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"
