To divide #(3x^2-2x+1)# by #(x-1)# using synthetic division, we should take following steps
One Write the coefficients of #x# in the dividend inside an upside-down division symbol.
#color(white)(1)|color(white)(X)3" "color(white)(X)-2color(white)(XXXX)1"#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
Two Put the divisor at the left, but for this put #x-1=0#, which gives #x=1#
#1|color(white)(X)3" "color(white)(X)-2color(white)(XXXX)1#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
Three Drop the first coefficient of the dividend below the division symbol.
#1|color(white)(X)3" "color(white)(X)-2color(white)(XXXX)1#
#color(white)(1)|" "color(white)(X)#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)3#
Four Multiply the result of #1xxcolor(blue)3=3#, here #1# comes from divisor, and put the product in the next column.
#1|color(white)(X)3" "color(white)(X)-2color(white)(XXXX)1#
#color(white)(1)|" "color(white)(XXXX)3#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)3#
Five Add down the column.
#1|color(white)(X)3" "color(white)(X)-2color(white)(XXXX)1#
#color(white)(1)|" "color(white)(XXXX)3#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)3color(white)(XXXx)color(blue)1#
Six Repeat Steps Four and Five until you can go no farther.
#1|color(white)(X)3" "color(white)(X)-2color(white)(XXXX)1#
#color(white)(1)|" "color(white)(XXXX)3color(white)(XXXX)1#
#" "stackrel("—————————————)#
#color(white)(1)|color(white)(X)color(blue)3color(white)(XXXx)color(blue)1color(white)(XXXX)color(red)2#
WE find that quotient is #(x+1)# and remainder is #2#.
Hence #(3x^2-2x+1)=color(blue)((3x+1))(x-1)+color(red)2#
