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

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Answer 1 · 2018-04-09T13:36:35

Quotient is $x^{2} - x - 6$ $x^2-x-6$ and remainder is $0$ $0$

$x^{3} - 7 x - 6$ $x^3-7x-6$

= $x^{3} + x^{2} - x^{2} - x - 6 x - 6$ $x^3+x^2-x^2-x-6x-6$

= $x^{2} \cdot (x + 1) - x \cdot (x + 1) - 6 \cdot (x + 1)$ $x^2*(x+1)-x*(x+1)-6*(x+1)$

= $(x^{2} - x - 6) \cdot (x + 1)$ $(x^2-x-6)*(x+1)$

Hence quotient is $x^{2} - x - 6$ $x^2-x-6$ and remainder is $0$ $0$

How do you divide (x^3 - 7x - 6)/(x+1)x3−7x−6x+1(x^3 - 7x - 6)/(x+1)?