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  "body": "\n## Statement\n\nLet $a, b, c, d, e \\in \\C$ with $a \\ne 0$.\nPut\n$$p = \\frac{8ac - 3b^2}{8a^2}, \\qquad q = \\frac{b^3 - 4abc + 8a^2 d}{8a^3}, \\qquad r = \\frac{16ab^2c - 3b^4 - 64a^2bd + 256a^3e}{256a^4}.$$\nThen $x$ is a [[root|00EHX6FWC0]] of $ax^4 + bx^3 + cx^2 + dx + e$ if and only if $y = x + \\frac{b}{4a}$ is a root of $y^4 + py^2 + qy + r$.\nAssume $q \\ne 0$.\nLet $m$ be a root of the resolvent cubic\n$$8m^3 + 8pm^2 + (2p^2 - 8r)m - q^2,$$\nand let $s$ be a square root of $2m$.\nThen $m \\ne 0$, $s \\ne 0$, and the four roots of $y^4 + py^2 + qy + r$, counted with multiplicity, are\n$$\\frac{s \\pm \\sqrt{-2p - 2m - 2q/s}}{2} \\qquad \\text{and} \\qquad \\frac{-s \\pm \\sqrt{-2p - 2m + 2q/s}}{2},$$\nwhere $\\sqrt{z}$ denotes either square root of $z$.\n\n## Remarks\n\nSince $q \\ne 0$, the constant term $-q^2$ of the resolvent cubic is nonzero, so $m \\ne 0$.\nThe resolvent cubic says exactly that $(y^2 + \\frac{p}{2} + m)^2 - (y^4 + py^2 + qy + r) = 2my^2 - qy + (m^2 + pm + \\frac{p^2}{4} - r)$ is the square $(sy - \\frac{q}{2s})^2$.\nSo the quartic factors as $(y^2 + sy + \\frac{p}{2} + m - \\frac{q}{2s})(y^2 - sy + \\frac{p}{2} + m + \\frac{q}{2s})$, and the quadratic formula gives the four roots.\nThe resolvent cubic itself is solved by the [[cubic formula|FB6J5WAT5V]], so all roots are radical expressions in the coefficients.\nIf $q = 0$, the equation $y^4 + py^2 + r = 0$ is a quadratic in $y^2$, and $y^2 = \\frac{-p \\pm \\sqrt{p^2 - 4r}}{2}$.\nExample: for $y^4 - 2y^2 + 8y - 3$, the value $m = 2$ is a root of $8m^3 - 16m^2 + 32m - 64$, so $s = 2$ and the factors are $y^2 + 2y - 1$ and $y^2 - 2y + 3$.\n\n## Explanations\n\n### L1 human\n\nThe quartic formula gives the roots of any equation of degree $4$ using arithmetic, square roots and cube roots.\nThe trick is to add a well-chosen quantity $m$ to both sides so that both become perfect squares.\nFinding $m$ needs a cubic equation, which the cubic formula solves.\n\n### L2 human\n\nThis method is due to Ferrari, and it shows that every quartic is [[solvable by radicals|J7YS0DA3HQ]].\nIn Galois terms, $S_4$ is solvable, with the normal subgroup $V_4$ of order $4$ and quotient $S_3$.\nIf $y_1, y_2, y_3, y_4$ are the roots, each root $m$ of the resolvent cubic has the form $\\frac{1}{2}(y_i + y_j)^2$, and $(y_1 + y_2)^2$ is fixed by $V_4$.\nThis is why a cubic appears.\nFor degree $5$ no such reduction exists, by the [[Abel–Ruffini theorem|163A2WQQJX]].\n",
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