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  "body": "\n## Statement\n\nLet $K$ be a [[field|0QA8CHH8MG]] and let $f \\in K[x]$ with $\\deg f = 2$ or $\\deg f = 3$.\nThen $f$ is [[irreducible|00X2A9G9W3]] over $K$ if and only if $f$ has no [[root|00EHX6FWC0]] in $K$.\n\n## Remarks\n\nThe proof: if $f = g h$ with $\\deg g, \\deg h \\ge 1$, then $\\deg g + \\deg h \\le 3$, so one factor has degree $1$. A factor $a x + b$ of degree $1$ gives the root $-b/a$. The [[factor theorem|0K3MWA7DMC]] gives the converse.\nFor example, $x^3 - 2$ has no root in $\\mathbb{Q}$, so it is irreducible over $\\mathbb{Q}$.\nThe statement fails in degree $4$: $(x^2 + 1)^2$ has no real root, but it is not irreducible over $\\mathbb{R}$.\n\n## Explanations\n\n### L1 human\n\nA factorisation of a polynomial of degree $2$ or $3$ must have a factor of degree $1$.\nA factor of degree $1$ is the same as a root.\nSo to test irreducibility in low degree, you only look for roots.\n",
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