How do you divide #(x^3-7x-6) / (x+1) #?

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How do you divide #(x^3-7x-6) / (x+1) #?

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Answer 1 · 2016-04-16T03:25:59

#(x^2(x+1)-x^2-7x-6)/(x+1)#
#=(x^2(x+1)-x(x+1)-6x-6)/(x+1)#
#=(x^2(x+1)-x(x+1)-6(x+1))/(x+1)#
#=(x^2(x+1))/(x+1)-(x(x+1))/(x+1)-(6(x+1))/(x+1)#
#=x^2-x-6#

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How do you divide #(x^3-7x-6) / (x+1) #?

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