Solving equations: reasoning doesn't work backwards?

Question

In doing my (high school) math homework, I came to an issue that doesn't make sense to me. Given an equation $0 = a_1 + a_2x + a_3x^2 + \dots$, we can multiply both sides by $x$ to obtain $0 = a_1x + a_2x^2 + a_3x^3 + \dots$ Since the second equation has a root at $x = 0$, why doesn't the first? Isn't multiplying both sides by $x$ a valid operation? Why don't all polynomials have roots at $0$?

Answer 1

Your title says it all. Some operations work backwards, but others don't! Here's the reason not every polynomial has a root at $0$: to go from $a_1x+a_2x^2+...$ to $a_1+a_2x+...$ you'd divide by $x$, right? But you can't always divide by $x$-why?-because $x$ could be zero, and you can't divide by zero. The right way to solve $a_1x+a_2x^2+...$ is instead to factor, $0=x(a_1+a_2x+...)$, which shows that the zeroes are either 0, or else any zero of $a_1+a_2x+...$.

There are other operations that can't be reversed. Consider the equation $3x=2$. Square both sides, to get $9x^2=4$. This has a solution, of course, at $x=2/3$, but it also has a solution at $x=-2/3$, which isn't a solution of the original equation! The problem here is similar: to undo your squaring, you'd take a square root, but a positive number has two square roots, so you can't possibly get back just your original equation. Many operations can be undone, of course-for instance, multiplying both sides of the equation by a number (other than zero!). These are called invertible operations, and you should always try to keep an eye out for moments when you apply non-invertible operations to your equations, since this is when you might get extra, false, solutions.