Number of roots in a polynomial over a field I have a polynomial of the form: $f(x)=x^3+a_1x^2+a_2x+a_3$, where $a_1$, $a_2$, $a_3$ are constants in field $K$, e.g. the real numbers, the rational numbers, finite fields etc.
I have now given 2 roots of the polynomial within field $K$; Also I know that for the two roots $r_1$ and $r_2$, we have  $\frac{\partial f}{\partial x}(r_1)\neq 0$ and $\frac{\partial f}{\partial x}(r_2)\neq 0$.
My questions are: Is this knowledge enough to say that neither $r_1$, nor $r_2$ are double roots?
E.g. for the real numbers it obviously holds because if $a$ is a double root I could write the function as
$f(x)=(x-a)^2*g(x)$
and this shows that the derivative $\frac{\partial f}{\partial x}(a)= 0$. But does this hold for any field? And if yes, is there a theorem or how can I see that?
And the second question? Is there a third root within the field $K$ under these circumstances? If yes, how can I see or prove it?
And the last question: Is it possible that there are more than 4 roots within a field? I assume no, but how do I prove that, is there a theorem for that?
I'd be happy about a reply - even if you can just answer one question of it, I'D be very happy about it...
Best,
John
 A: For any field $K$ the polynomial ring $K[x]$ is a Euclidean domain, with the degree as its Euclidean function. Let $f\in K[x]$ and $r\in K$. Then long division of $f$ by $x-r$ shows that there exist $g_r(x),h_r(x)\in K[x]$ such that
$$f(x)=(x-r)\cdot g_r(x)+h_r(x),$$
with $\deg h_r(x)<\deg(x-r)$, so $h_r(x)$ is constant. Plugging in $x=r$ shows that
$$f(r)=(r-r)\cdot g_r(r)+h_r(r)=h_r(r),$$
so if $r$ is a root of $f$ then $h_r(r)=f(r)=0$ and hence $f(x)=(x-r)\cdot g_r(x)$.
This shows that if $r_1$ and $r_2$ are distinct roots of $f(x)=x^3+a_1x^2+a_2x+a_3$, then
$$f(x)=(x-r_1)(x-r_2)g(x),$$
and comparing degrees shows that $\deg g=1$, so $g=x-r_3$ for some $r_3\in K$. If $r_3=r_1$ then
$$f(x)=(x-r_1)^2(x-r_2)\qquad\text{ so }\qquad f'(r_1)=0,$$
contradicting your assumption that $f'(r_1)\neq0$. Similarly $r_3\neq r_2$, so indeed $f$ does not have a double root.
This also shows that $f$ has no other roots; if $r\in K$ is a root of $f$ then $x-r$ divides $f$. We just saw that $f(x)=(x-r_1)(x-r_2)(x-r_3)$, so $r=r_i$ for some $i\in\{1,2,3\}$.
