $D\left(x\right)$. However, in the present paper, derivatives have been only used in the particular cases presented in Section 2.2 and Section 2.4.

The findings of this paper can be applied to the construction and analysis of finite fields of order $p^n$, based on a irreducible polynomial of degree n over a finite field of p elements.