By showing that $R/I$ is an integral domain, deduce that it's a field.

Let $$R = \mathbb{F}_3[x]$$ and let $$I$$ be the set $$\{ (x^2+1)p(x)|p(x) \in R \}$$ which is an ideal of R. By showing that $$R/I$$ is an integral domain, deduce that $$R/I$$ is a field.

I know that $$R/I$$ is commutative and that it's finite. I am having trouble proving that it has no zero-divisors.

If $$q(x),r(x)\in\mathbb F_3[x]$$ are such that $$[q(x]\times[r(x)]=0$$, then $$q(x)r(x)=(x^2+1)p(x)$$ for some $$p(x)\in\mathbb F_3[x]$$. So, $$x^2+1\mid q(x)r(x)$$. But $$x^2+1$$ is irreducible in $$\mathbb F_3[x]$$, and therefore, since it divides the producte $$q(x)r(x)$$, it divides one of its factors. But this means the $$[q(x)]$$ or $$[r(x)]$$ is equal to $$0$$ in your ring.