Let $\Bbb R$ be a field of all real numbers. Prove that $x+1$ is not a unit in $\Bbb R[x]$. [duplicate]

Let $$\Bbb R$$ be a field of all real numbers. Prove that $$x+1$$ is not a unit in $$\Bbb R[x]$$.

Attempts:

Suppose that $$x+1$$ is a unit in $$\Bbb R[x]$$. Then, there is exists $$g(x) \in \Bbb R[x]$$ such that $$(x+1)g(x) = 1_R$$.

Now, we have $$\begin{equation*} deg(x+1) + deg(g(x)) = deg((x+1)g(x)) = deg(1_R) = 0 \end{equation*}$$ Thus, we have \begin{align*} deg(x+1) + deg(g(x)) &= 0 \\ 1 + deg(g(x)) &= 0 \\ deg(g(x)) &= -1 \end{align*} which is a contradiction since $$deg(f(x))$$ must be non-negative integers for all $$f(x) \in \Bbb R[x]$$. Hence, $$x+1$$ is not a unit in $$\Bbb R[x]$$.

Is the above true?

Another way to see this is that $$\Bbb{R}[x]/(x+1)\cong \Bbb{R}$$ and hence $$(x+1)\ne \Bbb{R}[x]$$ so that $$x+1$$ is not a unit.