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 ..

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