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I have come across this question in a textbook

How many factors are there in the term $5ab(x+y)$? State what they are

It is being praised because it encourages thinking, which it does. However, I'm not sure if there is really a clear-cut answer.

One could argue that there are two factors $5ab$ and $(x+y)$. But then I've also imagine arguments for $5$, $a$, $b$ and $(x+y)$ each being factors giving 4 in total. However, the problem with that is that each of $a$ and $b$ could be further decomposed if they are not prime.

  1. What do you think? What definition would you give to your students of what a factor is in light of this question?

They also call $5ab(x+y)$ a term

For me, it looks like a two-term expression.

  1. What definition would you give to your students as to what a term is?
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    $\begingroup$ Does there need to be a clear-cut answer? Personally, I would think that the number of factors we identify here would depend on the context in which the problem is presented. For example, if the problem is to factor a polynomial, I see only one factor ($x+y$)---the fundamental theorem of algebra gives a correspondence between factors and roots, and (as a polynomial in $x$) this thing has only one root. On the other hand, in the context of number theory, we might consider there to be four factors (or more, depending on the primality of the terms!). $\endgroup$ – Xander Henderson Aug 3 at 13:34
  • $\begingroup$ No there doesn't, and that is what I am trying to gauge. However, it is good to see people's opinions of this and how they would explain this to secondary school kids. $\endgroup$ – PhysicsMathsLove Aug 3 at 15:10
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    $\begingroup$ I fear that "here's a question, what definitions would you use for it?" is putting things the wrong way around. Presumably the textbook itself has some kind of definition for "factor", what is it? Definitions can vary between papers, and we should inspect those carefully first. $\endgroup$ – Daniel R. Collins Aug 4 at 5:54
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There are at least 16 factors: anything of the form $5^ia^jb^k(x+y)^l$, where the exponents are all either 0 or 1. You could in fact say that there are infinitely many factors. For instance, I could say that the expression contains a factor of $5/7$, and also a factor of $7$. The only moral I can see to this example is that the normal way we talk about words like "factor" is very general, and if we want to get more specific, we need to use more specific definitions. For example, if you consider the field of polynomials in 4 variables with integer coefficients, then factoring becomes more well defined. If that's the intended moral of the exercise, then I guess that's OK. But that doesn't mean we should demand an exact definition of "factor" for all contexts. There can and should be many different definitions of a general, flexible term like this.

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    $\begingroup$ (+1) In addition to all that you said (and possibly implicitly implied by what you said), this type of question seems to serve no purpose in my opinion. I seriously doubt anything is gained from having students study and work "problems" like this. Of course, if this is intended as an open-ended type of question designed to promote the kind of discussion we're having, then it's fine, but in that case simply giving the question without this additional context is not fair play. $\endgroup$ – Dave L Renfro Aug 3 at 21:16
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    $\begingroup$ This was the first question in a textbook. I don't not particularly like it, but wondered if I was just unaware of a standard definition or just being annoying. There are teachers on Twitter explaining how great of a question it is and I think it is quite flawed in that it demands a static definition/classification of which of these are factors, when really classification should be decided by the context in which the problem has been set. $\endgroup$ – PhysicsMathsLove Aug 3 at 22:08
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I agree that it isn't a term. A term is a monomial and this can be simplified to 2 monomials. Certainly if I was collecting like terms, I would simplify this first.

As for factors, I would count: 5, a, b, and (x+y). The reason is that if I was looking for a common factor (if this was part of a long expression with other terms), I might factor out any one of these: 5, a, b, or (x+y). I might also factor out a combination of factors, such as 5a or ab. Since a and b are variables I wouldn't worry about their factorizations.

I would be curious to see if others agree.

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[I'm answering your questions in the order I usually introduce these words to beginning algebra students.]

  1. What definition would you give to your students as to what a term is?

In our in-house texts, we teach that terms are:

Parts of expressions separated by addition or subtraction symbols, except those between "grouping symbols" or in implied groups. [We then go on to identify grouping symbols as including all forms of parentheses, () [] {}; radical symbol, $\sqrt{\phantom{x}}$; fraction bar; exponents]

This would make $5ab(x+y)$ a single-term expression, as the addition is happening inside of parentheses.

  1. What definition would you give to your students of what a factor is in light of this question?

We use such problems in our texts, where the intent is for students to identify factors as:

Numbers or variables separated by multiplication symbols, whether explicitly-written (e.g. $\ast$, $\cdot$, $\times$) or implied (e.g. concatenation or integer exponents)

This would give four factors in your expression: $$5ab(x+y) \implies 5, a, b, x+y$$

This definition (my wording) is given to students immediately before introducing the concept of factoring an expression, such as $5a^3b(c+1)+7a^2b^2(c+1)$. They are taught to list out the factors in each term (there are two terms) so they can decide what can be factored (usually by expanding exponents and then circling or underlining the factors that match):

$$5a^3b(c+1)+7a^2b^2(c+1)=5\underline{aa}\phantom{.}a\phantom{.}\underline{b}\phantom{.}(\underline{c+1}) + 7\underline{aa}\phantom{.}\underline{b}\phantom{.}b\phantom{.}(\underline{c+1}) = a^2b(c+1)\left(5a+7b\right)$$

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  • $\begingroup$ Seems clunky to me. Why not follow the usual progression of monomials, standard form of monomial, polynomial, standard form of polynomial. Basically, you have a thing (polynomial, like a big organic molecule with blobs connected to other blobs, and when you break it, you break between blobs) that you try to break apart, where does it break (yes, where + and - are, but your definition is overly complicated and cumbersome in my opinion). So far Amy B's explanation is the simplest: a term is a monomial, it is made of factors. Counting them is less important than being able to manipulate them. $\endgroup$ – Rusty Core Aug 4 at 1:50
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    $\begingroup$ “ but your definition is overly complicated and cumbersome in my opinion” That’s cool with me — I was paraphrasing from a book I didn’t write. $\endgroup$ – Nick C Aug 4 at 1:55
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tl;dr Probably best to focus on having students draw abstract-syntax-trees. That said, an expression's terms are the addends of the top-level summation; if the explicit top-level operator isn't a summation, then we regard it as a single-addend summation, where the entire explicit expression is the only addend, and thus the only term.


Students might be better off focusing on abstract-syntax-trees.

Probably best to refocus the question on drawing the abstract syntax tree for the expression instead.

There are multiple correct answers. For example,

  *
 / \
5   *
   / \
  a   *
     / \
    b   +
       / \
      x   y

, which focuses on binary operators, or

   *
  / \
 /| |\
5 a b +
     / \
    x   y

, where there's a single, quaternary (or $n\text{-ary})$ product operator describable as a folding of the root binary operator from the first example with its subordinate product operators.

There're two most correct answers to this question:

  1. The factors are $\left\{5, \, ab\left(x+y\right)\right\}.$

  2. The factors are $\left\{5, \, a, \, b, \, \left(x+y\right)\right\}.$

The first answer's a tad more technical, as it avoids assuming properties like commutativity and associativity, making it more general if we're constructing math. The second answer'd probably be preferred if students are to assume commutativity and associativity.

Regardless, as long as students can demonstrate an understanding of the abstract-syntax-tree behind an expression, that'd seem sufficient. No need to push them toward adopting unnecessary notions of what a "term" ought to be taken as in this context.

Overall, it looks like a good question once we fix it up a bit.



Definition of "term".

As explained above, I don't think it's productive to get too focused on terminology. Seems to promote limited, inflexible understandings.

That said:

They also call $5ab(x+y)$ a term

For me, it looks like a two-term expression.

  1. What definition would you give to your students as to what a term is?

When you have any expression, e.g. $`` 5ab \left(x+y\right) " ,$ a term is a component of the top-level summation. In this case, the top-level summation is trivial, and the single term is simply $`` 5ab \left(x+y\right) " .$

In other words, if we draw the abstract-syntax-tree where there's a top-level-summation node, it's:

   +
   |
   *
  / \
 /| |\
5 a b +
     / \
    x   y

, where the the top-level $`` + "\text{-node}$ has a single element. Since the terms are the child nodes of the top-level-summation, that's just

   *
  / \
 /| |\
5 a b +
     / \
    x   y

, which in linear notation is $`` 5ab \left(x+y\right) " .$

This is a well-defined concept, so it's not really ambiguous. And the word "term" is common in math-related texts, so it would be a good thing for students to know.


Procedure for identifying an expression's terms.

To identify the terms in an expression:

  1. Write the expression as an abstract-syntax-tree.

  2. If the expression's top-level node isn't $`` + " ,$ append a higher-level $`` + "\text{-node}$ which has the prior top-level node as a child-element.

  3. The top-level node's child elements are the expression's terms.

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To further complicate matters, note that definitions in two of the answers (so far) either say or at least could be interpreted to say that $5a$ and $5ab$ and $ab$ and $ab(x+y)$ and $b(x+y)$ are also factors. And I've probably used "factor" to also include things like $5b$, for example if I wanted to cancel a factor from the numerator and denominator of a fraction.

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  • $\begingroup$ That is exactly what the definition in the book we use allows for. $\endgroup$ – Nick C Aug 3 at 17:37
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In my experience (German high school teacher):

A factor is an operand in a multiplication and I would identify 5, a, b and (x+y) as factors.

For "term", I'm aiming for a distinction between terms and equations. Thus a quite good definition of a term could be "Any legal sentence of mathematical symbols except equations". Depending on the previous knowledge, relations or other things might have to be excluded.

Another way to express this is that a term is something that describes a calculation in symbolic form.

Perhaps the only unambiguous way (but in my opinion, not helpful for "secondary-education") is to define a term with a context free grammar:

  • A number is a term.
  • A variable is a term.
  • If a and b are terms, a+b is a term.
  • ... [expand to known operations] ...
  • Nothing else is a term.
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    $\begingroup$ Can you clarify your distinction between "term" and "expression"? You surely can't mean that $\sqrt{x^2 + 1}$ is a term and $\sqrt{x}$ is not a term, right? $\endgroup$ – Justin Aug 3 at 20:29
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    $\begingroup$ Echoing @Justin : You seem to be using "term" as a synonym of "expression" and or "subexpression". Would you say "$\int_{a}^{b} \ \,\mathrm{d}t$" is a "term" in "$\int_{a}^{b} \; f(t) \,\mathrm{d}t$"? Would you say "$x \rightarrow 0$" is a "term" in "$\lim_{x \rightarrow 0} f(x)$"? $\endgroup$ – Eric Towers Aug 4 at 14:28
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    $\begingroup$ Yes, I think what you are calling a term is what most people just call an "expression." $\endgroup$ – Justin Aug 4 at 14:36
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    $\begingroup$ I have always learned that a+b is the sum of 2 terms but it is one expression. $\endgroup$ – Amy B Aug 4 at 18:42
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    $\begingroup$ Agreed with Amy B, a+b is not a term, it is a polynomial with two monomials (terms). $\endgroup$ – Rusty Core Aug 4 at 23:26

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