I tried writing this polynomial as a product of two polynomials $g, h$ of degree 2 and 1, respectively, and tried to arrive at a contradiction by multiplication of their coefficients. However, I arrived at an overwhelmingly large non linear condition, and gave up.
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2Since the $y$-degree of this polynomial is $2$, if it splits, then it must split as $(y-g(x))(y-h(x))$ for some $g, h \in \mathbb{R}[x]$. Can you finish from here? – Sameer Kailasa May 06 '19 at 05:41
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Same as I do here, using nonsquareness or Eisenstein, and UFD properties. – Bill Dubuque May 06 '19 at 18:46
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Yeah, in fact, there is a simple way to prove this result. At first, we can view the ring $\mathbb R[x,y]$ as a ring with one variable $Y$ over $\mathbb R[x]$, i.e., $\mathbb R[x, y] \cong \mathbb R[x][y]$. Note that $\mathbb R[x]$ is an integral domain and $x$ is an irreducible element in $\mathbb R[x]$. Hence, $x$ is a prime element in $\mathbb R[x]$. By $$x \mid x(x-1)(x+1), x^2 \not \mid y^2$$ and the generalization of Eisenstein's criterionenter link description here. Then we are done.
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