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

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