The method is very easy, but the process is a bit difficult to explain.
Follow the colours.
#(n^3+6n^2+4n-2)div(n+1) = ?????????#
#" (dividend) " div " (divisor)" = ("quotient")#
#color(magenta)("step 1:")# The dividend must be in descending powers of n.
#color(white)(xxxxxxxxxxx)n^3+6n^2+4n-2#
#color(white)(xxxxxxxxx) rArr color(magenta)(1" +6 +4 -2")#
In the dividing use only the numerical coefficients #color(magenta)("(top row)")darr#.
(If there are any missing, leave a space or fill in a zero).
#color(orange)("Step 2")#: Make the divisor = 0. # " " (n+1) = 0 rArr n = color(orange)(-1) " this goes outside the box"#
Step 3 : Begin the division - see details below....
#color(white)(xxxxx) | color(brown)(1)" "+6" "+4 " "-2 color(magenta)(" step 1 top row")#
#color(white)(x.x)color(orange)(-1) ""| darr " "color(red)(-1) " "color(blue)(-5) " "color(olive)(+1)#
#color(white)(xxxxxx) ul(" ")#
#color(white)(xxxxxxx) color(brown)(1) " "color(blue)(+5) " "color(olive)(-1)" "color(teal)(-1) larr " the remainder!"#
#color(white)(xxxx.xx)uarr " "uarr " "uarr#
#color(white)(xxxxxxx) n^2 " "n^1 " "n^0#
Dividing details
#"Bring down the " color(brown)( 1 ) " to below the line"#
#"multiply " color(orange)(-1) xx color(brown)(1) = color(red)(-1)#
#"Add " +6color(red)(-1) = color(blue)(+5)#
#"multiply " color(orange)(-1) xx color(blue)(+5) = color(blue)(-5)#
#"Add " 4 color(blue)( -5) = color(olive)(-1)#
#"multiply " color(orange)(-1) xxcolor(olive)(-1) = color(olive)(+1)#
#"Add " -2 +color(olive)(1) = color(teal)(-1)#
That's it Folks!
We have now found the numerical coefficients of the terms in the quotient (answer)
We divided an expression with #n^3# by an expression with #n#,
so the first term will be #n^3/n = n^2#
The last value is the remainder. In this case it is #color(teal)(-1)#
This means that #(n+1)# is not a factor of #n^3+6n^2+4n-2#
#(n^3+6n^2+4n-2) div(n+1) = n^2+5n -1" rem -1"#
