This question is at a very low level. Let $R$ be a commutative ring with unity. I am using the standard definition $\langle x \rangle = \{rx: r \in R\}$ for the ideal generated by $x$. It seems like the polynomial $p(x) = x^2 + x$ should not live in $\langle x \rangle$, since there is no $r \in R$ such that $p(x) = rx$. Does $\langle x \rangle$ automatically include all $\underline{linear}$ $\underline{ combinations}$ of $x$? Because the definition doesn't seem to imply that to me.
Thanks!
It is closed under linear combinations (not more), since you have the unit in $R$. Let $y,z \in <x>$, hence you have elements $r_y,r_z \in R$ s.t. $y=r_y x, z = r_z x$ holds. Since $R$ is additively closed also $r:=r_y + r_z \in R$ holds, so $y+z=(r_y+r_z)x=rx \in <x>$ holds. You need the unit, that also $x = 1 x \in <x>$ holds.
You have to be carefull what you are doing above. $<x>$ as definied as you did it, is not an ideal in the ring of polynomials over $R$, but for arbitrary $x \in R$ an ideal in $R$. Instead $<x> := \{ rx \ | \ r \in R[x]\}$ is an ideal in the ring of polynomials over $R$.
Assuming you are now working with a polynomial ring (to see not every linear combination is in the ideal): E.g. your example $p(x)=x^2+1$ has not a root 0 since $1 \neq 0$ but obviously every $f(x) \in <x>$, so $f(x)=rx (r \in R[X])$, has root $0$.
Edit: Since you changed your polynomial later on, here the answer to your edit: for $x \in R$ we have $p(x):=x^2+x=(x+1)x \in <x>$ since $x,1 \in R$ and hence $x+1 \in R$.