Question

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

Answer 1

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)`