Question on irreducible polynomials over $R=\frac{\mathbb{Q}[x]}{(x^2+1)}$. 
Let $R=\frac{\mathbb{Q}[x]}{(x^2+1)}$ then which of the polynomials are irreducible over $R$
$(A)$ $x^2+1$
$(B)$ $x^2+x+1$
$(C)$ $x^2-x+1$
$(D)$ $x^3+x^2+x+1$

My answer: So $R=\frac{\mathbb{Q}[x]}{(x^2+1)} \simeq \mathbb{Q}[i]$.
So  since the polynomials are of degree 3 or less we only need check for roots in $\mathbb{Q}[i]$. Now polynomials $(A)$ and $(D)$ have roots while $(B)$ and $(C)$ do not have roots in $\mathbb{Q}[i]$. Hence the answer is $(B)$ and $(C)$. Am I right?

Yes, you are right. 
Just to give you some sort of answer, you may note, say, for $(A)$ and $(B)$, that $i \in \mathbb{Q}[i]$ is a root of $x^2+1 \in (\mathbb{Q}[i])[x]$, while if you assume for some $a,b \in \mathbb{Q}$ that $a+bi$ is a root of $x^2+x+1 \in (\mathbb{Q}[i])[x]$, then $$(a^2-b^2+a+1)+i(2ab+b)=(a+bi)^2+(a+bi)+1=0$$ so that $$0=a^2-b^2+a+1=b(2a+1);$$ but then, either $b=0$, and we have $$a^2+a+1=0,$$ a contradiction ($a \in \mathbb{Q}$, but $x^2+x+1$ has no roots in $\mathbb{Q}$), or $a=-1/2$, and we have $$b^2-3/4=0,$$ which again, gives a contradiction ($b \in \mathbb{Q}$, but $x^2-3/4$ has no roots in $\mathbb{Q}$)
