How do you divide #(x^3+5x^2+7x+2)div(x+2)# using synthetic division?

1 Answer
Jan 11, 2017

Answer:

Quotient is #x^2+3x+1# and remainder is #0#

Explanation:

To divide #x^3+5x^2+7x+2# by #x+2#

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)7" "" "2#
#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)7" "" "2#
#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)7" "" "2#
#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)7" "" "2#
#color(white)(xx)|" "color(white)(Xx)-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)7" "" "2#
#color(white)(xx)|" "color(white)(Xx)-2#
#" "stackrel("—————————————)#
#color(white)(xx)|color(white)(X)color(blue)1color(white)(X11)color(red)3#

Six Repeat Steps Four and Five until you can go no farther.

#-2|color(white)(X)1" "color(white)(X)5color(white)(XX)7" "" "2#
#color(white)(xx)|" "color(white)(X)-2color(white)(X)-6color(white)(X)-2#
#" "stackrel("—————————————)#
#color(white)(xx)|color(white)(X)color(blue)1color(white)(X11)color(red)3color(white)(XX)color(red)1color(white)(XXX)color(red)0#

Hence, Quotient is #x^2+3x+1# and remainder is #0#.