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  "body": "\n## Proof\n\n1. Let $K$ be a [[field|0QA8CHH8MG]] and let $f \\in K[x]$ with $\\deg f \\in \\set{2, 3}$.\n2. We first find the [[units|019NFGFPR5]] of $K[x]$. If $u v = 1$ in $K[x]$, then $\\deg u + \\deg v = 0$ by the [[degree of a product|0T233AZ1GE]]. So $\\deg u = 0$, and $u$ is a nonzero constant.\n3. Conversely, a nonzero constant $c \\in K$ has the inverse $c^{-1} \\in K$. So the units of $K[x]$ are exactly the polynomials of degree $0$.\n4. Since $\\deg f \\ge 1$, the polynomial $f$ is nonzero and, by Step 3, it is not a unit.\n5. Suppose that $f$ has a [[root|00EHX6FWC0]] $c \\in K$. By the [[factor theorem|0K3MWA7DMC]], $f = (x - c) g$ for some $g \\in K[x]$.\n6. By the degree of a product, $\\deg g = \\deg f - 1 \\ge 1$. By Step 3, neither $x - c$ nor $g$ is a unit. So $f$ is not [[irreducible|00X2A9G9W3]].\n7. This proves: if $f$ is irreducible, then $f$ has no root in $K$.\n8. Now suppose that $f$ has no root in $K$. Let $f = g h$ with $g, h \\in K[x]$. We show that $g$ or $h$ is a unit.\n9. Since $f \\ne 0$, both $g$ and $h$ are nonzero. By the degree of a product, $\\deg g + \\deg h = \\deg f \\le 3$.\n10. Suppose that neither $g$ nor $h$ is a unit. By Step 3, $\\deg g \\ge 1$ and $\\deg h \\ge 1$.\n11. If both degrees were at least $2$, their sum would be at least $4$, which contradicts Step 9. So one factor, say $g$, has degree $1$.\n12. Write $g = a x + b$ with $a, b \\in K$ and $a \\ne 0$. Then $g(-b/a) = -b + b = 0$.\n13. [[Evaluation|0125QVNCFQ]] at $-b/a$ is a ring homomorphism. So $f(-b/a) = g(-b/a) \\, h(-b/a) = 0$, and $-b/a$ is a root of $f$ in $K$.\n14. This contradicts the assumption of Step 8. So $g$ or $h$ is a unit.\n15. With Step 4, $f$ is irreducible over $K$. Together with Step 7, $f$ is irreducible over $K$ if and only if it has no root in $K$.\n\n## Idea\n\nA factorisation into two factors of positive degree splits the degree $2$ or $3$ as a sum of two positive integers. One of them must be $1$, and a factor of degree $1$ gives a root. Conversely, a root $c$ gives the factor $x - c$.\n",
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