The Critical Points

$This\quad is\quad the\quad solution\quad I\quad used\quad in\quad solving\quad the\quad math\quad problem.\\ f\left( x \right) ={ \quad x }^{ 2 }-4x-1\\ f^{ \quad \prime }\left( x \right) =\quad 2x-4\\ 0=\quad 2x-4\\ 2x=4\\ x=2\\ \\ Using\quad x=2:\\ f(2)=\quad 4-8-1\\ f(2)=\quad -5\\ The\quad critical\quad point\quad of\quad f(x)\quad is\quad (2,-5).\\ \\ g(x)\quad =\quad { x }^{ 3 }-{ 9x }^{ 2 }+15x-5\\ g^{ \prime }\left( x \right) =\quad 3{ x }^{ 2 }-18x+15\\ 0=\quad (3x-3)(x-5)\\ x=1,\quad x=5\\ Using\quad x=1:\\ g(1)\quad =\quad 1-9+15-5\\ g(1)\quad =\quad 2\\ Using\quad x=5\\ g(5)\quad =\quad 125-225+75-5\\ g(5)\quad =\quad -30\\ The\quad critical\quad points\quad of\quad g(x)\quad are\quad (1,2)\quad and\quad (5,-30).\\ \\ Since\quad A\quad is\quad the\quad sum\quad of\quad all\quad x-coordinates\quad of\quad the\quad critical\quad points\quad of\quad f(x)\quad and\quad g(x),\quad \\ then:\\ A=2+1+5\\ A=\quad 8\\ Since\quad B\quad is\quad the\quad sum\quad of\quad all\quad y-coordinates\quad of\quad the\quad critical\quad points\quad of\quad f(x)\quad and\quad g(x),\\ then:\\ B=\quad -5+2-30\\ B=\quad -33\\ \\ For\quad the\quad zeroes\quad of\quad h(x):\\ h(x)=\frac { A{ x }^{ 2 } }{ 2 } \quad -(B-3)x\quad -(B+1)\\ h(x)=\quad { 4x }^{ 2 }\quad +36x\quad +32\\ 0={ \quad 4x }^{ 2 }\quad +36x\quad +32\\ 0=\quad (4x+4)(x+8)\\ x=\{ -1,-8\} \\ Since\quad the\quad zeroes\quad of\quad j(x)\quad are\quad the\quad double\quad of\quad the\quad zeroes\quad of\quad h(x),\quad \\ then\quad -2\quad and\quad -16\quad are\quad the\quad zeroes\quad of\quad j(x).\\ Let:\\ { r }_{ 1 }\quad =\quad 2\quad and\quad { r }_{ 2 }\quad =\quad -16\\ j(x)\quad =\quad (x-{ r }_{ 1 })(x-{ r }_{ 2 })\\ j(x)\quad =\quad (x+2)(x+16)\\ j(x)\quad =\quad { x }^{ 2 }\quad +18x\quad +32,\quad where\quad a=1,\quad b=18,\quad and\quad c=32.\\ For\quad a+b+c:\\ a+b+c=1+18+32\\ a+b+c=51\\ \\ Conjecture:\\ The\quad sum\quad of\quad a+b+c\quad of\quad the\quad standard\quad form\quad of\quad j(x)\quad is\quad 51.//$