{
  "index": "1942-A-2",
  "type": "ALG",
  "tag": [
    "ALG"
  ],
  "difficulty": "",
  "question": "2. If a polynomial \\( f(x) \\) is divided by \\( (x-a)^{2}(x-b) \\), where \\( a \\neq b \\), derive a formula for the remainder.",
  "solution": "First Solution. Since \\( f(x) \\) is divided by a cubic polynomial, the remainder \\( R(x) \\) will be of degree at most two in \\( x \\), say \\( R(x)=A x^{2}+B x+C \\). Then\n\\[\nf(x)=(x-a)^{2}(x-b) Q(x)+A x^{2}+B x+C\n\\]\nand\n\\[\n\\begin{array}{l}\nf^{\\prime}(x)=2(x-a)(x-b) Q(x)+(x-a)^{2} Q(x) \\\\\n\\quad+(x-a)^{2}(x-b) Q^{\\prime}(x)+2 A x+B .\n\\end{array}\n\\]\n\nFrom these relations one gets\n\\[\n\\begin{aligned}\nf(a) & =A a^{2}+B a+C \\\\\nf(b) & =A b^{2}+B b+C \\\\\nf^{\\prime}(a) & =2 A a+B\n\\end{aligned}\n\\]\n\nSolving for \\( A, B, C \\) one gets\n\\[\n\\begin{array}{l}\nA=\\frac{1}{(b-a)^{2}}\\left[f(b)-f(a)-(b-a) f^{\\prime}(a)\\right] \\\\\nB=\\frac{-1}{(b-a)^{2}}\\left[2 a(f(b)-f(a))-\\left(b^{2}-a^{2}\\right) f^{\\prime}(a)\\right] \\\\\nC=\\frac{1}{(b-a)^{2}}\\left[(b-a)^{2} f(a)+a^{2}(f(b)-f(a))+a b(a-b) f^{\\prime}(a)\\right] .\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nR(x)=\\frac{1}{(b-a)^{2}} & \\left\\{\\left(f(b)-f(a)-(b-a) f^{\\prime}(a)\\right) x^{2}\\right. \\\\\n& -\\left(2 a(f(b)-f(a))+\\left(b^{2}-a^{2}\\right) f^{\\prime}(a)\\right) x \\\\\n& \\left.+\\left((b-a)^{2} f(a)+a^{2}(f(b)-f(a))+a b(a-b) f^{\\prime}(a)\\right)\\right\\}\n\\end{aligned}\n\\]\n\nThis is easier to check if written in the form\n\\[\nR(x)=f(a)+\\frac{f(b)-f(a)}{(b-a)^{2}}(x-a)^{2}-\\frac{f^{\\prime}(a)}{b-a}(x-a)(x-b) .\n\\]\n\nSecond Solution. We can write\n\\[\nf(x)=f(a)+(x-a) f^{\\prime}(a)+(x-a)^{2} g(x)\n\\]\nwhere \\( g \\) is a polynomial, and \\( g(x)=g(b)+(x-b) h(x) \\) where \\( h \\) is a polynomial. Then\n\\[\nf(x)=f(a)+(x-a) f^{\\prime}(a)+(x-a)^{2} g(b)+(x-a)^{2}(x-b) h(x)\n\\]\nand the desired remainder is\n\\[\nf(a)+(x-a) f^{\\prime}(a)+(x-a)^{2} g(b)\n\\]\n\nSubstituting \\( b \\) for \\( x \\) in (1) gives us \\( g(b) \\) and the remainder is\n\\[\nf(a)+(x-a) f^{\\prime}(a)+\\frac{(x-a)^{2}}{(b-a)^{2}}\\left(f(b)-f(a)-(b-a) f^{\\prime}(a)\\right) .\n\\]",
  "vars": [
    "x"
  ],
  "params": [
    "a",
    "b",
    "f",
    "Q",
    "A",
    "B",
    "C",
    "R",
    "g",
    "h"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "x": "variable",
        "a": "firstconst",
        "b": "secondconst",
        "f": "polyfunc",
        "Q": "quotient",
        "A": "coefquadr",
        "B": "coeflinear",
        "C": "coefconst",
        "R": "remainder",
        "g": "auxpoly",
        "h": "helperpoly"
      },
      "question": "If a polynomial \\( polyfunc(variable) \\) is divided by \\( (variable-firstconst)^{2}(variable-secondconst) \\), where \\( firstconst \\neq secondconst \\), derive a formula for the remainder.",
      "solution": "First Solution. Since \\( polyfunc(variable) \\) is divided by a cubic polynomial, the remainder \\( remainder(variable) \\) will be of degree at most two in \\( variable \\), say \\( remainder(variable)=coefquadr\\, variable^{2}+coeflinear\\, variable+coefconst \\). Then\n\\[\npolyfunc(variable)=(variable-firstconst)^{2}(variable-secondconst)\\, quotient(variable)+coefquadr\\, variable^{2}+coeflinear\\, variable+coefconst\n\\]\nand\n\\[\n\\begin{array}{l}\npolyfunc^{\\prime}(variable)=2(variable-firstconst)(variable-secondconst)\\, quotient(variable)+(variable-firstconst)^{2}\\, quotient(variable) \\\\\n\\quad+(variable-firstconst)^{2}(variable-secondconst)\\, quotient^{\\prime}(variable)+2\\, coefquadr\\, variable+coeflinear .\n\\end{array}\n\\]\n\nFrom these relations one gets\n\\[\n\\begin{aligned}\npolyfunc(firstconst) & =coefquadr\\, firstconst^{2}+coeflinear\\, firstconst+coefconst \\\\\npolyfunc(secondconst) & =coefquadr\\, secondconst^{2}+coeflinear\\, secondconst+coefconst \\\\\npolyfunc^{\\prime}(firstconst) & =2\\, coefquadr\\, firstconst+coeflinear\n\\end{aligned}\n\\]\n\nSolving for \\( coefquadr, coeflinear, coefconst \\) one gets\n\\[\n\\begin{array}{l}\ncoefquadr=\\dfrac{1}{(secondconst-firstconst)^{2}}\\left[polyfunc(secondconst)-polyfunc(firstconst)-(secondconst-firstconst)\\, polyfunc^{\\prime}(firstconst)\\right] \\\\\ncoeflinear=\\dfrac{-1}{(secondconst-firstconst)^{2}}\\left[2\\, firstconst\\left(polyfunc(secondconst)-polyfunc(firstconst)\\right)-\\left(secondconst^{2}-firstconst^{2}\\right) polyfunc^{\\prime}(firstconst)\\right] \\\\\ncoefconst=\\dfrac{1}{(secondconst-firstconst)^{2}}\\left[(secondconst-firstconst)^{2} polyfunc(firstconst)+firstconst^{2}\\left(polyfunc(secondconst)-polyfunc(firstconst)\\right)+firstconst\\, secondconst(firstconst-secondconst)\\, polyfunc^{\\prime}(firstconst)\\right] .\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nremainder(variable)=\\dfrac{1}{(secondconst-firstconst)^{2}} & \\left\\{\\left(polyfunc(secondconst)-polyfunc(firstconst)-(secondconst-firstconst)\\, polyfunc^{\\prime}(firstconst)\\right) variable^{2}\\right. \\\\\n& -\\left(2\\, firstconst\\left(polyfunc(secondconst)-polyfunc(firstconst)\\right)+\\left(secondconst^{2}-firstconst^{2}\\right) polyfunc^{\\prime}(firstconst)\\right) variable \\\\\n& \\left.+\\left((secondconst-firstconst)^{2} polyfunc(firstconst)+firstconst^{2}\\left(polyfunc(secondconst)-polyfunc(firstconst)\\right)+firstconst\\, secondconst(firstconst-secondconst)\\, polyfunc^{\\prime}(firstconst)\\right)\\right\\}\n\\end{aligned}\n\\]\n\nThis is easier to check if written in the form\n\\[\nremainder(variable)=polyfunc(firstconst)+\\frac{polyfunc(secondconst)-polyfunc(firstconst)}{(secondconst-firstconst)^{2}}(variable-firstconst)^{2}-\\frac{polyfunc^{\\prime}(firstconst)}{secondconst-firstconst}(variable-firstconst)(variable-secondconst) .\n\\]\n\nSecond Solution. We can write\n\\[\npolyfunc(variable)=polyfunc(firstconst)+(variable-firstconst)\\, polyfunc^{\\prime}(firstconst)+(variable-firstconst)^{2}\\, auxpoly(variable)\n\\]\nwhere \\( auxpoly \\) is a polynomial, and \\( auxpoly(variable)=auxpoly(secondconst)+(variable-secondconst)\\, helperpoly(variable) \\) where \\( helperpoly \\) is a polynomial. Then\n\\[\npolyfunc(variable)=polyfunc(firstconst)+(variable-firstconst)\\, polyfunc^{\\prime}(firstconst)+(variable-firstconst)^{2}\\, auxpoly(secondconst)+(variable-firstconst)^{2}(variable-secondconst)\\, helperpoly(variable)\n\\]\nand the desired remainder is\n\\[\npolyfunc(firstconst)+(variable-firstconst)\\, polyfunc^{\\prime}(firstconst)+(variable-firstconst)^{2}\\, auxpoly(secondconst)\n\\]\n\nSubstituting \\( secondconst \\) for \\( variable \\) in (1) gives us \\( auxpoly(secondconst) \\) and the remainder is\n\\[\npolyfunc(firstconst)+(variable-firstconst)\\, polyfunc^{\\prime}(firstconst)+\\frac{(variable-firstconst)^{2}}{(secondconst-firstconst)^{2}}\\left(polyfunc(secondconst)-polyfunc(firstconst)-(secondconst-firstconst)\\, polyfunc^{\\prime}(firstconst)\\right) .\n\\]"
    },
    "descriptive_long_confusing": {
      "map": {
        "x": "grapefruit",
        "a": "lemongrass",
        "b": "butterscotch",
        "f": "rhinoceros",
        "Q": "turpentine",
        "A": "pineapples",
        "B": "watermelon",
        "C": "blacksmith",
        "R": "cinematics",
        "g": "parchment",
        "h": "sugarcane"
      },
      "question": "2. If a polynomial \\( rhinoceros(grapefruit) \\) is divided by \\( (grapefruit-lemongrass)^{2}(grapefruit-butterscotch) \\), where \\( lemongrass \\neq butterscotch \\), derive a formula for the remainder.",
      "solution": "First Solution. Since \\( rhinoceros(grapefruit) \\) is divided by a cubic polynomial, the remainder \\( cinematics(grapefruit) \\) will be of degree at most two in grapefruit, say \\( cinematics(grapefruit)=pineapples grapefruit^{2}+watermelon grapefruit+blacksmith \\). Then\n\\[\nrhinoceros(grapefruit)=(grapefruit-lemongrass)^{2}(grapefruit-butterscotch) turpentine(grapefruit)+pineapples grapefruit^{2}+watermelon grapefruit+blacksmith\n\\]\n\nand\n\\[\n\\begin{array}{l}\nrhinoceros^{\\prime}(grapefruit)=2(grapefruit-lemongrass)(grapefruit-butterscotch) turpentine(grapefruit)+(grapefruit-lemongrass)^{2} turpentine(grapefruit) \\\\\n\\quad+(grapefruit-lemongrass)^{2}(grapefruit-butterscotch) turpentine^{\\prime}(grapefruit)+2 pineapples grapefruit+watermelon .\n\\end{array}\n\\]\n\nFrom these relations one gets\n\\[\n\\begin{aligned}\nrhinoceros(lemongrass) & =pineapples lemongrass^{2}+watermelon lemongrass+blacksmith \\\\\nrhinoceros(butterscotch) & =pineapples butterscotch^{2}+watermelon butterscotch+blacksmith \\\\\nrhinoceros^{\\prime}(lemongrass) & =2 pineapples lemongrass+watermelon\n\\end{aligned}\n\\]\n\nSolving for \\( pineapples, watermelon, blacksmith \\) one gets\n\\[\n\\begin{array}{l}\npineapples=\\frac{1}{(butterscotch-lemongrass)^{2}}\\left[rhinoceros(butterscotch)-rhinoceros(lemongrass)-(butterscotch-lemongrass) rhinoceros^{\\prime}(lemongrass)\\right] \\\\\nwatermelon=\\frac{-1}{(butterscotch-lemongrass)^{2}}\\left[2 \\,lemongrass\\,(rhinoceros(butterscotch)-rhinoceros(lemongrass))-\\left(butterscotch^{2}-lemongrass^{2}\\right) rhinoceros^{\\prime}(lemongrass)\\right] \\\\\nblacksmith=\\frac{1}{(butterscotch-lemongrass)^{2}}\\left[(butterscotch-lemongrass)^{2} rhinoceros(lemongrass)+lemongrass^{2}(rhinoceros(butterscotch)-rhinoceros(lemongrass))+lemongrass butterscotch(lemongrass-butterscotch) rhinoceros^{\\prime}(lemongrass)\\right] .\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\ncinematics(grapefruit)=\\frac{1}{(butterscotch-lemongrass)^{2}} & \\left\\{\\left(rhinoceros(butterscotch)-rhinoceros(lemongrass)-(butterscotch-lemongrass) rhinoceros^{\\prime}(lemongrass)\\right) grapefruit^{2}\\right. \\\\\n& -\\left(2 \\, lemongrass\\, (rhinoceros(butterscotch)-rhinoceros(lemongrass))+\\left(butterscotch^{2}-lemongrass^{2}\\right) rhinoceros^{\\prime}(lemongrass)\\right) grapefruit \\\\\n& \\left.+\\left((butterscotch-lemongrass)^{2} rhinoceros(lemongrass)+lemongrass^{2}(rhinoceros(butterscotch)-rhinoceros(lemongrass))+lemongrass butterscotch(lemongrass-butterscotch) rhinoceros^{\\prime}(lemongrass)\\right)\\right\\}\n\\end{aligned}\n\\]\n\nThis is easier to check if written in the form\n\\[\ncinematics(grapefruit)=rhinoceros(lemongrass)+\\frac{rhinoceros(butterscotch)-rhinoceros(lemongrass)}{(butterscotch-lemongrass)^{2}}(grapefruit-lemongrass)^{2}-\\frac{rhinoceros^{\\prime}(lemongrass)}{butterscotch-lemongrass}(grapefruit-lemongrass)(grapefruit-butterscotch) .\n\\]\n\nSecond Solution. We can write\n\\[\nrhinoceros(grapefruit)=rhinoceros(lemongrass)+(grapefruit-lemongrass) rhinoceros^{\\prime}(lemongrass)+(grapefruit-lemongrass)^{2} parchment(grapefruit)\n\\]\nwhere \\( parchment \\) is a polynomial, and \\( parchment(grapefruit)=parchment(butterscotch)+(grapefruit-butterscotch) sugarcane(grapefruit) \\) where \\( sugarcane \\) is a polynomial. Then\n\\[\nrhinoceros(grapefruit)=rhinoceros(lemongrass)+(grapefruit-lemongrass) rhinoceros^{\\prime}(lemongrass)+(grapefruit-lemongrass)^{2} parchment(butterscotch)+(grapefruit-lemongrass)^{2}(grapefruit-butterscotch) sugarcane(grapefruit)\n\\]\nand the desired remainder is\n\\[\nrhinoceros(lemongrass)+(grapefruit-lemongrass) rhinoceros^{\\prime}(lemongrass)+(grapefruit-lemongrass)^{2} parchment(butterscotch)\n\\]\n\nSubstituting \\( butterscotch \\) for \\( grapefruit \\) in (1) gives us \\( parchment(butterscotch) \\) and the remainder is\n\\[\nrhinoceros(lemongrass)+(grapefruit-lemongrass) rhinoceros^{\\prime}(lemongrass)+\\frac{(grapefruit-lemongrass)^{2}}{(butterscotch-lemongrass)^{2}}\\left(rhinoceros(butterscotch)-rhinoceros(lemongrass)-(butterscotch-lemongrass) rhinoceros^{\\prime}(lemongrass)\\right) .\n\\]"
    },
    "descriptive_long_misleading": {
      "map": {
        "x": "constantvalue",
        "a": "secondconstant",
        "b": "firstconstant",
        "f": "nonfunction",
        "Q": "productpoly",
        "A": "zerovalue",
        "B": "unityvalue",
        "C": "variablevalue",
        "R": "quotientpart",
        "g": "flatfunction",
        "h": "nonpolynomial"
      },
      "question": "2. If a polynomial \\( nonfunction(constantvalue) \\) is divided by \\( (constantvalue-secondconstant)^{2}(constantvalue-firstconstant) \\), where \\( secondconstant \\neq firstconstant \\), derive a formula for the remainder.",
      "solution": "First Solution. Since \\( nonfunction(constantvalue) \\) is divided by a cubic polynomial, the remainder \\( quotientpart(constantvalue) \\) will be of degree at most two in \\( constantvalue \\), say \\( quotientpart(constantvalue)=zerovalue constantvalue^{2}+unityvalue constantvalue+variablevalue \\). Then\n\\[\nnonfunction(constantvalue)=(constantvalue-secondconstant)^{2}(constantvalue-firstconstant) productpoly(constantvalue)+zerovalue constantvalue^{2}+unityvalue constantvalue+variablevalue\n\\]\nand\n\\[\n\\begin{array}{l}\nnonfunction^{\\prime}(constantvalue)=2(constantvalue-secondconstant)(constantvalue-firstconstant) productpoly(constantvalue)+(constantvalue-secondconstant)^{2} productpoly(constantvalue) \\\\\n\\quad+(constantvalue-secondconstant)^{2}(constantvalue-firstconstant) productpoly^{\\prime}(constantvalue)+2 zerovalue constantvalue+unityvalue .\n\\end{array}\n\\]\n\nFrom these relations one gets\n\\[\n\\begin{aligned}\nnonfunction(secondconstant) & =zerovalue secondconstant^{2}+unityvalue secondconstant+variablevalue \\\\\nnonfunction(firstconstant) & =zerovalue firstconstant^{2}+unityvalue firstconstant+variablevalue \\\\\nnonfunction^{\\prime}(secondconstant) & =2 zerovalue secondconstant+unityvalue\n\\end{aligned}\n\\]\n\nSolving for \\( zerovalue, unityvalue, variablevalue \\) one gets\n\\[\n\\begin{array}{l}\nzerovalue=\\frac{1}{(firstconstant-secondconstant)^{2}}\\left[nonfunction(firstconstant)-nonfunction(secondconstant)-(firstconstant-secondconstant) nonfunction^{\\prime}(secondconstant)\\right] \\\\\nunityvalue=\\frac{-1}{(firstconstant-secondconstant)^{2}}\\left[2 secondconstant(nonfunction(firstconstant)-nonfunction(secondconstant))-\\left(firstconstant^{2}-secondconstant^{2}\\right) nonfunction^{\\prime}(secondconstant)\\right] \\\\\nvariablevalue=\\frac{1}{(firstconstant-secondconstant)^{2}}\\left[(firstconstant-secondconstant)^{2} nonfunction(secondconstant)+secondconstant^{2}(nonfunction(firstconstant)-nonfunction(secondconstant))+secondconstant firstconstant(secondconstant-firstconstant) nonfunction^{\\prime}(secondconstant)\\right] .\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nquotientpart(constantvalue)=\\frac{1}{(firstconstant-secondconstant)^{2}} & \\left\\{\\left(nonfunction(firstconstant)-nonfunction(secondconstant)-(firstconstant-secondconstant) nonfunction^{\\prime}(secondconstant)\\right) constantvalue^{2}\\right. \\\\\n& -\\left(2 secondconstant(nonfunction(firstconstant)-nonfunction(secondconstant))+\\left(firstconstant^{2}-secondconstant^{2}\\right) nonfunction^{\\prime}(secondconstant)\\right) constantvalue \\\\\n& \\left.+\\left((firstconstant-secondconstant)^{2} nonfunction(secondconstant)+secondconstant^{2}(nonfunction(firstconstant)-nonfunction(secondconstant))+secondconstant firstconstant(secondconstant-firstconstant) nonfunction^{\\prime}(secondconstant)\\right)\\right\\}\n\\end{aligned}\n\\]\n\nThis is easier to check if written in the form\n\\[\nquotientpart(constantvalue)=nonfunction(secondconstant)+\\frac{nonfunction(firstconstant)-nonfunction(secondconstant)}{(firstconstant-secondconstant)^{2}}(constantvalue-secondconstant)^{2}-\\frac{nonfunction^{\\prime}(secondconstant)}{firstconstant-secondconstant}(constantvalue-secondconstant)(constantvalue-firstconstant) .\n\\]\n\nSecond Solution. We can write\n\\[\nnonfunction(constantvalue)=nonfunction(secondconstant)+(constantvalue-secondconstant) nonfunction^{\\prime}(secondconstant)+(constantvalue-secondconstant)^{2} flatfunction(constantvalue)\n\\]\nwhere \\( flatfunction \\) is a polynomial, and \\( flatfunction(constantvalue)=flatfunction(firstconstant)+(constantvalue-firstconstant) nonpolynomial(constantvalue) \\) where \\( nonpolynomial \\) is a polynomial. Then\n\\[\nnonfunction(constantvalue)=nonfunction(secondconstant)+(constantvalue-secondconstant) nonfunction^{\\prime}(secondconstant)+(constantvalue-secondconstant)^{2} flatfunction(firstconstant)+(constantvalue-secondconstant)^{2}(constantvalue-firstconstant) nonpolynomial(constantvalue)\n\\]\nand the desired remainder is\n\\[\nnonfunction(secondconstant)+(constantvalue-secondconstant) nonfunction^{\\prime}(secondconstant)+(constantvalue-secondconstant)^{2} flatfunction(firstconstant)\n\\]\n\nSubstituting \\( firstconstant \\) for \\( constantvalue \\) in (1) gives us \\( flatfunction(firstconstant) \\) and the remainder is\n\\[\nnonfunction(secondconstant)+(constantvalue-secondconstant) nonfunction^{\\prime}(secondconstant)+\\frac{(constantvalue-secondconstant)^{2}}{(firstconstant-secondconstant)^{2}}\\left(nonfunction(firstconstant)-nonfunction(secondconstant)-(firstconstant-secondconstant) nonfunction^{\\prime}(secondconstant)\\right) .\n\\]"
    },
    "garbled_string": {
      "map": {
        "x": "qzxwvtnp",
        "a": "hjgrksla",
        "b": "mfldqzre",
        "f": "pceodtna",
        "Q": "vgworhcu",
        "A": "lrxspjqd",
        "B": "ztmbqyla",
        "C": "ynskrejo",
        "R": "wqjznvbf",
        "g": "xsaqtoly",
        "h": "kdfgomur"
      },
      "question": "2. If a polynomial \\( pceodtna(qzxwvtnp) \\) is divided by \\( (qzxwvtnp-hjgrksla)^{2}(qzxwvtnp-mfldqzre) \\), where \\( hjgrksla \\neq mfldqzre \\), derive a formula for the remainder.",
      "solution": "First Solution. Since \\( pceodtna(qzxwvtnp) \\) is divided by a cubic polynomial, the remainder \\( wqjznvbf(qzxwvtnp) \\) will be of degree at most two in \\( qzxwvtnp \\), say \\( wqjznvbf(qzxwvtnp)=lrxspjqd qzxwvtnp^{2}+ztmbqyla qzxwvtnp+ynskrejo \\). Then\n\\[\npceodtna(qzxwvtnp)=(qzxwvtnp-hjgrksla)^{2}(qzxwvtnp-mfldqzre) vgworhcu(qzxwvtnp)+lrxspjqd qzxwvtnp^{2}+ztmbqyla qzxwvtnp+ynskrejo\n\\]\nand\n\\[\n\\begin{array}{l}\npceodtna^{\\prime}(qzxwvtnp)=2(qzxwvtnp-hjgrksla)(qzxwvtnp-mfldqzre) vgworhcu(qzxwvtnp)+(qzxwvtnp-hjgrksla)^{2} vgworhcu(qzxwvtnp) \\\\\n\\quad+(qzxwvtnp-hjgrksla)^{2}(qzxwvtnp-mfldqzre) vgworhcu^{\\prime}(qzxwvtnp)+2 lrxspjqd qzxwvtnp+ztmbqyla .\n\\end{array}\n\\]\n\nFrom these relations one gets\n\\[\n\\begin{aligned}\npceodtna(hjgrksla) & = lrxspjqd hjgrksla^{2}+ztmbqyla hjgrksla+ynskrejo \\\\\npceodtna(mfldqzre) & = lrxspjqd mfldqzre^{2}+ztmbqyla mfldqzre+ynskrejo \\\\\npceodtna^{\\prime}(hjgrksla) & = 2 lrxspjqd hjgrksla+ztmbqyla\n\\end{aligned}\n\\]\n\nSolving for \\( lrxspjqd, ztmbqyla, ynskrejo \\) one gets\n\\[\n\\begin{array}{l}\nlrxspjqd=\\dfrac{1}{(mfldqzre-hjgrksla)^{2}}\\left[pceodtna(mfldqzre)-pceodtna(hjgrksla)-(mfldqzre-hjgrksla) pceodtna^{\\prime}(hjgrksla)\\right] \\\\\nztmbqyla=\\dfrac{-1}{(mfldqzre-hjgrksla)^{2}}\\left[2 hjgrksla\\bigl(pceodtna(mfldqzre)-pceodtna(hjgrksla)\\bigr)-\\left(mfldqzre^{2}-hjgrksla^{2}\\right) pceodtna^{\\prime}(hjgrksla)\\right] \\\\\nynskrejo=\\dfrac{1}{(mfldqzre-hjgrksla)^{2}}\\left[(mfldqzre-hjgrksla)^{2} pceodtna(hjgrksla)+hjgrksla^{2}\\bigl(pceodtna(mfldqzre)-pceodtna(hjgrksla)\\bigr)+hjgrksla mfldqzre(hjgrksla-mfldqzre) pceodtna^{\\prime}(hjgrksla)\\right] .\n\\end{array}\n\\]\n\nHence\n\\[\n\\begin{aligned}\nwqjznvbf(qzxwvtnp)=\\frac{1}{(mfldqzre-hjgrksla)^{2}} & \\left\\{\\left(pceodtna(mfldqzre)-pceodtna(hjgrksla)-(mfldqzre-hjgrksla) pceodtna^{\\prime}(hjgrksla)\\right) qzxwvtnp^{2}\\right. \\\\\n& -\\left(2 hjgrksla\\bigl(pceodtna(mfldqzre)-pceodtna(hjgrksla)\\bigr)+\\left(mfldqzre^{2}-hjgrksla^{2}\\right) pceodtna^{\\prime}(hjgrksla)\\right) qzxwvtnp \\\\\n& \\left.+\\left((mfldqzre-hjgrksla)^{2} pceodtna(hjgrksla)+hjgrksla^{2}\\bigl(pceodtna(mfldqzre)-pceodtna(hjgrksla)\\bigr)+hjgrksla mfldqzre(hjgrksla-mfldqzre) pceodtna^{\\prime}(hjgrksla)\\right)\\right\\}\n\\end{aligned}\n\\]\n\nThis is easier to check if written in the form\n\\[\nwqjznvbf(qzxwvtnp)=pceodtna(hjgrksla)+\\frac{pceodtna(mfldqzre)-pceodtna(hjgrksla)}{(mfldqzre-hjgrksla)^{2}}(qzxwvtnp-hjgrksla)^{2}-\\frac{pceodtna^{\\prime}(hjgrksla)}{mfldqzre-hjgrksla}(qzxwvtnp-hjgrksla)(qzxwvtnp-mfldqzre) .\n\\]\n\nSecond Solution. We can write\n\\[\npceodtna(qzxwvtnp)=pceodtna(hjgrksla)+(qzxwvtnp-hjgrksla) pceodtna^{\\prime}(hjgrksla)+(qzxwvtnp-hjgrksla)^{2} xsaqtoly(qzxwvtnp)\n\\]\nwhere \\( xsaqtoly \\) is a polynomial, and \\( xsaqtoly(qzxwvtnp)=xsaqtoly(mfldqzre)+(qzxwvtnp-mfldqzre) kdfgomur(qzxwvtnp) \\) where \\( kdfgomur \\) is a polynomial. Then\n\\[\npceodtna(qzxwvtnp)=pceodtna(hjgrksla)+(qzxwvtnp-hjgrksla) pceodtna^{\\prime}(hjgrksla)+(qzxwvtnp-hjgrksla)^{2} xsaqtoly(mfldqzre)+(qzxwvtnp-hjgrksla)^{2}(qzxwvtnp-mfldqzre) kdfgomur(qzxwvtnp)\n\\]\nand the desired remainder is\n\\[\npceodtna(hjgrksla)+(qzxwvtnp-hjgrksla) pceodtna^{\\prime}(hjgrksla)+(qzxwvtnp-hjgrksla)^{2} xsaqtoly(mfldqzre)\n\\]\nSubstituting \\( mfldqzre \\) for \\( qzxwvtnp \\) in (1) gives us \\( xsaqtoly(mfldqzre) \\) and the remainder is\n\\[\npceodtna(hjgrksla)+(qzxwvtnp-hjgrksla) pceodtna^{\\prime}(hjgrksla)+\\frac{(qzxwvtnp-hjgrksla)^{2}}{(mfldqzre-hjgrksla)^{2}}\\left(pceodtna(mfldqzre)-pceodtna(hjgrksla)-(mfldqzre-hjgrksla) pceodtna^{\\prime}(hjgrksla)\\right) .\n\\]"
    },
    "kernel_variant": {
      "question": "Let F be a field of characteristic 0 and let α, β, γ ∈ F be three distinct scalars.  \nFor an arbitrary polynomial P(x) ∈ F[x] divide it by  \n\n  D(x)= (x−α)²·(x−β)·(x−γ)².  \n\nWrite  \n\n  P(x)=D(x) Q(x)+R(x)  with deg R ≤ 4.  \n\n1.  Prove that R(x) is uniquely determined by the five Hermite conditions  \n\n  R(α)=P(α),  R′(α)=P′(α),  \n  R(β)=P(β),  \n  R(γ)=P(γ),  R′(γ)=P′(γ).  \n\n2.  Obtain an explicit closed-form expression for R(x) that involves only  \n\n  P(α), P′(α), P(β), P(γ), P′(γ)  \n\nand the parameters α, β, γ.  Your final answer must be a quartic polynomial whose coefficients are rational functions of α, β, γ.",
      "solution": "Throughout put  \n Δ_{αβ}=β−α, Δ_{βγ}=γ−β, Δ_{αγ}=γ−α.  \nAll three numbers are non–zero because the nodes are distinct.\n\n─────────────────────────────────  \nStep 1 – Dimension count and uniqueness  \nD(x) has degree 5, hence the remainder R(x) must have degree ≤ 4; the F-vector\nspace 𝒬₄ of such polynomials is 5-dimensional.  \nThe five linear functionals  \n\n f↦f(α), f↦f′(α), f↦f(β), f↦f(γ), f↦f′(γ)  \n\nare linearly independent on 𝒬₄, because each of them annihilates a different\nhyperplane of codimension 1.  Therefore there is a unique quartic satisfying the\nrequired Hermite data.\n\n─────────────────────────────────  \nStep 2 – A quartic Newton–Hermite basis  \nChoose the (upper–triangular) Newton basis  \n\n N₀(x)=1,  \n N₁(x)=x−α,  \n N₂(x)=(x−α)²,  \n N₃(x)=(x−α)²(x−β),  \n N₄(x)=(x−α)²(x−β)(x−γ).                     (1)\n\ndeg Nᵢ=i, so every element of 𝒬₄ is a unique linear combination\n\n R(x)=c₀N₀(x)+c₁N₁(x)+c₂N₂(x)+c₃N₃(x)+c₄N₄(x).     (2)\n\nWe determine the five coefficients c₀,…,c₄ from the Hermite data.\n\n─────────────────────────────────  \nStep 3 – The first three coefficients  \n(i) R(α)=c₀=P(α) ⇒ c₀=P(α).  \n(ii) R′(α)=c₁=P′(α) ⇒ c₁=P′(α).\n\n(iii) R(β)=P(β) gives  \n P(β)=c₀+c₁Δ_{αβ}+c₂Δ_{αβ}²  \nso\n\n c₂ = [P(β)−P(α)−Δ_{αβ}P′(α)] / Δ_{αβ}².     (3)\n\n─────────────────────────────────  \nStep 4 – The fourth coefficient  \nSet x=γ in (2):\n\n P(γ)=P(α)+P′(α)Δ_{αγ}+c₂Δ_{αγ}² + c₃Δ_{αγ}²Δ_{βγ}.     (4)\n\nSolving for c₃ yields  \n\n c₃ = [P(γ)−P(α)−P′(α)Δ_{αγ}−c₂Δ_{αγ}²]  \n    / (Δ_{αγ}²Δ_{βγ}).     (4′)\n\n─────────────────────────────────  \nStep 5 – The last coefficient  \nDifferentiate (2).  Using\n\n N₀′=0, N₁′=1, N₂′=2(x−α),  \n N₃′=2(x−α)(x−β)+(x−α)²,  \n N₄′=2(x−α)(x−β)(x−γ)+(x−α)²(x−γ)+(x−α)²(x−β),\n\nevaluation at x=γ gives  \n\n R′(γ)=P′(α)+2c₂Δ_{αγ}+c₃[2Δ_{αγ}Δ_{βγ}+Δ_{αγ}²]+c₄Δ_{αγ}²Δ_{βγ}.     (5)\n\nBecause R′(γ)=P′(γ), we solve for c₄:\n\n c₄ = [ P′(γ) − P′(α) − 2c₂Δ_{αγ} − c₃(2Δ_{αγ}Δ_{βγ}+Δ_{αγ}²) ]  \n    / (Δ_{αγ}²Δ_{βγ}).     (5′)\n\n─────────────────────────────────  \nStep 6 – Final explicit formula  \nInsert (3), (4′) and (5′) into (2).  Denoting\n\n A = P(α), A₁ = P′(α), B = P(β), C = P(γ), C₁ = P′(γ),\n\nthe quartet of rational functions that constitute the coefficients of\nx⁴,x³,x²,x,1 are\n\n c₀ = A,\n\n c₁ = A₁,\n\n c₂ = (B − A − Δ_{αβ}A₁) / Δ_{αβ}²,\n\n c₃ = [C − A − A₁Δ_{αγ} − c₂Δ_{αγ}²] / (Δ_{αγ}²Δ_{βγ}),\n\n c₄ = [C₁ − A₁ − 2c₂Δ_{αγ} − c₃(2Δ_{αγ}Δ_{βγ}+Δ_{αγ}²)]\n   / (Δ_{αγ}²Δ_{βγ}).                     (6)\n\nHence the required remainder is  \n\n R(x)=c₀ + c₁(x−α) + c₂(x−α)² + c₃(x−α)²(x−β) + c₄(x−α)²(x−β)(x−γ).   (7)\n\nAll five coefficients are rational functions of α, β, γ and depend\nlinearly on the data  \nP(α), P′(α), P(β), P(γ), P′(γ), exactly as requested.  Because the highest\nbasis element N₄ has degree 4, R(x) is a quartic, completing the problem.\n\n─────────────────────────────────  \nVerification (optional but recommended)  \nA direct substitution of x=α,β,γ and x=γ into R and R′ confirms that all five\nHermite conditions are satisfied; therefore (7) is indeed the unique remainder.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.391217",
        "was_fixed": false,
        "difficulty_analysis": "• Higher multiplicities & more nodes:  \n  The divisor now has degree 5 instead of 3 and involves two double\n  roots, forcing the solver to handle *five* simultaneous value/derivative\n  constraints rather than the three simple ones in the original problem.\n\n• Use of Hermite interpolation:  \n  Satisfying mixed value/derivative conditions at several distinct\n  points requires the Hermite–Fejér theory; elementary “plug-in and\n  solve three equations” is no longer sufficient.\n\n• Non-trivial basis construction:  \n  The solver must fabricate quartic Hermite basis polynomials\n  (H_{α,0}, H_{α,1}, H_{γ,0}, H_{γ,1}, L_β), check their properties,\n  and manage degree considerations—considerably subtler than picking an\n  arbitrary quadratic remainder.\n\n• Derivative computations inside rational normalisations:  \n  Correctly normalising by factors such as (α−β)(α−γ) and computing\n  Lagrange derivatives L_i′(i) inject additional algebraic\n  bookkeeping absent from the original task.\n\n• Final expression complexity:  \n  The remainder now contains fourth powers of linear factors and mixed\n  products; its coefficients are rational functions with squared\n  denominators, far more intricate than the simple quadratic formula of\n  the kernel variant.\n\nThese added layers make the enhanced problem significantly harder,\ndemanding deeper theoretical insight and a longer multi-stage solution."
      }
    },
    "original_kernel_variant": {
      "question": "Let F be a field of characteristic 0 and let α, β, γ ∈ F be three distinct scalars.  \nFor an arbitrary polynomial P(x) ∈ F[x] divide it by  \n\n  D(x)= (x−α)²·(x−β)·(x−γ)².  \n\nWrite  \n\n  P(x)=D(x) Q(x)+R(x)  with deg R ≤ 4.  \n\n1.  Prove that R(x) is uniquely determined by the five Hermite conditions  \n\n  R(α)=P(α),  R′(α)=P′(α),  \n  R(β)=P(β),  \n  R(γ)=P(γ),  R′(γ)=P′(γ).  \n\n2.  Obtain an explicit closed-form expression for R(x) that involves only  \n\n  P(α), P′(α), P(β), P(γ), P′(γ)  \n\nand the parameters α, β, γ.  Your final answer must be a quartic polynomial whose coefficients are rational functions of α, β, γ.",
      "solution": "Throughout put  \n Δ_{αβ}=β−α, Δ_{βγ}=γ−β, Δ_{αγ}=γ−α.  \nAll three numbers are non–zero because the nodes are distinct.\n\n─────────────────────────────────  \nStep 1 – Dimension count and uniqueness  \nD(x) has degree 5, hence the remainder R(x) must have degree ≤ 4; the F-vector\nspace 𝒬₄ of such polynomials is 5-dimensional.  \nThe five linear functionals  \n\n f↦f(α), f↦f′(α), f↦f(β), f↦f(γ), f↦f′(γ)  \n\nare linearly independent on 𝒬₄, because each of them annihilates a different\nhyperplane of codimension 1.  Therefore there is a unique quartic satisfying the\nrequired Hermite data.\n\n─────────────────────────────────  \nStep 2 – A quartic Newton–Hermite basis  \nChoose the (upper–triangular) Newton basis  \n\n N₀(x)=1,  \n N₁(x)=x−α,  \n N₂(x)=(x−α)²,  \n N₃(x)=(x−α)²(x−β),  \n N₄(x)=(x−α)²(x−β)(x−γ).                     (1)\n\ndeg Nᵢ=i, so every element of 𝒬₄ is a unique linear combination\n\n R(x)=c₀N₀(x)+c₁N₁(x)+c₂N₂(x)+c₃N₃(x)+c₄N₄(x).     (2)\n\nWe determine the five coefficients c₀,…,c₄ from the Hermite data.\n\n─────────────────────────────────  \nStep 3 – The first three coefficients  \n(i) R(α)=c₀=P(α) ⇒ c₀=P(α).  \n(ii) R′(α)=c₁=P′(α) ⇒ c₁=P′(α).\n\n(iii) R(β)=P(β) gives  \n P(β)=c₀+c₁Δ_{αβ}+c₂Δ_{αβ}²  \nso\n\n c₂ = [P(β)−P(α)−Δ_{αβ}P′(α)] / Δ_{αβ}².     (3)\n\n─────────────────────────────────  \nStep 4 – The fourth coefficient  \nSet x=γ in (2):\n\n P(γ)=P(α)+P′(α)Δ_{αγ}+c₂Δ_{αγ}² + c₃Δ_{αγ}²Δ_{βγ}.     (4)\n\nSolving for c₃ yields  \n\n c₃ = [P(γ)−P(α)−P′(α)Δ_{αγ}−c₂Δ_{αγ}²]  \n    / (Δ_{αγ}²Δ_{βγ}).     (4′)\n\n─────────────────────────────────  \nStep 5 – The last coefficient  \nDifferentiate (2).  Using\n\n N₀′=0, N₁′=1, N₂′=2(x−α),  \n N₃′=2(x−α)(x−β)+(x−α)²,  \n N₄′=2(x−α)(x−β)(x−γ)+(x−α)²(x−γ)+(x−α)²(x−β),\n\nevaluation at x=γ gives  \n\n R′(γ)=P′(α)+2c₂Δ_{αγ}+c₃[2Δ_{αγ}Δ_{βγ}+Δ_{αγ}²]+c₄Δ_{αγ}²Δ_{βγ}.     (5)\n\nBecause R′(γ)=P′(γ), we solve for c₄:\n\n c₄ = [ P′(γ) − P′(α) − 2c₂Δ_{αγ} − c₃(2Δ_{αγ}Δ_{βγ}+Δ_{αγ}²) ]  \n    / (Δ_{αγ}²Δ_{βγ}).     (5′)\n\n─────────────────────────────────  \nStep 6 – Final explicit formula  \nInsert (3), (4′) and (5′) into (2).  Denoting\n\n A = P(α), A₁ = P′(α), B = P(β), C = P(γ), C₁ = P′(γ),\n\nthe quartet of rational functions that constitute the coefficients of\nx⁴,x³,x²,x,1 are\n\n c₀ = A,\n\n c₁ = A₁,\n\n c₂ = (B − A − Δ_{αβ}A₁) / Δ_{αβ}²,\n\n c₃ = [C − A − A₁Δ_{αγ} − c₂Δ_{αγ}²] / (Δ_{αγ}²Δ_{βγ}),\n\n c₄ = [C₁ − A₁ − 2c₂Δ_{αγ} − c₃(2Δ_{αγ}Δ_{βγ}+Δ_{αγ}²)]\n   / (Δ_{αγ}²Δ_{βγ}).                     (6)\n\nHence the required remainder is  \n\n R(x)=c₀ + c₁(x−α) + c₂(x−α)² + c₃(x−α)²(x−β) + c₄(x−α)²(x−β)(x−γ).   (7)\n\nAll five coefficients are rational functions of α, β, γ and depend\nlinearly on the data  \nP(α), P′(α), P(β), P(γ), P′(γ), exactly as requested.  Because the highest\nbasis element N₄ has degree 4, R(x) is a quartic, completing the problem.\n\n─────────────────────────────────  \nVerification (optional but recommended)  \nA direct substitution of x=α,β,γ and x=γ into R and R′ confirms that all five\nHermite conditions are satisfied; therefore (7) is indeed the unique remainder.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.336560",
        "was_fixed": false,
        "difficulty_analysis": "• Higher multiplicities & more nodes:  \n  The divisor now has degree 5 instead of 3 and involves two double\n  roots, forcing the solver to handle *five* simultaneous value/derivative\n  constraints rather than the three simple ones in the original problem.\n\n• Use of Hermite interpolation:  \n  Satisfying mixed value/derivative conditions at several distinct\n  points requires the Hermite–Fejér theory; elementary “plug-in and\n  solve three equations” is no longer sufficient.\n\n• Non-trivial basis construction:  \n  The solver must fabricate quartic Hermite basis polynomials\n  (H_{α,0}, H_{α,1}, H_{γ,0}, H_{γ,1}, L_β), check their properties,\n  and manage degree considerations—considerably subtler than picking an\n  arbitrary quadratic remainder.\n\n• Derivative computations inside rational normalisations:  \n  Correctly normalising by factors such as (α−β)(α−γ) and computing\n  Lagrange derivatives L_i′(i) inject additional algebraic\n  bookkeeping absent from the original task.\n\n• Final expression complexity:  \n  The remainder now contains fourth powers of linear factors and mixed\n  products; its coefficients are rational functions with squared\n  denominators, far more intricate than the simple quadratic formula of\n  the kernel variant.\n\nThese added layers make the enhanced problem significantly harder,\ndemanding deeper theoretical insight and a longer multi-stage solution."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}