this came up in class yesterday and I feel like my explanation could have been more clear/rigorous. The students were given the task of finding the zeros of the following equation $$6x^2 = 12x$$ and one of the students did $$\frac{6x^2}{6x}=\frac{12x}{6x}$$ $$x = 2$$ which is a valid solution but this method eliminates the other solution of $$ x = 0$$ When the student brought it up, I explained to the student that if $x = 0, 2$ and we divide by $6x$ there is a possibility that we would be dividing by 0 which is undefined. The student, very reasonably, responded "Well, obviously I didn't know that zero was an answer when I was doing the problem". The student understands why we can't divide by 0 but is still struggling with how that connects to dividing by $x$. I went on to explain that by dividing by $x$ you are "dropping a solution" because the problem, which was quadratic, is now linear. Again, this didn't seem to click with the student. Does anyone have maybe an axiom/law/theorem that I can show the student to give a rigorous reason as to why you can't just divide by $x$?

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    $\begingroup$ Because the best way to solve this is to factor: $6x^2 - 12x = x(6x-12) =0$. Since the product of two numbers is zero if and only if at least one of them is zero, you check for both possibilities. This is much better than dividing by things you don't know. :) $\endgroup$ – Mark Fantini Jan 16 '15 at 16:36
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    $\begingroup$ I have the same issue when helping student with trig manipulations. When we divide by Sin(x), do we have to note that zero degrees is not allowed, and recheck along the way? The question is great, and it opens the door to many other similar situations. $\endgroup$ – JTP - Apologise to Monica Jan 16 '15 at 17:33
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    $\begingroup$ @Richard - actually I think it is a mistake to push them toward that method as a way of avoiding this error. It makes them inflexible. You want to graduate algebra students who can handle (say) the equation $x-2=2(x-2)$ either by getting everything to one side or by reasoning that since division by $x-2$ leads to the absurdity that $1=2$, it must have been that $x-2=0$. (Or in any other logically sound way.) Students who habitually push everything to one side before thinking about the specifics are at a disadvantage in many situations where another move is more expedient. $\endgroup$ – benblumsmith Jan 18 '15 at 17:34
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    $\begingroup$ @MarkFantini - a propos of what I said to Richard, I disagree that factoring is the best solution. It's certainly a fine solution, but the solution suggested by Aeryk below is just as logically sound and for many people more natural. Factoring is great, but it rubs me wrong to push students toward it as a way to avoid making them have to think about the subtlety that the OP brings up (and that Aeryk and Henry Towsner beautifully address). $\endgroup$ – benblumsmith Jan 18 '15 at 17:39
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    $\begingroup$ @Richard - This probably isn't a matter that we can resolve in this format, since it gets into much bigger questions, but I would like to register my continued disagreement. I used to teach students of that level (9th grade in the US) and at the time I did it similarly - i.e. encouraged them to take particular approaches that I knew avoided certain difficulties - but now I think that was a mistake. I've come to believe that teaching students a rigid version of algebra that prescribes particular solution methods is fundamentally misleading them about the nature of algebra. $\endgroup$ – benblumsmith Apr 3 '15 at 12:24

Just before dividing, you can reason "Either $x=0$ or I can divide by $x$." This creates two separate cases to be analyzed.

This works for dividing by anything. You want to divide by $\sin(x)$? You need to make two cases: $\sin(x) \neq 0$ and $\sin(x)=0$. And then analyze each independently.


I think your student pointed out the key issue: "Well, obviously I didn't know that zero was an answer when I was doing the problem". That's exactly right: at the time the student was dividing, they didn't know whether or not x is 0, which means they didn't know whether or not it made sense to divide by x.

If you know you can't divide by 0, you know not to divide by things which might be 0, because you don't know whether or not it's vaild.


Most experienced mathematicians will immediately identify $x = 0$ as a solution to this equation, and then do the same division (factoring) as your student in order to obtain the other solution at $x=2$. The only difference is that the experienced mathematician has already done due diligence with respect to the possibility of erasing a root/dividing by zero.

I am all for dividing both sides by $6x$ here, with the caveat that when dividing by an unknown quantity you are responsible for first checking whether that unknown quantity is (or might be) zero. Your student is most of the way there - all that's left is to push the conclusion C from information A and B which they already seem to have.

A) I don't know what $x$ is (or $6x$, as is the case here)
B) I'm 'not allowed' to divide by zero

C) I don't know whether or not I'm allowed to divide by $x$ (because it might be zero)

There are useful things to think about in this neighborhood as well. A similar situation where the variable-including-divisor being equal to zero represents the only solution, then we end up producing an absurdity. Eg, $$6x = 4x$$ $$\frac{6x}{x} = \frac{4x}{x}$$ $$6 = 4$$ If we trust our teacher, or the textbook, to have provided us with a logically sound equation in the first place, then the absurdity reached here tells us that we've broken some rules in our working with it. Since the only thing we did was to divide by $x$, we can guess that dividing by $x$ introduced the absurdity. Since we know that dividing both sides of an equation by a quantity is fair game for any value other zero, we now know that $x=0$, etc...

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    $\begingroup$ I strongly agree that it is entirely good to cultivate the intuition of dividing/cancelling... but/and to learn that there can be "gotchas" at the end, requiring some mop-up. But to see that possibility of simplification is very good, and should not be discouraged at all! :) $\endgroup$ – paul garrett Jan 17 '15 at 16:34

I think it is important to emphasize Mark Fantini's remark: factoring is different from dividing, and factoring is the way to get a complete solution. I would also suggest that the problem be rearranged and then factored to give 6x(x-2)=0, or even 2*3*x*(x-2)=0. Then one can see that one of 2 , 3, x, or x-2 must be zero.

When one is comfortable with using the factoring method, then one can consider "shortcuts" by jumping from the given equation to saying "Suppose 6x is not 0: then divide..."

Gerhard "Shortcuts Can Cut One Short" Paseman, 2015.01.16

  • $\begingroup$ You could also contrast dividing by x and dividing by 6. That might help in showing that x^2 = 2x has more than just the solution x=2. Gerhard "Bigger Steps For Bigger Feet" Paseman, 2015.01.16 $\endgroup$ – Gerhard Paseman Jan 16 '15 at 18:18

Firstly, division must obey rules regardless of what you are dividing. If the division is performed over real numbers, then you have to respect the rules for division of real numbers, that is no division by $0$. In other structures $0$ is not the only thing that you cannot "divide".

Secondly, when $x$ is a variable it should be made crystal clear what it refers to. If it can refer to any arbitrary object, then division by $x$ is meaningless. If it can refer to a real number, whether known or unknown, whether zero or not, then division by $x$ must again obey real number rules.

Thirdly, the rules must not be given empty and without meaning. For real numbers, multiplication corresponds to scaling or stretching something, and division is to undo that. If you multiply by zero, you are scaling something until it is gone; zero size. You can never recover the original size, so division by zero is not allowed.

Fourthly, and only after the first three points are made absolutely clear, then you can give some examples to show that indeed division by zero can lead to nonsense.

Fail to solve $x = 2x$ by dividing by $x$!

Fail to solve $x^2 = 0$ by dividing by $x$, twice!

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    $\begingroup$ very succinct explanation! $\endgroup$ – celeriko May 6 '15 at 15:50

Frankly, I think your reasoning is the best. You initially have a quadratic, thus you should expect to find two solutions (although you may not). By dividing by $x$, you get only a single solution: so what is the other? You clearly cannot find it by dividing by $x$, but you can find it by either factoring or using the quadratic equation (which should be discouraged in this case).

Really this is about pattern recognition. The student should recognize when they have:

$$ ax^2 + bx = 0 $$

That they should factor out $x$ to get: $x(ax + b) = 0$ and then set each factor to zero to solve: $x = 0$ and $ax + b = 0 \rightarrow x = -\frac{b}{a}$. The second part is recognizing that when you have something like $ax^2 + bx + c = dx^2 + ex + f$ that they should solve by setting the equation to zero, collecting like terms, and then attempting to factor (using the quadratic equation as a last resort).

If they can understand factoring and that this is the correct approach to solving such equations then they should have less trouble generalizing to higher degree polynomials (i.e. ones that don't have a nice formula like the quadratic formula).

The rationale for this is 1) the fact that polynomial equations, in general, have multiple solutions (so you should either find them or justify why there are fewer) and 2) we really only "know" how to solve linear equations so we must somehow convert our problem into a set of linear equations to be solved and the best way to do that is to attempt to factor (but explain that this is not always possible/"easy").

Granted, dividing by $x$ is one way to get a linear equation, but since we see that it's quadratic we should find two linear equations...

...which could be redundant like:

$$ x^2 = 2x - 1 \\ x^2 - 2x + 1 = 0 \\ (x - 1)^2 = 0 \\ (x - 1)(x - 1) = 0 \\ x - 1 = 0 \\ x - 1 = 0 $$

We got our two linear equations, but they were identical/gave the same solution.


It comes up frequently in solving trigonometric equations such as:


Students often divide by $sin x$. I find myself using this line a lot:

"If you're ever tempted to divide both sides by a variable expression, then what you probably need to do is use addition or subtraction to get everything on one side equal to $0$ on the other side."

Sometimes I explain why, and sometimes I don't. If I'm teaching a class, I will certainly explain why. If I'm tutoring a student who I see for just one hour a week, it depends on whether I feel we have time to go into it.

When I do explain why, I say, like the OP says, that you can't divide by a variable expression because it might be zero, and we can't divide by zero. If a student were to say, as in the OP's experience, "well obviously I didn't know that zero was an answer when I was doing the problem," I might point out that this is a little like saying, "well obviously I didn't know the gun was loaded when I fired it." Any risk of disaster is reason enough for caution.

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    $\begingroup$ This doesn't seem to answer the question in the OP. $\endgroup$ – mweiss May 5 '15 at 3:16
  • $\begingroup$ Good point, @mweiss. I edited the post. $\endgroup$ – Frank Newman May 9 '15 at 11:43

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