Principal ideal domain - polynomial ring

Question

Let $f(x)=x^2+3x+5,\ \ g(x)=x+1\in\mathbb{R}[x]$. Show that $1\in f(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]$ and hence that $f(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]=\mathbb{R}[x]$.

My idea is to use the division algorithm to get $x^2+3x+5=(x+1)(x+2)+3$, so $f(x)=(x+2)g(x)+3$. Hence $f(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]=(x+2)g(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]=g(x)\mathbb{R}[x](x+3)$ (I'm not sure this is allowed).

How do we show $1$ is an element of this and how do we then deduce the required result?

Answer 1

Hint:

Your work shows that $3\in f(x)\mathbb{R}[x]+g(x)\mathbb{R}[x]$, since rearranging yields $$ 3=1\cdot(x^2+3x+5)+(-(x+2))\cdot(x+1)=1\cdot f(x)+(-(x+2))\cdot g(x) $$ and clearly $1,-(x+2)\in\mathbb{R}[x]$.

Now, note that your coefficients are allowed to come from $\mathbb{R}$, not just $\mathbb{N}$ -- how could you modify the polynomials above (that is, $1$ and $-(x+2)$) to get $1$ as a result?