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

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

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Answer 1 · 2016-04-16T03:25:59

$\frac{x^{2} (x + 1) - x^{2} - 7 x - 6}{x + 1}$ $(x^2(x+1)-x^2-7x-6)/(x+1)$
$= \frac{x^{2} (x + 1) - x (x + 1) - 6 x - 6}{x + 1}$ $=(x^2(x+1)-x(x+1)-6x-6)/(x+1)$
$= \frac{x^{2} (x + 1) - x (x + 1) - 6 (x + 1)}{x + 1}$ $=(x^2(x+1)-x(x+1)-6(x+1))/(x+1)$
$= \frac{x^{2} (x + 1)}{x + 1} - \frac{x (x + 1)}{x + 1} - \frac{6 (x + 1)}{x + 1}$ $=(x^2(x+1))/(x+1)-(x(x+1))/(x+1)-(6(x+1))/(x+1)$
$= x^{2} - x - 6$ $=x^2-x-6$

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

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