showing expressions are equivalents using roots?

Question

A little background:

Today I saw an answer to a question, in the answer José Carlos Santos said that the roots of $x^5-1$ are $1$, $e^{\pm\frac{2\pi i}5}$, and $e^{\pm\frac{4\pi i}5}$. afterwards he claimed that $$x^5-1=(x-1)\left(x-e^{\frac{2\pi i}5}\right)\left(x-e^{\frac{-2\pi i}5}\right)\left(x-e^{\frac{4\pi i}5}\right)\left(x-e^{\frac{-4\pi i}5}\right)$$this can be checked to be true, i can also see that this method( multiply $(x-$the root) for all roots)is not always true, any function divide by $x$ will have the $x=0$ be undefined, and constructing expression like this create a polynomial hence $x=0$ will exists.

my question is under what condition this method works? and why in those conditions?

I don't understand your objection. That (complex) factoring of $x^5-1$ is entirely correct. Why do you doubt it? — lulu, Jan 01 '18 at 20:15
What's not always true? Any monic polynomial $P(x)$ can be written as the product of $x - r_j$ where $r_j$ are its roots, counted by multiplicity. — Robert Israel, Jan 01 '18 at 20:25
I also don't understand what your question is. Can you clarify? — Allawonder, Jan 01 '18 at 21:02

score 1 · Accepted Answer · answered Jan 01 '18 at 20:51

If you have a polynomial like $x^2-x$ where there is a factor $x$ you can write this as $$x^2-x=(x-0)(x-1)$$ where the roots are $0$ and $1$. Normally we would write $$x^2-x=x(x-1)$$.

If we have $$x(x-1)=0$$we have either $x=0$ or $x=1$, and we don't need to divide by zero (or by $x$ which might be zero). We just use that if the product of two (real, complex) numbers is zero, then one of the original numbers is zero. This works in any context where we don't have"zero-divisors".

If we were working modulo $6$, which has zero divisors, we would have $$3\times 2=6\equiv 0$$ so it is important to check.

showing expressions are equivalents using roots?

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