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

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Answer 1 · 2018-04-09T13:36:35

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

$x^3-7x-6$

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

= $x^2*(x+1)-x*(x+1)-6*(x+1)$

= $(x^2-x-6)*(x+1)$

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

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