How do you divide #(3x ^ { 3} + 12x ^ { 2} - 21x - 23) \div ( x + 5)#?

1 Answer
Jul 5, 2017

Answer:

#3x^2-3x-6# #r# #7#

Explanation:

Firstly, you divide #3x^3# by #x# to get #3x^2#. Then you do #3x^2(x+5)=3x^2+15x^2#.

#3x^3+12x^2-3x^3-15x^2=-3x^2#, the first value of the quptient is #3x^2#.

Now you do #(-3x^2)/x=-3x#. Then you do #-3x(x+5)=-3x^2-15x#.
#(3x^2-21x)-(-3x^2-15x)=-21x+15x=-6x#. #-3x# is the second part of the quotient.

Now you do #(-6x)/x=-6#. Then you do #-6(x+5)=-6x-30#.
#(-6x-23)-(-6x-30)=-23+30=7#. #-6# is the last part of the quotient. #7# is the remainder.

Visual guide:

#" "3x^2-3x-6# #r# #7#
#(x+5)"/"(3x^3+12x^2-21x-23)#
#" "-(3x^3+15x^2)#
#" "(0-3x^2)#
#" "-(-3x^2-15x)#
#" "(0-6x)#
#" "-(-6x-30)#
#" "(0+7)#