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#