How do you long divide $(x^3-7x-6) div (x+1)$ ?

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Answer 1 · 2016-05-23T04:05:52

The nominator can be factored as

$x^3-7x-6=x^3-6x-x-6= x^3-x-6(x+1)= x(x-1)(x+1)-6(x+1)= (x+1)*[x(x-1)+6]$

Hence

$(x^3-7x-6)/(x+1)=(x+1)*[x(x-1)+6]/(x+1)= x*(x-1)+6=x^2-x+6$

How do you long divide (x^3-7x-6) div (x+1) (x^3-7x-6) div (x+1) ?