How do you long divide # (x^2+4x+4) / (x+2) #?

1 Answer
Jul 16, 2015

It is a challenge to format and there are various ways of writing the details. Here's the one that I learned and teach (using a standard textbook in the US):

Explanation:

To divide # (x^2 +4x + 4)/(x+2)#

Write the divisor, #x+2# on the left. Then the dividend, #x^2+4x+4# under the division thing. (I don't know its name.)

#" " " "##-----#
#x+2 )# #x^2# #+4x# #+4#

Look at the first terms on the divisor #x+2# and the dividend #x^2+4x+4#. What do we need to multiply #x# by, to get #x^2#? We need to multiply by #x#. Write that #x# on the top line:

#" " " " " " "x#
#" " " "##-----#
#x+2 )# #x^2# #+4x# #+4#

Now multiply #x# times the divisor, #x+2#, to get #x^2+2x# and write that under the dividend.

#" " " " " " "# #x#
#" " " "##-----#
#x+2 )# #x^2##" "# #+4x##" "# #+4#
#" " " " "" "# #x^2# #" " ##+2x#
#" " " "##-----#

Now we need to subtract #x^2+2x# from the dividend. (You may find it simpler to change the signs and add.)

#" " " " " " "# #x#
#" " " "##-----#
#x+2 )# #x^2# #" "##+4x# #" "##+4#
#" " " "# #color(red)(-)x^2color(red)(-)2x#
#" " " "##-----#
#" "" "" "" "" " # #2x##" "# #+4#

Now, look at the first terms of the divisor and the expression on the last line. What do we need to multiply #x# (the first term of the divisor) by to get #2x# (the first term on the last line)? We need to multiply by #2#
Write #+2# on the top line, then multiply #2# times the divisor #x+2#, to get #2x+4# and write it underneath.

#" " " " " " "# #x##" "# #+2#
#" " " "##-----#
#x+2 )# #x^2# #" "##+4x# #" "##+4#
#" " " " # #color(red)(-)x^2color(red)(-)2x#
#" " " "##------#
#" " " " " "" "# #2x##" "# #+4#
#" " " " " " " "# #2x##" "# #+4#
#" " " "##------#

Now subtract (change the signs and add), to get:

#" " " " " " "# #x##" "# #+2#
#" " " "##-----#
#x+2 )# #x^2# #" "##+4x# #" "##+4#
#" " " " # #color(red)(-)x^2color(red)(-)2x#
#" " " "##------#
#" " " " " "" "# #2x##" "# #+4#
#" " " " " "# #color(red)(-)2x##" "# #color(red)(-)4#
#" " " "##------#
#" " " " " " " " " "" " " " # #0#

The quotient is #x+2# and the remainder is #0#.