How do you divide #(2x^3+ 3 x^2-4x-2)/(x+2) #?

1 Answer
Mar 12, 2018

#=(2x^2-x-2)+2/(x+2)#

Explanation:

We know that,
Dividend#-:# Divisor = quotient and remainder.In short,
Dividend = Divisor #xx# quotient + remainder.
Now,divisor is (x+2),and
dividend#=2x^3+3x^2-4x-2##=2x^3+4x^2-x^2-2x-2x-4+2#
#=2x^2(x+2)-x(x+2)-2(x+2)+color(red)(2)#
#=(x+2)(2x^2-x-2)+2#
i.e.quotient=#(2x^2-x-2)#,and remainder=2.
So,
#2x^3+3x^2-4x-2=(x+2)(2x^2-x-2)+2#
Hence,
#(2x^3+3x^2-4x-2)/(x+2)=[(x+2)(2x^2-x-2)+2]/(x+2)#
#=((x+2)(2x^2-x-2))/(x+2)+2/(x+2)#
#=(2x^2-x-2)+color(red)(2/(x+2)#