Consider $p(x)= x^3 - 6x +2$.
I first proved that it's irreducible over $\mathbb{Q}$. Then I concluded that $<p(x)>$ is a maximal ideal so that $\frac{\mathbb{Q}[x]}{<p(x)>}$ is a field.
After that, I was asked to show that $\frac{\mathbb{Q}[x]}{<p(x)>}$ has a solution of $p(x)= x^3 - 6x +2 =0$.
My prof. answered this way:
Let $<p(x)> =I$,
$p(x)\in I$ so $p(x) +I = I$ i.e. $x^3 - 6x +2 +I = 0+I$, so
$(x+I)^3 - 6(x+I) +2+I = I$ so $x+I$ is a root of $p(x)$ in $\frac{\mathbb{Q}[x]}{I}$. How did he add $I$ to the equation?? I got stuck at this point and tried alot to understand it alone :( Thanks for any help.