The equation

$$2t^2 + t^3 = t^4$$

is satisfied by $t = 0$

But if you cancel a $t^2$ on both sides, making it $$2 + t = t^2$$ $t = 0$ is no longer a solution.

What gives? I thought nothing really changed, so the same solutions should apply.


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    $\begingroup$ About why the unthinking "cancel" idea is harmful: math.stackexchange.com/questions/1505354/…. $\endgroup$ Nov 6, 2015 at 7:29
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    $\begingroup$ @Martín-BlasPérezPinilla and here. $\endgroup$
    – Eff
    Nov 6, 2015 at 11:43
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    $\begingroup$ The equation $2t=t$ is satisfied by $t=0$. But if you cancel a $t$ on both sides, making it $2=1$ ... $\endgroup$
    – JRN
    Nov 6, 2015 at 13:17
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    $\begingroup$ Of course you may cancel, but that is only if and not iff. $\endgroup$ Nov 6, 2015 at 13:46
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    $\begingroup$ Related: "$1\times0=2\times0$ is true. Canceling the $0$, we get $1=2$, which is no longer true! What gives?" The flaw in this is very similar to yours — you can't divide by $t^2$ if $t=0$. $\endgroup$ Nov 8, 2015 at 22:13

6 Answers 6


When you cancel out $t$ on both sides you are assuming $t\neq0$.

To be rigorous you need to divide into two cases,

  1. $t\neq0$, then you can cancel out $t$.
  2. $t=0$, then you cannot cancel out $t$ as division by $0$ is not allowed.

I think the answer lies in the steps you ignored, that is, the "cancelling out" process:

"Cancelling out" with few extra intermediate steps usually ignored:

$$2t^2+t^3-t^4=0\implies t^2(2+t-t^2)=0$$

$$\implies t^2=0\text{ or }(2+t-t^2)=0$$.


You must guard against division by zero. When you "cancel out $t^2$", you are dividing both sides of your equation by $t^2$. Consequently, the correct form of inference is

$$\begin{align*} 2t^2+t^3 &= t^4 & & \\ 2 + t &= t^2 \qquad \text{ or } \qquad t^2 = 0. \end{align*}$$

The resulting left choice has the solution set $\{-1,2\}$ and the resulting right choice has the solution set $\{0,0\}$, which together are the solution set of the first equation.

Note that the same thing happens in reverse. Multiplying both sides of an equation by zero can result in craziness. We can agree that $1 \neq 2$, but this does not mean $0 \cdot 1 \neq 0 \cdot 2$ because both sides of this are zero. It can be harder to see when one is doing this when using a more complicated expression that is only sometimes zero. For instance,

$$\begin{align*} 2 + t &= t^2 \\ t^2(2 + t) &= t^4 \qquad \text{ and } \qquad t^2 \neq 0 \\ 2t^2+t^3 &= t^4 \qquad \text{ and } \qquad t^2 \neq 0 \end{align*}$$

The left choice has solution set $\{-1,0,0,2\}$ and the right choice has solution set $\{0,0\}$. The (set) difference of these is $\{-1,2\}$, the solution set of the first equation.

None of this is hard to see when multiplying or dividing by constants. Either the constant is nonzero and everything works or the constant is zero and everyone can see that something bogus is going on. However, when we're not just using constants, a little more care is needed.

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    $\begingroup$ +1 for being the only answer to say that cancelling is really just division and it can be dangerous to apply it without understanding that. $\endgroup$
    – wchargin
    Nov 8, 2015 at 22:48

Write your equation as $$t^4 -t^3 - 2t^2 = 0,$$ factoring out $t$ to get $$t^2(t^2 - t - 2) = 0.$$ Factoring once more, we have $$t^2(t- 2)(t+1) = 0$$ so the roots are $-1,0,0,2$.

If you leave out the $t^2$ factor you lose the $0,0$ roots.

  • $\begingroup$ That's a different way of looking at it. Interesting $\endgroup$
    – Cruncher
    Nov 6, 2015 at 16:33
  • $\begingroup$ Yes, this is the way to do it. Put it in standard form first (move the whole polynomial to the left of the equal sign, with the highest-ordered term on the left, then the next-highest-ordered term, etc.), then factor it. $\endgroup$
    – Mathieu K.
    Nov 8, 2015 at 7:39
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    $\begingroup$ And so, basically, you didn't "cancel out" the t²; rather, you divided it from both sides: that is, you factored it out. Factoring out any root leaves an equation that doesn't have that root (as long as it's not a double root like your t = 0 is). For instance, if instead we had factored out (t + 1) to get (t + 1)(t³ - 2t) = 0 and then divided out the (t + 1), t = -1 is no longer a solution of t³ - 2t = 0—plug it in to see for yourself. $\endgroup$
    – Mathieu K.
    Nov 8, 2015 at 7:43
  • $\begingroup$ @Mathieu K - thanks. I got rid of the word "cancel" , factoring out is more insightful. $\endgroup$
    – user186104
    Nov 8, 2015 at 14:14

No such thing as cancelling actually exists. What you do in those moments is to divide both sides of an equation by the same value. This means, cancelling like this:

$2t^2 + t^3 = t^4$ to $2t + t^2 = t^3$

Is actually a shortcut for:

$2t^2 + t^3 = t^4$ to $(2t^2 + t^3)/t = t^4/t$ to $2t + t^2 = t^3$

And you can only do that by assuming $t \neq 0$, so you discard that solution by doing such operation. Perhaps $0$ would still be a valid solution if the exponents are $> 2$ in each term (so $0$ would still be a valid root for them) but you should never assume that.

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    $\begingroup$ Be careful when you say canceling doesn't exist. It most definitely exists when working over any integral domain. (See Wikipedia.) $\endgroup$
    – gebruiker
    Nov 8, 2015 at 22:02

If $ t = 0 $, then dividing both sides by $ t^2 $ would mean dividing by zero.


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