= Solution
{c}
Regard $\mathbb R[X,Y]$ as the <polynomial ring> $(\mathbb R[X])[Y]$. The <coefficient> <ring> $\mathbb R[X]$ is a <Euclidean domain>, hence a <principal ideal domain>. The element $X$ is a <prime element>: the <quotient ring> $\mathbb R[X]/(X)\cong\mathbb R$ is an <integral domain>.
The <monic polynomial> $Y^2+X$ has its other <coefficients> divisible by $X$, while $X^2\nmid X$. Part (a)(i), with $X$ playing the role of $p$ and $Y$ the <indeterminate>, proves that $Y^2+X$ is an <irreducible polynomial>.
Now use the <automorphism of a polynomial ring>
$$
\tau:\mathbb R[X,Y]\longrightarrow\mathbb R[X,Y],\qquad
\tau(X)=X,\quad\tau(Y)=Y+1.
$$
Its inverse fixes $X$ and sends $Y$ to $Y-1$. A <ring isomorphism> preserves <units> and factorizations, hence preserves <irreducible elements>. Since
$$
\tau(Y^2+X)=(Y+1)^2+X=Y^2+2Y+1+X,
$$
we obtain
$$
\boxed{Y^2+X\text{ and }Y^2+2Y+1+X\text{ are irreducible in }\mathbb R[X,Y]}.
$$
Applying the <Eisenstein criterion> directly to the second expression with the <prime element> $X$ would fail, since $X\nmid2$; the change of <indeterminate> supplies the required argument.
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