0
$\begingroup$

Here are three situations in which students might try to apply the factor theorem.

  1. Proving that $x + 1$ is a factor of the polynomial $x^3 + x + 2$ can be done using the factor theorem by showing that the polynomial vanishes when $ x = -1$.

  2. If we want to prove that $a + b$ is a factor of $a^3 + b^3 $, the same approach can be used by replacing $a$ with $-b$.

  3. Can we use the factor theorem to check whether there exist positive integer pairs $(a,b)$ such that $a^2 + b^2 + ab$ is divisible by $a + b$?

In case (1) we can say $x + 1$ is a factor for any value of $x$, and in case (2) also we can say $a + b$ is a factor for any integer pair $(a,b)$. But how can we explain situations when $a$ and $b$ both have the same sign, considering that we substitute $a = - b$ to obtain the factor?

I think in case (3) we cannot replace $a$ by $- b$ because both are positive values, and the factor theorem is totally irrelevant because we need to find particular values for $a$ and $b$.

What would be the importance of showing students this limitation of the factor theorem? How do we help students achieve better understanding of the concepts involved, rather than only being able to answer questions of common standard patterns?

$\endgroup$
5
  • 5
    $\begingroup$ This sounds like a question about mathematics that you are not sure about, rather than a question about teaching maths, and therefore the question belongs on mathematics.se. As for the question itself, I don't really get your misconception (but I think there is one). But I think your question could do with more clarity. $\endgroup$ Commented Aug 11, 2023 at 10:07
  • $\begingroup$ @AdamRubinson may be the way I asked it one can get that kind of feeling. Actually if I want to clear my doubts I have better platforms to raise this issue. Therefore I will edit the question so that there is no need to answer the issue but discuss about the importance of giving this kind of questions which need much understanding of concepts to answer comparatively . $\endgroup$ Commented Aug 11, 2023 at 12:50
  • 4
    $\begingroup$ The confusion in case 3 is similar to thinking that because the polynomial $x^2 + 1$ cannot be factored over the integers, that this must mean that $x^2 + 1$ is prime (since it cannot be factored "for any $x$"). This is very wrong. $\endgroup$ Commented Aug 11, 2023 at 19:33
  • 1
    $\begingroup$ You wrote that in case (3) both are positive, but I see nothing about positivity in the specification of case (3). Did you mean "positive integer" where you wrote "integer"? $\endgroup$ Commented Aug 12, 2023 at 15:31
  • $\begingroup$ @AndreasBlass Thanks for pointing out, I included the term in case (3) $\endgroup$ Commented Aug 12, 2023 at 15:41

1 Answer 1

5
$\begingroup$

An idea with pedagogical importance here is that students often have trouble distinguishing between objects that look identical but are mathematically distinct, and these frequently appear in school algebra when students work with polynomial expressions. These expressions can represent (1) expressions involving arbitrary numbers, (2) formal expressions, and (3) functions. Mathematicians commonly refer to objects of the second type as polynomials and objects of the third type as polynomial functions.

As one example, Steven Gubkin brought up how students might think that the polynomial $x^2+1$ is prime/irreducible. This is true if we consider $x^2+1$ as an element of $\mathbb{R}[x]$, the ring of polynomials in $x$ that have real coefficients, but it is not true if we consider it as an element of $\mathbb{C}[x]$, the ring of polynomials in $x$ that have complex coefficients.

I'll first address your questions about case (2) and case (3) by carefully distinguishing between polynomials and expressions involving arbitrary integers. Anyone who is only interested in the pedagogical discussion can skip the next section.


In your cases (2) and (3), the expressions $a+b$, $a^3+b^3$, and $a^2+ab+b^2$ represent both

  • Polynomials with indeterminate $a$ and constant $b$ (an element of the underlying ring), and
  • Integers given two arbitrary integers $a$ and $b$.

To help distinguish these, instead of $a$ and $b$, I'll use $X$ and $c$ when talking about polynomials, and I'll use $n$ and $m$ when talking about integers.

The factor theorem is a theorem about polynomials. It states that $X-r$ is a factor of the polynomial $P(X)$ if and only if $r$ is a root of $P(X)$, i.e. $P(r)=0$.

Given the polynomial $P(X)=X^3+c^3$, since $P(-c)=0$, we know that $X+c$ is a factor of $X^3+c^3$. This means that $X^3+c^3$ is the product of $X+c$ and some polynomial $Q(X)$. Specifically, $$(X+c)(X^2-cX+c^2)=X^3+c^3.$$

So far, no integers have been involved. Now, if we replace $X$ and $c$ with arbitrary integers $n$ and $m$, then we get a general relationship between integers, $$ (n+m)(n^2-nm+m^2) = n^3 + m^3, $$ which we could also call an identity. This identity tells us that, for any integers $n$ and $m$, $n^3+m^3$ is the product of $n+m$ and some integer, and therefore $n+m$ divides $n^3 + m^3.$

Sometimes zero is not allowed as a divisor, in which case we can restrict our conclusion that $n+m$ divides $n^3+m^3$ to integers $n$ and $m$ such that $n+m\neq0$. There's no need for a similar restriction on the statement that $X+c$ divides $X^3+c^3$, because $X+c$ is not the zero polynomial. However if $n$ and $-m$ have the same value, then $n+m$ is the integer zero.

To your question about case (2), notice that nowhere in this argument did we assume that $n$ and $m$ have different signs. We took a relationship between polynomials that the factor theorem helped us to establish, and we used it to deduce a general relationship between integers. If we wanted to, we could restrict it to a general relationship between positive integers: for any positive integers $n$ and $m$, $n+m$ divides $n^3+m^3$.

In case (3), we are asked to find specific relationships between integers. The factor theorem tells us that, as polynomials, $X+c$ is not a factor of $X^2+cX+c^2$. This means that we cannot use the same argument we used above to deduce a general relationship/identity involving integers. However, there may be non-general integer relationships that have this form. And indeed, $2+2$ divides $2^2+2\cdot2+2^2.$


There's a conversation to be had about when students should be introduced to the level of abstraction at which polynomials and expressions involving arbitrary numbers are seen as distinct objects. I think expressions like $a^3+b^3$ should first be introduced in the context of generalized arithmetic and not formal algebra or functions, which Hung-Hsi Wu makes an argument for in this article. But also, prior to introducing the factor theorem in his textbook Algebra and Geometry, Wu says,

Up to this point, the algebra we have been doing may be called generalized arithmetic, because the symbols employed in all the expressions we have come across all stand for numbers. It follows that all our computations have been with numbers. This kind of algebra is more or less the algebra of the period from al-Khwarizmi (780–850) to 1800. After 1800, the work of Gauss, Abel, Galois, and others slowly transformed algebra into an abstract study of structure. School algebra, at some point, must also give an introduction to such abstract considerations. The study of polynomials provides the most natural venue to showcase the transition from algebra as generalized arithmetic to formal algebra. (p. 175).

I would argue that students need to be aware of this transition in order to flexibly solve problems and construct arguments using the factor theorem, and in turn, the factor theorem motivates the transition. At the secondary level, students don't need the machinery and language of polynomial rings, but they need to have a sense that the statement "the polynomial $x+1$ is a factor of the polynomial $x^3+x+2$" is of a completely different kind than the statement "given an integer $a$, $a+1$ is a factor of $a^3+a+2$."

$\endgroup$
1
  • $\begingroup$ Thank you very much for sensing the importance of this issue in the very beginning and sharing your expert knowledge in order to give a start for a fruitful discussion . I think you have tried your best to cover almost everything stated in my post. $\endgroup$ Commented Aug 14, 2023 at 1:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.