Why can't we multiply both sides of $x-1=0$ by $x$ to create another solution $x=0$? Consider the equation $x-1=0$. The obvious solution is $x=1$. However, what is stopping us from creating more solutions by multiplying both sides by arbitrary values? For example:
$$ \begin{align}
x-1 & = 0 \\
x\cdot(x-1) &= 0\cdot x \\
x\cdot(x-1) &= 0 \end{align} $$
If we substitute $x=0$ into this equation now, it gives us a solution. But 0 is not a solution to $x-1=0$. Does multiplying both sides by $x$ imply that there is a restriction $x\neq0$ or something? Why?
We could(?) go further with this, for example, multiply by $(x-2)$ giving $(x-2)(x-1)=0$, meaning $x=2$ is now an (incorrect) solution.
 A: Taking your question, every one of your steps is valid.
$$\begin{equation}x - 1 = 0\end{equation}$$
has one solution, $x = 1$, and
$$x(x-1) = 0$$
has two solutions, $x = 0$ and $x = 1$.
Where this goes wrong is in assuming that this holds for the original equation: you are implicitly saying
$$\begin{aligned}0(0 - 1) &= 0\\
\therefore \, 0-1&=\frac{0}{0} =0\end{aligned}$$
However, this fails to hold, as division by zero is undefined. It is this undefined division by zero which lets you accomplish this for any polynomial $p(x)$, to which you wish to add the solution $x = c$:
$$\begin{aligned} p(x) &= 0\\
(x -c)p(x) &= 0\\
(c - c)p(c)&=0\\
p(c) &= \frac{0}{c-c} = \frac{0}{0} \neq 0\end{aligned}$$
A: The point is that the set of solutions to an equation does not change if you apply a one-to-one function to both sides of the equation. In going from $A = B$ to $f(A) = f(B)$ in the solution process the logic always goes in reverse: if $x$ is a solution of $f(A) = f(B)$, then $x$ must be a solution of $A=B$.  Multiplication by a nonzero number is one-to-one: if $ab = ad$ and $a \ne 0$ then $b = d$.  But multiplication by $0$ is not one-to-one, as $0 \times (anything) = 0 \times (any\ other\ thing)$.  So when you
multiply both sides of your equation by something that could be $0$, you can introduce spurious solutions, as in your example.
A: Given functions $~f(x), g(x),~$ for a particular value $~x_0,~$ assume that $f$ and $g$ are well defined at $~x_0.$
Then,
$$f(x_0) = 0 \implies \left[f(x_0) \times g(x_0)\right] = 0.$$
However, the converse is not generally true.
That is,
$$\left[f(x_0) \times g(x_0)\right] = 0 ~~\text{does not imply}~~ f(x_0) = 0.$$
A: We have two equations that $x$ must satisfy:

The equation you obtained, $x(x-1)=0$
AND
The given equation, $x-1=0$

So $x=1$ is the only solution.
