How do you divide #(-2x^3+7x^2+9x+15)/(3x-1) #?

1 Answer
Tony B
Jun 24, 2016

#=>-2/3x^2+19/2x+100/27+505/(81x-27)#

Explanation:

#" "color(white)(..)-2x^3+7x^2+9x+15#
#" "color(magenta)(-2/3x^2)(3x-1) ->color(white)(.)ul( -2x^3+2/3x^2)" "larr" Subtract"#
#" "color(white)(.)0+19/3x^2+9x+15#
#" "color(magenta)(+19/9x)(3x-1)-> " "color(white)(.)ul(+19/3x^2-19/9x)larr" Subtract"#
#" "color(white)(..)0+100/9x+15#
#" "color(magenta)(+100/27)(3x-1)->" " ul(+100/9x-100/27)"Subt."#
#" remainder "rarr 0 color(white)(....) color(magenta)(+505/27)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(-2x^3+7x^2+9x+15)/(3x-1) = color(magenta)(-2/3x^2+19/9x+100/27+[505/27-:(3x-1)]#

#=>-2/3x^2+19/2x+100/27+505/(81x-27)#

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