I'm working on a few problems of a similar type. I'm pretty confident I got the right answer but I'm not sure how to provide satisfiable enough proof that the answer is correct. I'd be glad if you could point out any "holes" in my arguments, if there are any. Thanks.
Let's assume the problem is as follows:
Find a group isomorphic to the group $(G, \cdot)/H$, where $$G = \left \{ \begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \mid q \in \mathbb{Q} \setminus\{0\}\,,\,\, r \in \mathbb{R}\,,\,\, f,g \in \mathbb{R}[x]\,,\,\, g(5) = 0 \right \} \text{,}$$ $$H = \left \{ \begin{pmatrix} 1 & 0 & 0 \\ r & 1 & 0 \\ f & g & 1 \end{pmatrix} \mid r \in \mathbb{R}\,,\,\, f,g \in \mathbb{R}[x]\,,\,\, g(5) = f(5) = 0 \right \}\text{.} $$
My hypothesis is that $(G, \cdot)/H \cong (K, \times)$ where $(K, \times) = (\mathbb{Q}^{*}, \cdot) \times (\mathbb{R}, +)$. Let's assume an surjective homomorphism $\alpha : (G, \cdot) \longrightarrow K$ is defined by the following formula: $$\alpha\left ( \begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \right ) = (q, f(5))\text{.}$$
To prove $\alpha$ is a surjective homomorphisum we show:
$\alpha$ is a homomorphism: $$\alpha\left ( \begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ \bar{r} & \bar{q} & 0 \\ \bar{f} & \bar{g} & 1 \end{pmatrix} \right ) = \alpha\left ( \begin{pmatrix} 1 & 0 & 0 \\ r+q\bar{r} & q\bar{q} & 0 \\ f + g\bar{r} + \bar{f} & g\bar{q}+\bar{g} & 1 \end{pmatrix} \right) = (q\bar{q},(f+g\bar{r}+\bar{f})(5)) = (q\bar{q}, (f+\bar{f})(5))$$ $$\alpha\left( \begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \right)\times \alpha\left( \begin{pmatrix} 1 & 0 & 0 \\ \bar{r} & \bar{q} & 0 \\ \bar{f} & \bar{g} & 1 \end{pmatrix} \right)= (q, f(5)) \times (\bar{q},\bar{f}(5)) = (q\bar{q}, f(5) + \bar{f}(5)) = (q\bar{q}, (f + \bar{f})(5))$$
$\alpha$ is surjective: given any $(q', r') \in K$ we put $q = q'$ and $f = x + (r' - 5)$.
Now we show that $H = Ker(\alpha)$: $$\begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \in Ker(\alpha) \Longleftrightarrow \alpha \left( \begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \right)= (1,0) \Longleftrightarrow q = 1 \wedge f(5) = 0 \Longleftrightarrow \begin{pmatrix} 1 & 0 & 0 \\ r & q & 0 \\ f & g & 1 \end{pmatrix} \in H $$ Thus, as per the first isomorphism theorem, $img(\alpha) \cong (G,\cdot)/H$. As $\alpha$ is surjective $img(\alpha) = K$, which proves our initial hypothesis $(G, \cdot)/H \cong (K, \times)$.