What does $\frac{x^{3} - 7 x - 6}{x + 1}$ $(x^3-7x-6) / (x+1)$ equal?

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What does $\frac{x^{3} - 7 x - 6}{x + 1}$ $(x^3-7x-6) / (x+1)$ equal?

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Answer 1 · 2015-11-02T17:22:58

Simplify $f (x) = \frac{x^{3} - 7 x - 6}{x + 1}$ $f(x) = (x^3 - 7x - 6)/(x + 1)$

Ans: $x^{2} - x - 6$ $x^2 - x - 6$

Explanation:

First, factor the numerator.
Since (- 1) is a real root, then one factor is (x + 1)
$(x^{3} - 7 x - 6) = (x + 1) (x^{2} - x - 6) .$ $(x^3 - 7x - 6) = (x + 1)(x^2 - x - 6).$
Therefor,
$f (x) = \frac{(x + 1) (x^{2} - x - 6)}{x + 1} = x^{2} - x - 6$ $f(x) = ((x + 1)(x^2 - x - 6))/(x + 1) = x^2 - x - 6$

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What does $\frac{x^{3} - 7 x - 6}{x + 1}$ $(x^3-7x-6) / (x+1)$ equal?

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