Let $R$ be a ring with unity. Then show that: $\dfrac{R[x]}{\langle x \rangle}\cong R$.
I have defined a homomorphism : $\phi:\mathbb R[x]\to \mathbb R$ such that $\phi( a_0+a_1x+\cdots)=a_0$.
Further I got $\ker(\phi) = \{f(x) \in R[x] \mid \phi(f(x)) = 0 \}$ which is
$\ker(\phi) = \{xg(x) \in R[x] \mid g(x) \in R[x]\}$.
My question is how $\langle x \rangle = \{xg(x) \in R[x] \mid g(x) \in R[x]\}$.
Let \begin{align*} \Phi:R[X]&\longrightarrow R\\ p(X)&\longmapsto p(0). \end{align*} I let you prove that it's a ring morphism. The surjectivity is clear. Now, as you said, $\ker(Phi)=\{p(X)\in R[X]\mid p(0)=0\}.$
Observe that $$p(X)=a_0+a_1X+...+a_nX^n\in \ker(Phi)\iff p(X)=a_1X+...+a_nX^n=X(a_1+...+a_nX^n)\iff \exists g(X)\in R[X]\mid p(X)=Xg(X)\in \left<X\right>.$$
Therefore, $$\ker(Phi)=\left<X\right>.$$