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   "source": [
    "# Solving a Quartic Equation\n",
    "\n",
    "Ferrari's method first translates a quartic to\n",
    "\n",
    "$$\n",
    "y^4+py^2+qy+r=0.\n",
    "$$\n",
    "\n",
    "It then chooses a resolvent parameter $u$ so that the polynomial\n",
    "factors into two quadratics. The condition on $u$ is the cubic\n",
    "\n",
    "$$\n",
    "u(p+u)^2-4ru-q^2=0.\n",
    "$$\n"
   ]
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   "source": [
    "## Three structural cases\n",
    "\n",
    "A biquadratic ($q=0$) needs only two quadratic solves. A general\n",
    "depressed quartic uses the resolvent cubic. Finally,\n",
    "$x=y-b/4$ removes the cubic term from a monic quartic; a non-monic\n",
    "polynomial is normalized first.\n"
   ]
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  {
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   "id": "905d616d",
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   "source": [
    "declare symbol x, y, u: MathValue\n",
    "\n",
    "def solveBiquadratic\n",
    "  (p : MathValue)\n",
    "  (q : MathValue)\n",
    "  : (MathValue, MathValue, MathValue, MathValue) :=\n",
    "  let (s1, s2) := qF' 1 p q\n",
    "      (r1, r2) := qF' 1 0 (- s1)\n",
    "      (r3, r4) := qF' 1 0 (- s2)\n",
    "   in (r1, r2, r3, r4)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "04d22c92",
   "metadata": {},
   "source": [
    "## Biquadratic shortcut\n",
    "\n",
    "For $x^4-5x^2+4=0$, the substitution $t=x^2$ yields\n",
    "$(t-1)(t-4)=0$.  Solving first for $t$ and then taking the two\n",
    "square roots of each value gives all four roots.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "79e4b59d",
   "metadata": {},
   "outputs": [
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     "data": {
      "text/html": [
       "$(2, -2, 1, -1)$"
      ]
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   "source": [
    "solveBiquadratic (-5) 4\n"
   ]
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   "id": "321c1efc",
   "metadata": {},
   "source": [
    "## Ferrari's general branch\n",
    "\n",
    "Consider the already-depressed polynomial\n",
    "\n",
    "$$\n",
    "y^4-15y^2-10y+24\n",
    "  =(y+3)(y+2)(y-1)(y-4).\n",
    "$$\n",
    "\n",
    "Here $(p,q,r)=(-15,-10,24)$, so this is genuinely outside the\n",
    "biquadratic case.  Its resolvent has the convenient root $u=1$.\n"
   ]
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   "cell_type": "code",
   "execution_count": 3,
   "id": "b197dfb2",
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   "outputs": [],
   "source": [
    "def p : MathValue := -15\n",
    "def q : MathValue := -10\n",
    "def r : MathValue := 24\n",
    "def chosenU : MathValue := 1\n",
    "\n",
    "def resolvent (u : MathValue) : MathValue :=\n",
    "  u * (p + u)^2 - 4 * r * u - q^2\n",
    "\n",
    "def factorPlus : MathValue :=\n",
    "  y^2 + (p + chosenU) / 2\n",
    "    + sqrt chosenU * (y - q / (2 * chosenU))\n",
    "\n",
    "def factorMinus : MathValue :=\n",
    "  y^2 + (p + chosenU) / 2\n",
    "    - sqrt chosenU * (y - q / (2 * chosenU))\n"
   ]
  },
  {
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   "execution_count": 4,
   "id": "8400ad4c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "$0$"
      ]
     },
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   ],
   "source": [
    "resolvent chosenU\n"
   ]
  },
  {
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   "execution_count": 5,
   "id": "34aa89fd",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "$(y^{2} + y - 2, y^{2} - y - 12)$"
      ]
     },
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     "output_type": "display_data"
    }
   ],
   "source": [
    "(factorPlus, factorMinus)\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "id": "9ea1631f",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "$((1, -2), (4, -3))$"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "(qF factorPlus y, qF factorMinus y)\n"
   ]
  },
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   "cell_type": "markdown",
   "id": "caed1bd8",
   "metadata": {},
   "source": [
    "## Why the factorization works\n",
    "\n",
    "Once a root $u$ of the resolvent is chosen, the depressed quartic is\n",
    "split into\n",
    "\n",
    "$$\n",
    "y^2+\\frac{p+u}{2}\n",
    "  \\pm\\sqrt{u}\\left(y-\\frac{q}{2u}\\right)=0.\n",
    "$$\n",
    "\n",
    "The two calls to the quadratic solver return all four roots.\n"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "36b0ae0e",
   "metadata": {},
   "source": [
    "## Takeaway\n",
    "\n",
    "The two executable paths mirror Ferrari's proof: the biquadratic case\n",
    "reduces immediately to quadratic equations, while the nonzero-linear\n",
    "case uses one resolvent root to expose two quadratic factors.  Egison\n",
    "keeps the exact factorization and all four roots symbolic.\n"
   ]
  }
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