Solved a quartic equation by dividing I was asked to solve: $$x^4+2x^3-22x^2+2x+1 = 0$$
Without using differential calculus (Newton's Method).
My Progress: Dividing by $x^2$, I get:
$$x^2+2x-22+\frac{2}{x}+\frac{1}{x^2} = 0$$
$$x^2 +\frac{1}{x^2} +2x +\frac{2}{x} -22 = 0$$
$$(x+\frac{1}{x})^2 +2(x+\frac{1}{x}) - 24 = 0$$
let $k = (x+\frac{1}{x})$
$$k^2+2k-24 = 0$$
$$(k+6)(k-4) = 0$$
$k = -6$ or $k = 4$
Putting $4 = x + \frac{1}{x} $ leads to
$x^2 − 4x + 1 = 0$ as above, and putting $−6 = x + \frac{1}{x}$ leads to $x^2 + 6x + 1 = 0.$
Never mind the answer that I get after that. The big question that I have is that I divided by $x^2$ at the top. My maths teachers always told me that one should never divide by a variable but I did it and got 4 different roots for this question. Why am I not allowed to divide by a variable and when is it possible that I can divide by a variable? It made me solve this equation.
 A: The reason why, in general, it is considered bad practice to divide by a variable, is because you may lose out on solutions, when whatever you divide by happens to be zero. It is however perfectly valid if you can ensure you are dividing by a nonzero quantity.
For example: solving $x(x-1)=x(2x-3)$. Naively one might divide by $x$ on both sides, and get that $x-1=2x-3\implies x=2$. However they would be implicitly assuming that $x\neq0$ (as you cannot divide by zero). One way to deal with this, for this example, is to bring all terms onto the same side and factor (which is usually prefered), however it would be equally valid to consider two cases instead.
Case 1: $x=0$. In this case, the equation is satisfied, so we have a valid solution.
Case 2: $x\neq0$. In this case, we can divide both sides by $x$, because it is nonzero, and proceed as before.
So in general, if it is inconvenient to factor, you can just consider the two cases when what you want to divide by is zero, and when it is nonzero. In your case, it is pretty easy to check that $x=0$ does not satisfy the equation, so it is perfectly valid to divide by the variable $x$ in this case, without any additional casework.
A: The teacher is half right and half wrong.
First you must check if the constant term in the polynomial equals zero.
If so, then zero is a root and you can factor a general degree of x^n where n is a natural number.
Then you can solve both factors.
If the constant term is not zero, then zero cannot be a root, like in your example where the constant term equals one.
In this case you can divide the equation by the variable without any problems with undefined expressions.
The teacher just wanted to make things simpler for the rest of the class, yet your solution is mathematically valid. Please tell the teacher about what I typed.
