Explanations for why someone cannot divide by $x-4$ for $x(x-4)=x(x-4)(x-5)$ A student divides both sides by $x-4$ and lost a solution $x=4$. How could you explain to the student that they are not allowed to divide by $x-4$
Here is the problem:

$x(x-4)=x(x-4)(x-5)$

I am having a hard time putting this in words for some reason. We know if the student divides by $x-4$ on both sides, not only do they lose a solution but technically they are dividing by $0$.
Does anyone have any other explanations
 A: You can divide by $x-4$ provided $x\neq 4$. For $x=4$ you need an another case.
The thinking goes as follows: I can divide by a number not equal to zero. I divide by $x-4$. When $x-4\neq0$ I can do that. When $x-4=0$ I check if $x=4$ is a solution.

Eventually you can show them this method
$$x(x-4)=x(x-4)(x-5)$$
$$x(x-4)-x(x-4)(x-5)=0$$
$$x(x-4)(1-(x-5))=0$$
$$-x(x-4)(x-6)=0\implies x\in\{0,4,6\}$$
A: They can divide by $x-4$. But the rest of their work following that division must include the restriction that $x \neq 4$:
$$x(x-4) = x(x-4)(x-5) \\
\implies x = x(x-5); x \neq 4 \\
\vdots
$$
You lose a solution because $x=4$ is a solution (by inspection) and $x \neq 4$ is our restriction in this line of work.
A: The actual process of solving equations is simply a chain of biconditionals. 
For example, $3x^2 + 3x = 9x \Longleftrightarrow 3x^2 - 6x = 0  \Longleftrightarrow 3x(x - 2) = 0 \Longleftrightarrow x = 2, 0$
In your example, the "chain of biconditionals" is broken. 
It is not the case that $x(x-4)=x(x-4)(x-5)  \Longleftrightarrow x = x(x-5)$. This fails for $x=4$, because division by zero is not permitted. 
For the biconditional to hold, we must alter as follows: $x(x-4)=x(x-4)(x-5) \land x \neq 4 \Longleftrightarrow x = x(x-5)$
For the same reason, squaring both sides of an equation often leads to imprecise results, because the biconditional does not hold (the square function is not injective). 
A: Reorganizing things to one side, you have:
$$x(x-4)-x(x-4)(x-5)=0$$
Factoring yields:
$$x(x-4)(1-x+5)=0$$
If you have $ab=0$ this is true if and only if $a=0$ or $b=0$.
More generally, if $a_1a_2a_3\dots a_n=0$ this is true if and only if at least one of $a_1,a_2,a_3,\dots$ are zero.
So, the expression has solutions $x=0$ or $(x-4)=0$ or $(1-x+5)=0$
Simplifying, we have solutions $0,4,6$
A: Before dividing by $x-4$, they should split the search for solutions
into two parts: (i) $x-4 = 0$ and (ii) $x-4 \neq 0$.
In the first case, we quickly check that $x=4$ is a solution and add that
to our bag.
In the second case, since $x \neq 4$ we can happily divide across by $x-4$ to get $x = x (x-5)$ and continue the search (keeping in mind that we now know that $x \neq 4$, not that it matters here).
