A question about root of a polynomial If we plug in $x$ into a polynomial and we get the value of $0$ as a result, can we be certain that $x$ is the root of the polynomial? 
If that is the case, why in this Wikipedia article, it says that

The factor theorem states that a polynomial $f(x)$  has a factor $(x-k)$ if and only if $f(k)=0$ or that $k$ is a root.

Does it mean that $k$ may not be a root even when $f(k)= 0$?
Thank you!
 A: There is no difference between saying that $f(k)=0$ and $k$ is a root of $f$; the latter is defined as the former.
Looking through previous revisions of the article you'll see that the "or $k$ is a root" bit didn't used to exist at all; someone must have put it there thinking they were being helpful. Maybe better would have been to put "i.e. $k$ is a root" or something like that. [Update: I've edited the article.]
A: If $x_0$ is a solution to $f(x_0) = 0$, so $x_0$ is a root of the polynomial $f(x)$.
If the polynomial has a factor of $(x -k)$ $$f(x) = (x-k)g(x)$$ where $g(x)$ is some polynomial, then clearly when we evaluate $f$ at $k$ we get $$f(k) = (k - k)g(k) = 0g(k) = 0$$ 

The factor theorem is basically telling us that we can factorise any nth degree  polynomial in the form $$F(x) = \prod_{i = 1}^{n} (x - a_i)$$
Where the $a_i$ are the roots of $F(x)$. So $F(x)$ is zero when $x = a_i$ for one of the $a_i$ and further, if $F(k) = 0$ then we have that $k = a_i$ for one of the $a_i$.
For example, you might wish to solve $x^2 + 6x + 5 = 0$ by factorising as follows $$x^2 + 6x + 5 = (x + 3)(x + 2) = 0$$
So the Factor Theorem tells us that the roots of our polynomial, i.e. the solution of our equation, are (only!) $x = -3$ and $x = -2$
A: If $f$ is a polynomial, we can use polynomial long division to write $f(x) = g(x)(x-k)+m$ for some polynomial $g$ and some number $m$.
$f(k) = m$, so $k$ is a root of $f$ if and only if $m=0$, but $m=0$ means that $f(x)=g(x)(x-k)$ and hence $f$ has a factor $(x-k)$.
