Let us divide #x^3-5x^2+px+6# by #x+2# by synthetic division
One Write the coefficients of #x# in the dividend inside an upside-down division symbol.
#color(white)(1)|color(white)(X)1" "color(white)(X)-5color(white)(XX)p" "" "6#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
Two As #x+2=0# gives #x=-2# put #-2# at the left.
#-2|color(white)(X)1" "color(white)(X)-5color(white)(XX)p" "" "6#
#color(white)(xx)|" "color(white)(XX)#
#" "stackrel("—————————————)#
Three Drop the first coefficient of the dividend below the division symbol.
#-2|color(white)(X)1" "color(white)(X)-5color(white)(XX)p" "" "6#
#color(white)(xx)|" "color(white)(X)#
#" "stackrel("—————————————)#
#color(white)(xx)|color(white)(X)color(red)1#
Four Multiply the result by the constant, and put the product in the next column.
#-2|color(white)(X)1" "color(white)(X)-5color(white)(XX)p" "" "6#
#color(white)(xx)|" "color(white)(xxX)-2#
#" "stackrel("—————————————)#
#color(white)(xx)|color(white)(X)color(blue)1#
Five Add down the column.
#-2|color(white)(X)1" "color(white)(X)-5color(white)(XX)p" "" "6#
#color(white)(xx)|" "color(white)(xXX)-2#
#" "stackrel("—————————————)#
#color(white)(xx)|color(white)(X)color(blue)1color(white)(X11)color(red)-7#
Six Repeat Steps Four and Five until you can go no farther.
#-2|color(white)(X)1" "color(white)(X)-5color(white)(XX)p" "" "color(white)(XX)6#
#color(white)(xx)|" "color(white)(XXx)-2color(white)(xx)14color(white)(X)-2p-28#
#" "stackrel("—————------------————————)#
#color(white)(xx)|color(white)(X)color(blue)1color(white)(XX)color(red)-7color(white)(XX)color(red)(p+14)color(white)()color(red)((-2p-22))#
Hence, Quotient is #x^2-7x+p+14# and remainder is #-2p-22#.
We can also work out remainder using remainder theorem, which gives remainder as #f(-2)# i.e.
#f(-2)=(-2)^3-5(-2)^2+p(-2)+6=-8-20-2p+6=-2p-22#
