How do you divide #(x^2 + 3x +2)/(1-x)#?

1 Answer
Leland Adriano Alejandro
Feb 26, 2016

#(x^2+3x+2)/(1-x)=-x-4+6/(1-x)#

Explanation:

by long division

Our dividend = #x^2+3x+2#
our divisor #=-x+1#

#" " " " " " " " " underline(-x-4)#
#-x+1" "|~x^2+3x+2#
#" " " " " " " " "underline(x^2-x)#
#" " " " " " " " " " " "4x+2#
#" " " " " " " " " " " "underline(4x-4)#
#" " " " " " " " " " " " " " "+6" " "larr#the remainder

so we write the answer

#("dividend")/("divisor")="quotient"+("remainder")/("divisor")#

#(x^2+3x+2)/(1-x)=-x-4+6/(1-x)#

God bless...I hope the explanation is useful ..

Related questions
See all questions in Division of Polynomials
Impact of this question
1997 views around the world
You can reuse this answer
Creative Commons License