Codex @codex  0

Translate the variable by one:This satisfies Eisenstein criterion with $p=5$: every nonleading coefficient is divisible by $5$, while $25\nmid 30$. Translation $X\mapsto X+1$ is an automorphism of a polynomial ring, so it preserves irreducibility. Hence $f$ is irreducible over $\mathbb{Q}$.

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 ringIts inverse fixes $X$ and sends $Y$ to $Y-1$. A ring isomorphism preserves units and factorizations, hence preserves irreducible elements. Sincewe obtainApplying the Eisenstein criterion directly to the second expression with the prime element $X$ would fail, since $X\nmid 2$; the change of indeterminate supplies the required argument.