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

1 Answer
Guillaume L.
Jun 25, 2018

#(2x³+4x²+2x+2)/(x²+4x-2)=2x-4+(22x-6)/(x²+4x-2)#

Explanation:

#(2x³+4x²+2x+2)/(x²+4x-2)#
#=2(x³+2x²+x+1)/(x²+4x-2)#
#=2(x³+4x²-2x-2x²+3x+1)/(x²+4x-2)#
#=2x(cancel((x²+4x-2)/(x²+4x-2)))^(=1)+2(-2x²+3x+1)/(x²+4x-2)#
#=2x+2(-2x²-8x+4+11x-3)/(x²+4x-2)#
#=2x+4(cancel((-x²-4x+2)/(x²+4x-2)))^(=-1)+(22x-6)/(x²+4x-2)#
#=2x-4+(22x-6)/(x²+4x-2)#
\0/ here's our answer !

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