In high school mathematics, we come across the topic of factorisation of polynomials with integer or rational coefficients.

Usually, we only consider factorisation of polynomials of low degrees. Checking whether or not a polynomial of higher degree such as X ³⁷ − 27X ¹¹ + 3 is factorisable turns out to be a difficult problem.

Although there are some methods available in higher mathematics to deal with such problems, the tools and techniques of high school mathematics seem to be of little use for addressing such problems.

In this article, we discuss the factorisation of a particular infinite family of polynomials of arbitrary degree.

Statement of the problem

In [1], it is shown that for any integer n ≥ 1 and distinct integers a₁, a₂,. .. , a_n, the polynomial

is not factorisable. It is interesting to note that if we consider a variant of this polynomial, namely

then the polynomial is sometimes factorisable.
For instance we have:

However, keeping +1 in place of −1, we can show that a large number of such polynomials are not factorisable. More precisely, we prove the following theorem.

Theorem 1. Let n ≥ 1 be an integer and let a₁, a₂,. . ., a_n be distinct odd integers. Then the polynomial

is not factorisable over the integers.

Remark 1. The presence of the factor (X − 2) in Theorem 1 is crucial, otherwise the following serves as a counter-example:

is factorisable over the integers. Let f(X)= g(X)h(X) for some two polynomials g and h with integer coefficients. We note that

Consequently, we have

Since g and h are polynomials with integer coefficients, g(2), h(2), g(a₁), h(a₁), . .., g(a_n), h(a_n) are all integers. Therefore, from (4),

Let P(X)= g(X) − h(X). Then P is a polynomial with integer coefficients, and P(2)= P(a_i)= 0 for all

i ∈ {1,. .. , n}. Now,

This means that P(X) is a polynomial of degree < n + 1, having at least n + 1 distinct zeros (namely, 2, a₁,. .. , a_n). Hence P(X) is identically zero, which implies that g(X)= h(X).

Therefore, the relation f(X)= g(X)h(X) becomes f(X)= (g(X))².

Let g(0)= k. Then from the equation f(X)= (g(X))², we get f(0)= (g(0))² = k². That is,

Since the a_i are all odd integers, k is odd. Hence k2 − 1 = (k − 1)(k + 1) is divisible by 8 (as it is the product of two consecutive even integers). This implies that

is divisible by 8, which is not possible as all the a_iare odd.

This contradiction shows that the stated factorisation is not possible.

This completes the proof of Theorem 1.

Reference

[1] A. Engel, Problem-Solving Strategies, Springer.

About the author

Siddhartha Sankar Chattopadhyay is a retired Mathematics teacher from Bidhannagar Government High School, Salt Lake, Kolkata. He is actively associated with the Kolkata-based ​‘Association for Improvement of Mathematics Teaching’ for 30 years.

He has participated in seminars and workshops for school students organised by Jagadis Bose National Science Talent Search over the past 15 years as a resource person, and has been engaged in spreading Mathematics Education across West Bengal by organising seminars and workshops.

He may be contacted at 1959ssc@​gmail.​com

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