"Why $x^2+3xy^2+4xy+7x^2y$ can not be simplified? Why can these terms not be simplified?" I would like an explanation that is understandable by 8th-grade students. The only proof I know (based on linear Independence of powers of any variable) is not elementary at all.

P.S. I don't use "simplification" as a mechanical and by-the-law procedure. The question is deeper. For example "why $x^2+3xy^2+4xy+7x^2y \neq x^3-3xy+9x^2y^3$?". Why can't we reduce the number of mononomials?

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    $\begingroup$ @MichaelE2 Well, when they asked so, I asked them back "Let's simplify it. What's your suggestion?" and disproved their suggestions by giving $x$ and $y$ appropriate values. I'm not looking for a proof. An explanation is enough. $\endgroup$
    – Behzad
    Dec 15, 2015 at 18:44
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    $\begingroup$ Part of the reason this is a hard question to respond to is because your counterexample seems unnecessarily complicated and arbitrary. What is the crux of the problem? Would you be satisfied with an explanation of why $4x + 2x^2$ can't be "simplified" to a single term? Is it important that there be two variables in the expression? If so, would you be satisfied with an explanation of why $4x + 3y$ can't be combined into a single term? Or do there need to be higher-degree combinations? In other words, can you delineate for what kinds of expressions you think students require an explanation? $\endgroup$
    – mweiss
    Dec 15, 2015 at 19:44
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    $\begingroup$ If such a simplification existed for some expression, you could always move all terms to one side of the equation, so your question is equivalent to: "Why is a non-zero polynomial never identical to zero?" $\endgroup$ Dec 16, 2015 at 1:33
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    $\begingroup$ By the way I think this is a great question. Something that is taken as granted in school turns out not to be so obvious. $\endgroup$ Dec 16, 2015 at 1:46
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    $\begingroup$ @DagOskarMadsen Well, it is obvious, but not to a grade schooler who doesn't understand that polynomials form an infinite dimensional vector space whose basis is given by monomial products of the indeterminants- that's the point. $\endgroup$
    – Andrew
    Dec 18, 2015 at 14:01

12 Answers 12


Since you want to negate an implicit universal quantifier, what you have to do is nothing else than showing that there is a counter example to any simplification equality one could come up with. Instead of a proof, at this stage in one's curriculum giving a strong conviction based on a meaningful insight is probably the best possible expectation. The best idea I can come up with is to use one of the rigorous proofs to cook up a counter-example which shows the idea.

For example, take the proof by using asymptotic behavior of monomials as $x$ and $y$ go to infinity. Why $x^2+3xy^2+4xy+7x^2y \neq x^3−3xy+9x^2y^3$? Let's take $x=1$ and $y=1000$. Then clearly $x^2+3xy^2+4xy+7x^2y$ is about $3$ million, while $x^3−3xy+9x^2y^3$ is roughly $9$ billion. After a few cases, the idea appears and a smart kid may be able to construct the counter-examples him or herself, and see why such a simplification cannot happen.

The point is that the proof of the independence of monomial functions is really simple; it is only quite cumbersome to write out precisely.

As a side point, note that this proof relies on a somewhat subtle property of real numbers (their order), for a good reason: the statement fails in finite fields, even for one-variable polynomials. If you don't know about finite fields, let's say they are sets of "numbers" with rules for addition and multiplication that have the same basic properties as for usual numbers, but with only finitely many numbers. The point of this side note is to stress that your desire to prove the non-simplification statement is really sane: this is by no way obvious, and cannot be proved just by applying the usual rules of computation (factorization, distributivity, etc.) which also holds in finite fields.

As a side side note, I think this side note is a good example why it is nice for a teacher to have a much higher education than the level at which one teaches: it helps distinguishing subtle things from trivial ones and gives perspective.

  • $\begingroup$ Thanks. 1. In side note, do you mean something like $2x+3y+5x=3y$ in $\mathbb{}Z_5$ or have a more complicated example in mind? 2. While proving independence of monomials, isn't it necessary to use the fact that every polynomial of one variable has finitely many roots? $\endgroup$
    – Behzad
    Dec 15, 2015 at 21:51
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    $\begingroup$ @Behzad An example would be $x^5=x$ (as functions) in $\mathbb F_5$. $\endgroup$ Dec 16, 2015 at 1:38
  • $\begingroup$ @Behzad wrt 2. I do not see a necessity for this. When there is only one variable $x$, taking $x$ large enough shows independence; if there are two variables $x,y$, then for each value of y, taking $x$ large enough reduces independence to the vanishing of polynomials in $y$ whose coefficients are coefficients from the original linear combination, and that's it. $\endgroup$ Dec 16, 2015 at 15:38
  • $\begingroup$ That proof indeed needs the order of the real numbers, but wouldn't the statement still hold over any field with infinitely many elements? So I would say that is actually the subtle point here. $\endgroup$ Oct 29, 2023 at 20:48

I start by asking them how they would order wall paper or floor tile. "Square Feet". Then we talk about the third dimension, height. This gives us volume as opposed to just area.

I take a step back and ask what the distance is to school. Is its units 'square feet'? Of course not. It's measured in feet or miles.

I then ask how they'd add 12 feet to 9 square feet. At that point they start to see the question is absurd. My 2000 sq ft house and the 2000 ft I walk to the end of my street can't be combined in any way. But if your house is 2500 sq ft, I can add them, and say that combined, we'd have 4500 sq ft of living space.

It's then a small move to explain how X, x^2, and x^3 represent length, area and volume.


The "high level" answer is that a polynomial algebra is a free commutative algebra generated by the indeterminates. But this would probably not be a very satisfying answer to 8th graders.

A possible discussion starts with the cliche that "you cannot add apples and oranges." And you cannot "add lengths and weights." But you can multiply them! After all, we do use "foot$*$pound" as a unit in physics (at least in the US), and so we could just as well talk about "apple$*$orange" if we wanted to. So we can combine $x$ and $y$ into a single unit $x*y=xy$. But $x+y$ is just kind of stuck out there with no place to go.


Let's say we have this expression: $$2 + 3 + x + 2x + 2x^2 + 2x^2$$

First of all, we know that we can combine $2 + 3$ as $5$, but let's take a closer look.

We're going to break the $2$ and the $3$ up into smaller pieces. To make things easy, let's do this so that all of the pieces that we wind up with are exactly the same size: $1$ is an easy number to work with, so that's the size that our pieces should be.

The $2$ breaks up in to

$$1 + 1$$

and the $3$ breaks up in to

$$1 + 1 + 1$$

If we put these groups of pieces together and get ready to add them, we have this:

$$1 + 1 + 1 + 1 + 1$$

I count five $1$'s, which means that our

$$2 + 3$$

which became

$$1 + 1 + 1 + 1 + 1$$

is equal to $5$.

If you take things down to the most basic level, this is how addition works. You break things up into units and then you count up the number of those units that you have.

Now let's do the same thing with $x + 2x$, but this time our smallest piece will be $x$.

$$x + 2x$$


$$x + x + x$$

I count three $x$'s, and we write that result as $3x$.

Next up is our $2x^2 + 2x^2$. Our pieces are going to be $x^2$ for this part.

$$2x^2 + 2x^2$$


$$x^2 + x^2 + x^2 + x^2$$

I count four $x^2$'s, and we write that result at $4x^2$.

Let's put our parts together...

$$5 + 3x + 4x^2$$

Can we combine, or add together, $3x$ and $4x^2$? In order for it to work, we'd need to be able to break each of them up into some identical parts, or units, and then count how many units that we have.

We've seen that we can break the second thing up into $x$'s. We've seen that we can break the third thing up into $x^2$'s. Unfortunately, these units don't match. Maybe we can keep breaking things up until we get units that match.

I can break $x$ up into

$\frac 12 x + \frac 12 x$

and I can break $x^2$ up into

$\frac 12 x^2 + \frac 12 x^2$

but we're still not going to find a match.

We could break these things up in to any size parts that we want, meaning that we could choose any unit size that we want to break up these terms in to, but we'll never wind up with a unit from $3x$ that matches a unit from $4x^2$. If we can't do that, we can't add them together.

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    $\begingroup$ There must be something more going on, since the statement is false in finite fields. See Benoît Kloeckner's answer. $\endgroup$ Dec 16, 2015 at 12:38
  • $\begingroup$ Finite fields? I thought OP wanted to help frustrated eight graders with basic algebra. Most of these answers seem like they belong in math.stackexchange, not matheducators.stackexchange. $\endgroup$
    – Jason
    Dec 16, 2015 at 14:02
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    $\begingroup$ I am only saying that some of the simplest arguments cannot be logically sound since they are contradicted in other (admittedly more abstract) mathematical settings. $\endgroup$ Dec 16, 2015 at 14:33
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    $\begingroup$ The reasoning proposed here is not invalidated by finite fields, as it uses division by integers which may be zero in a given finite field. However, this answer really assume that $x$ and $x^2$ are fundamentally not interchangeable, which is basically what is asked to be explained. Also, the "breaking into parts" point of view may provoke misconceptions, as not all numbers are rationals. If the kid has been exposed to non-integer square roots, or is to be exposed to them, it may conflict with what he or she will be or has been told. $\endgroup$ Dec 16, 2015 at 15:44
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    $\begingroup$ To make my criticism of this answer clearer: a good answer should make clear the difference between $x+x^2$ not simplifying, and $\sin^2 x + \cos^2 x$ simplifying to $1$. Short of this, the core point has not been explained. $\endgroup$ Dec 16, 2015 at 15:49

We can "simplify" the sum of two or more terms, combining them into single term, if they are each numerical multiples of the same algebraic expression.

Example 1

We can simplify $2x^2y + 5x^2y$ because $2x^2y$ and $5x^2y$ are numerical multiples of the same algebraic expression, namely $x^2y$.

For any given combination of values for $x$ and $y$, you will always get the same result for $x^2y$. If $x=2$ and $y=3$, you will always get $x^2y= 12$. If $x=3$ and $y=7$, we will always get $x^2y=63$. So, for whatever values $x$ and $y$ may have, $x^2y$ will be a common factor in the expression $2x^2y + 5x^2y$.

Therefore, we can write:

$2x^2y + 5x^2y=x^2y(2+5)=(2+5)x^2y= 7x^2y$

Example 2

We cannot simplify $2x^2y + 5x^3y$ because $2x^2y$ and $5x^3y$ are not numerical multiples of the same algebraic expression. Notice that the exponents are different.

If $x=2$ and $y=1$, we have different values for $x^2y$ and $x^3y$, namely $x^2y=4$ and $x^3y=8$. So, there is no common factor in the expression $2x^2y + 5x^3y$.

  • $\begingroup$ Thanks, but that's not the point! The question is not the law or mechanical procedure of simplification. $\endgroup$
    – Behzad
    Dec 15, 2015 at 18:53
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    $\begingroup$ Do they understand common factors? In expressions that can be simplified in this way, the common algebraic expression is like a common factor. You would be using the distribution rule in reverse: $ax + bx = x(a+b)$. If $a$ and $b$ are numbers, we can calculate their sum, say $c$. Then we would have $ax+bx=cx$. In the example you gave, the algebraic expressions are all different, i.e. they need not always have the same value. $\endgroup$
    – user6104
    Dec 15, 2015 at 19:06
  • $\begingroup$ I understand and appreciate your answer, but that does not solve the problem. Please read the "P.S." in the post. $\endgroup$
    – Behzad
    Dec 15, 2015 at 19:15
  • $\begingroup$ Expanding on your counter-example idea then, when we say that $x^2+3xy^2+4xy+7x^2y = x^3-3xy+9x^2y^3$, we mean that this is true for every combination of $x$ and $y$. It is true for $x=0$ and $y=0$, but it is false for $x=2$ and $y=0$. So, this equation is not always true. Then explain that identical algebraic expressions in $x$ and $y$ will always give the same value for every combination of $x$ and $y$, and can therefore be used as a common factor. $\endgroup$
    – user6104
    Dec 15, 2015 at 19:54
  • $\begingroup$ See edit to my answer. $\endgroup$
    – user6104
    Dec 15, 2015 at 20:05

The main reason why you can or can't do things to algebraic expressions is because you can or can't do those things with numerical expressions. The rules for manipulating algebraic expressions are exactly the same rules for manipulating numeric expressions, with some extra restrictions introduced by the fact that you don't know what some of the numbers are (or that they represent all possible numbers).

I suggest talking through the ways that you can rearrange expressions made entirely of numbers without changing their values. A good start is to talk through different strategies to calculate expressions where "like terms" will make it easier to do the calculation, such as $3 + 2\times3 + 3\times7$.

Then you can give students an expression and ask them to rearrange it so that the "answer" is the same. Get many different students to talk through their rearrangement and why it is the same. After that, give them several different expressions which are similar and ask them to decide if they have the same answer without actually calculating the final result.

Then you could take the same activities above and replace copies of the same number with a letter and ask the same question. This might help to explain what you can and can't do with algebraic expressions, including the inability to add unlike terms.


I am going to disagree (slightly) with Benoit Kloeckner's observation that the OP's implicit claim is false in finite fields. If we are going to allow ourselves the sophistication to look at fields other than $\mathbb{R}$, we should also oblige ourselves to distinguish between a polynomial and a polynomial function.

Give any ring $R$, a polynomial in $R$ is an element of the ring $R[x]$. A polynomial can be written uniquely as a finite sum of terms of the form $a_n x^n$, where $a_n \in R$ and $x$ is a formal variable. A polynomial function is what you get when you interpret a polynomial as a function on $R$ in the obvious way. A polynomial is an expression, but a polynomial function is (at the formal level) a set of ordered pairs. Every polynomial naturally induces a polynomial function, but they remain different kinds of objects.

When we talk about "simplifying polynomials" we are referring to operations in $R[x]$. For example, the fact that $(x-1)(x+1)(x+2)$ can be simplified to $x^3+2x^2-x-2$ is a consequence of the way the ring operations are defined in $R[x]$ (assuming $R$ is $\mathbb{Z}$, $\mathbb{R}$, or some other ring in which the symbols $1$ and $2$ have a natural interpretation). The fact that $x^3+2x^2-x-2$ cannot be further reduced to an expression with fewer terms is also true in $R[x]$ for any such $R$. Now, it turns out (surprisingly!) that if $R=\mathbb{Z}/(3)$ this polynomial induces the exact same function as does $2x^2-2$.

But (and here is where my disagreement with Benoit lies) despite the fact that $x^3+2x^2-x-2$ and $2x^2-2$ are identical when considered as functions over $\mathbb{Z}/(3)$, they are still different polynomials. They have different degrees, different number of terms, and different leading coefficients, and they generate different ideals.

So while it is true that $t^3+2t^2-t-2 = t^2-2$ may be true for all $t \in \mathbb{Z}/(3)$, I would not describe the replacement of the left-hand side by the right-hand side as "simplification".

Having gone through all of these preliminaries, let's go back to the OP's original question, which we can now tease apart into two different questions:

  1. Why is it that "unlike terms" in a polynomial can't be combined into a single term?

  2. Why is it that when working over the real numbers, two different polynomials never induce the same function?

To answer the first question, I would turn it back onto the asker (or his students): Why do you think they should be combined? What do you think they should be combined into? The fact that like terms (e.g. $5x^2y + 3 x^2y$) can be combined (continuing the example, into $8x^2y$) depends on the distributive property, but the OP does not want an explanation that has to do with formal properties. So that means we first need an informal explanation for why you can combine like terms; then we can try to see why the same reasoning does not work for unlike terms.

Informally, "combining like terms" (as in our example $5x^2y = 3x^2y = 8 x^2y$) can be explained by noting that five things plus three things equals eight things, regardless of what the 'things' are. Five bags of 100 marbles, plus three bags of 100 marbles, add up to 8 bags of 100 marbles. Five 12-packs of bottled water plus three 12-packs of bottled water add up to eight 12-packs of bottled water. In this case, the things are expressions of the form $x^2y$. If you have five of them and three of them, then you have eight of them.

But if you have unlike things -- say, five bags of 100 marbles and three 12-packs of bottled water -- what can you combine them into? There are eight "somethings", but the "somethings" cannot be given a simple name. Likewise, if you want to combine $5x^2y+3xy^2$, there may be a strong temptation to want to say that there are 8 "somethings", but how can you say what the "somethings" are?

That, at least, is how I would informally address the formal question of why unlike terms in a polynomial cannot be combined.

Which brings us to the second question: Why is it that two different polynomials over the reals never induce the same function? As has been noted, this is a rather special property of the reals. It is false when working over any finite ring; it is true when working over any infinite ring that is also an integral domain. The most general conditions under which this property holds are given in the answers to this question on MathOverflow.

  • $\begingroup$ Well but for abstract polynomials the answer to your question 1 is "because they are defined that way" and if one asks for a motivation for the definition I think one of the strongest ones is the old school algebra over real numbers, where the distinction between polynomials and polynomial functions is redundant. $\endgroup$ Oct 29, 2023 at 20:55

Thank you for clarifying that by "simplify", you mean "reduce the number of mononomials".

It seems that you are defining "like terms" to have enough in common that you could use the distributive property to combine them into a single mononomial. Thus, the answer is obvious:

"Unlike terms" do not have enough in common to use the distributive property to factor out all the variables, and combine the constants into a single constant expression. Thus, you cannot combine two unlike terms into a single mononomial.

  • $\begingroup$ No, that's not the point. I don't use "simplification" as a mechanical and by-the-law process. The question is "Why $x^2+3xy^2+4xy+7x^2y \neq x^3-3xy+9x^2y^3$?" (for example!). Why can't we do "anything" to this polynomial? $\endgroup$
    – Behzad
    Dec 15, 2015 at 18:49

To the "Why is $...\neq ...$": just find a counter example.

The simplification you are referring to seems to be the distibutive law " backwards". Do your students know this already? Without this law, you probably have to go with the mechanical approach.

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    $\begingroup$ Note that the non-simplification cannot be reduced to the distributive law and its companion (see the side note in my answer). $\endgroup$ Dec 15, 2015 at 21:08

Suppose $x$ and $y$ represent lengths, then terms can represent geometric shapes and explain why unlike terms cannot be added together.

What do you get when you add a square and two more identical squares? Three squares. So $x^2+2x^2=3x^2$

What do you get when you add a square and a rectangle? A Shape that cannot necessarily be defined in terms of either the original shapes. And so $x^2+xy$ cannot be simplified into one term. Although in this case it could make a new rectangle with length $x$ and width $x+y$, so $x(x+y)$.

What do you get when you add a square ($x^2$) plus three rectangular prisms ($xy^2$) plus four rectangles ($xy$) plus 7 different rectangular prisms ($x^2y$)?


James Brennan explains Simplifying Algebraic Expressions in the following way:

By “simplifying” an algebraic expression, we mean writing it in the most compact or efficient manner, without changing the value of the expression. This mainly involves collecting like terms, which means that we add together anything that can be added together. The rule here is that only like terms can be added together. Like terms are those terms which contain the same powers of same variables. They can have different coefficients, but that is the only difference.

Examples: $3x$, $x$, and $–2x$ are like terms.

$2x^2$, $–5x^2$, and are like terms.

$xy^2$, $3y^2 x$, and $3xy^2$ are like terms.

$xy^2$ and $x^2 y$ are NOT like terms, because the same variable is not raised to the same power.

Combining Like terms

Combining like terms is permitted because of the distributive law. For example, $3x^2 + 5x^2 = (3 + 5)x^2 = 8x^2$

What happened here is that the distributive law was used in reverse—we “undistributed” a common factor of x2 from each term. The way to think about this operation is that if you have three x-squareds, and then you get five more x-squareds, you will then have eight x-squareds.

Example: $x^2 + 2x + 3x^2 + 2 + 4x + 7$
Starting with the highest power of $x$, we see that there are four x-squareds in all $(1x^2 + 3x^2)$. Then we collect the first powers of $x$, and see that there are six of them $(2x + 4x)$. The only thing left is the constants $2 + 7 = 9$. Putting this all together we get

$ x^2 + 2x + 3x^2 + 2 + 4x + 7 = 4x^2 + 6x + 9$

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    $\begingroup$ This entire answer is plagiarised from here. $\endgroup$
    – ArtOfCode
    Dec 24, 2015 at 12:53
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    $\begingroup$ While it can be fine and helpful to reproduce content that exists elsewhere on the internet here, one must make sure to attribute it and should also make an effort to see if this reproduction is allowed. Please do not contribute unattributed content; it can cause problems in various ways. $\endgroup$
    – quid
    Dec 24, 2015 at 13:32

If we examine the distributive property; we will see it does not apply to your algebraic expression:


If the terms 'ab' and 'ac' have a common factor of 'a' then the sum of the two terms is "factorised" by the following steps:

  1. Remove the common factor 'a' from each expression ie 'ab and ac' become 'b and c'.

  2. Sum 'b and c' together: b + c and put parentheses to indicate consolidation: (b + c)

  3. multiply the term (b+c) with the "common factor" 'a': a(b+c)

Thus $ab + ac = a(b+c)$.

In summary; the consolidation of the sum of two terms multiplied with their common factor outside the parentheses as opposed to attaching each term with said common factor can be justified using the definition of multiplication:

Let $a = 2; b,c \in \mathbb{R} \mid$

$$2b + 2c = \sum_{i=1}^{2} b + \sum_{i=1}^{2} c = b + b + c + c$$

Apply the commutative property of addition:

$$b + b + c + c = b + c + b + c$$

Apply the associative property of addition:

$$b + c + b + c = (b + c) + (b + c)$$

But we know:

$$2(b + c) = \sum_{i=1}^{2} (b + c) = (b + c) + (b + c)$$


$$2(b + c) = b + c + b + c = 2b + 2c$$

Therefore: $2(b + c) = 2b + 2c$

If you attempt to apply these steps to any two of the terms from your expression you will find that the sums will not consolidate like they do here.


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