Is $\mathbb R[x]/(x^2-2)=\mathbb R$? 
Is $\mathbb R[x]/(x^2-2)=\mathbb R$?

I am reading Artin's Algebra and trying to analyze quotient rings. In one example he shows that $\mathbb Z[i]/(i-2) \cong \mathbb Z/5\mathbb Z$, example $11.4.5$. Very cool.
Now, I am trying to do a similar analyzing to see what $\mathbb R[x]/(x^2-2)$ is.
We know $x^2-2$ is reducible and $x^2-2=(x-\sqrt 2)(x+\sqrt 2)$. So, is it true I that I can write $\mathbb R[x]/(x^2-2)=\mathbb R[x]/(x-\sqrt 2)(x+\sqrt 2)$?
Also, I presume that since the canoncial map $\mathbb R[x] \to \mathbb R[x]/(x^2-2)$ where $x \mapsto \sqrt 2$, then we can write $\mathbb R[x]/(x^2-2)=\mathbb R[\sqrt 2]$. But since $\sqrt 2 \in \mathbb R$, then $\mathbb R[\sqrt 2]=\mathbb R$

Take the homomorphism:
$$\mathbb R[x]\to \mathbb R\times\mathbb R\\
p(x)\mapsto (p(\sqrt{2}),p(-\sqrt{2}))$$
The kernel of this homomorphism is the set of $p$ such that $p(\sqrt{2})=p(-\sqrt{2})=0$, which means the $p$ which are divisible by $x^2-2$. 
So the factor map: $\mathbb R[x]/(x^2-2)\to\mathbb R\times\mathbb R.$ is one-to-one.
Now you just have to prove that this map is onto.
