How do you divide `(x^3-3x-2) / (x+8)` using polynomial long division?

Using different variables but the approach is the same; have a look at:`" "` https://socratic.org/s/auC4VyMH

Solution without as much explanation as shown on the hyperlinked page.

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`" "x^2-8x+61`
`x+8" "bar("| "x^3+0x^2-3x-2)`
`" "color(blue)(underline(x^3+8x^2 )" " larrx^2(x+8) " subtract")`
`" "color(brown)(0-8x^2-3x-2" " larr "Remainder")`
`" "color(blue)(underline(-8x^2-64x )" "larr -8x(x+8)" subtract")`
`" "color(brown)(0+61x-2" "larr" Remainder")`
`" "color(blue)(underline(61x+488)" "larr" "61(x+8)" subtract")`
`" "color(brown)(0-490" "larr" Remainder")`

`color(brown)("remainder "->-490/(x+8))`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`color(blue)((x^3-3x-2)/(x+8)" " =" " x^2-8x+61 -490/(x+8))`.....(1)

'@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
'@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
`color(magenta)("Check")`
`x^2-8x+61-490/(x+8)`

`underline(" "x+8)" "larr" multiply"`
`x^3-8x+61x+0-(490x)/(x+8)" "larr" multiplied by "x`

`underline(" "+8x^2-64x+488+0-3920/(x+8))" "larr" multiplied by 8"`
`color(green)(x^3+0x^2-3x+488-(490x)/(x+8)-3920/(x+8))`...(2)

'........................................

Consider `-(490x)/(x+8)-3920/(x+8)`

`(-490(x-8))/(x-8)=-490`
'...............................................
So equation (2) becomes

`color(green)(x^3+0x^2-3x+488-490`

`color(green)(x^3-3x-2)`

`color(red)("Solution is correct")`