Q:

Definition of algebraically closed field.

A field F is algebraically closed if every non constant polynomial in F[x] has a root in F. 
Is this the right definition? I am wondering if only one root in F and the rest of the roots not in F can also be considered 'algebraically closed'? 
I remember reading somewhere that if a root is in F then all roots are in F. Is this a special case of something? If so, could someone enlighten me on what it is?
Thank you

A:

1) The definition is right. 
2) No. That not the same. 
3) That's wrong. Consider $x(x^2+1)$ over the reals, which is not algebraically closed.

A:

Yes, every non-constant polynomial needs to have a root in the field; equivalently, it factors completely over the field.  The second statement is not saying the same thing, because if only one root $\alpha$ is in $\mathrm{F}$, then we can factor out $(x-\alpha)$ to obtain a new (non-constant) polynomial, which must also have a root in $\mathrm{F}$.
The third statement is maybe something you read about quadratics; certainly if one root of a quadratic over $\mathrm{F}$ is in $\mathrm{F}$, then the other root is as well.  But this statement is not true in general.

