How do you long divide #(9x^3 + 3x^2 + 24x)/(3x^2 + 1)#?

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1 Answer
Tony B
Jun 19, 2016

Due to formatting issues I have my own way of showing this on Socratic

#x+1+(23-1)/(3x^2+1)#

Explanation:

I have included a place holder for the missing constant. I also insert place holders for missing terms as they occur to make the maths line up properly.

#(9x^3+3x^2+24x+0) -: (3x^2+1)#

#" "9x^3+3x^2+24x+0#
#color(magenta)(+x)(3x^2+1)" "->ul(3x^3+0x^2+x)" "larr" subtract"#
#" "0+3x^2+23x+0#
#color(magenta)(+1)(3x^2+1)" "->" "ul(3x^2+color(white)(.)0x+1)" "larr" subtract"#
#" "color(magenta)(0+23x-1)" " larr" remainder"#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(9x^3+3x^2+24x+0) -: (3x^2+1) = color(magenta)(x+1+(23-1)/(3x^2+1)#

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