Question

What is the completely factored form of f(x)=x^3-7x^2+2x+4

Answer 1

`Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)`

`Steps:`

`x^3-7x^2+2x+4`

`Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm (1,\:2,\:4)/(1),\\(1)/(1)\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)(x^3-7x^2+2x+4)/(x-1)`

`(x^3-7x^2+2x+4)/(x-1)`

`\mathrm{Divide}\:(x^3-7x^2+2x+4)/(x-1),\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}(x^3)/(x),\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\(x^3-7x^2+2x+4)/(x-1)=x^2+(-6x^2+2x+4)/(x-1)`

`\mathrm{Divide}\:(-6x^2+2x+4)/(x-1),\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}(-6x^2)/(x)=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\(-6x^2+2x+4)/(x-1)=-6x+(-4x+4)/(x-1)`

`\mathrm{Divide}\:(-4x+4)/(x-1),\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}(-4x)/(x)=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\(-4x+4)/(x-1)=-4`

`\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}`

`\mathrm{Hope\:This\:Helps!!!}`

`\mathrm{-Austint1414}`