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This question shows how dividing both sides of an equation by some $f(x)$ may eliminate some solutions, namely $f(x)=0$. Naturally, all examples admit $f(x)=0$ as a solution to prove the point.

I tried to find a simple example of an equation that could be solved by dividing both sides by some $f(x)$ but where $f(x)=0$ was not a solution, and failed miserably. Sure, I can divide both sides of, let's say, $x^2-1=0$ by $x$ ($x=0$ is not a solution), but that doesn't help me solve the equation.

I started wondering if actually the equations that can be solved (or at least simplified) by dividing both sides by some $f(x)$ were precisely those where $f(x)=0$ is a solution: by removing a solution, the division reduces the equation to a simpler form. This is particularly obvious with this example given in that question's accepted answer:

$$(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)=0.$$

By successively dividing by $x-1$, $x-2$ and so on, the equation becomes simpler as the solutions are removed, until there's no solution left ($1=0$).

However, both the accepted answer and the quote in the question itself say that $f(x)=0$ may be a solution, which I also understand as it may not be one.

So, are there equations where dividing by some $f(x)$ significantly improves the equation resolution, without $f(x)=0$ being a solution?

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Well, assuming that we are looking for solutions over the reals, we can solve say $$(x^2+1)(x+1)=0$$ by dividing both sides by $x^2+1$ to find that the only solution is $x=-1$. However, we still have to check that $f(x)=x^2+1=0$ has no solutions in the reals (easy to see because $f(x)\ge1$ for every $x$). Is this the sort of thing you are looking for?

For a slightly more complicated example, we can take the equation

$$\sin(2x)=0$$ to be solved on the interval $(0,\pi)$. Since $\sin(2x)=2\sin(x)\cos(x)$, we can divide both sides by $2\sin(x)$ (since $\sin(x)$ is positive on the interval in question) to find the solutions are exactly where $\cos(x)=0$. This would give us exactly one solution at $x=\frac{\pi}{2}$.

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  • $\begingroup$ Thanks, that's exactly what I was looking for! I somehow assumed that f(x)=0 is a solution was an "absolute" statement, while it depends on the field and/or the interval you're working on. I tend to forget that math has more than the real numbers :) $\endgroup$ – OxTaz Jan 13 '15 at 15:19
  • $\begingroup$ @OxTaz Even working within the complex numbers, we still have equations like $(x+1)e^x=0$. (Which may or may not lead you to wonder if there's a set bigger than $\Bbb C$ in which $e^x=0$ has a solution. If you try to construct one, you'd need to seriously alter the laws of arithmetic, or else end up with $e^xe^{-x}=1$$\implies$$0e^{-x}=1$$\implies$$0=1$.) $\endgroup$ – Akiva Weinberger Sep 7 '15 at 4:09
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Well, it depends. In general, $g(x)$ divides $f(x)$ if and only if $g(x) = 0$ is a root of the equation. This happens when you are working over any field, because if $g(x)$ divides $f(x)$, then $f(x) = q(x)g(x)$, so $f(x)$ can only be zero by the null factor law if $g(x) = 0$ or $q(x) = 0$. However, $g(x)=0$ may not have any solutions in the field you are working in. Take for example $f(x) = (x^2 + 1)(x-2)$. In this case, $x^2+1$ divides $f(x)$, but it gives no solutions if you only want to find the real roots. If however you are working over an algebraically closed field, like $\mathbb C$, then every factor of $f(x)$ also gives a zero of $f(x)$.

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  • $\begingroup$ +1 for emphasizing the difference between the general case, which seems to confirm my intuition, and the particular field you are working in. $\endgroup$ – OxTaz Jan 13 '15 at 15:24
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Consider this differential equation:

$\frac{df}{dx} = kf$

You'd have to divide both sides by $f$ to get all your $f$ terms on the same side, right?

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