How do you divide $\frac{x^{3} - 2 x - 3}{x^{2} - x - 6}$ $(x^3 - 2x - 3)/( x^2 - x - 6)$ ?

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How do you divide $\frac{x^{3} - 2 x - 3}{x^{2} - x - 6}$ $(x^3 - 2x - 3)/( x^2 - x - 6)$ ?

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Answer 1 · 2016-04-22T17:16:51

I tried long division to get:
$\frac{x^{3} - 2 x - 3}{x^{2} - x - 6} = (x^{2} - x - 6) (x + 1) + (5 x + 3)$ $(x^3-2x-3)/(x^2-x-6)=(x^2-x-6)(x+1)+(5x+3)$

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How do you divide $\frac{x^{3} - 2 x - 3}{x^{2} - x - 6}$ $(x^3 - 2x - 3)/( x^2 - x - 6)$ ?

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How do you divide (x^3 - 2x - 3)/( x^2 - x - 6)x3−2x−3x2−x−6(x^3 - 2x - 3)/( x^2 - x - 6)?

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How do you divide $\frac{x^{3} - 2 x - 3}{x^{2} - x - 6}$ $(x^3 - 2x - 3)/( x^2 - x - 6)$ ?