How do you divide $\frac{6 x^{2} - 4 x - 24}{2 x + 2}$ $( 6x^2 - 4x - 24 )/(2x+2)$ ?

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How do you divide $\frac{6 x^{2} - 4 x - 24}{2 x + 2}$ $( 6x^2 - 4x - 24 )/(2x+2)$ ?

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Answer 1 · 2015-12-28T22:02:47

$\frac{6 x^{2} - 4 x - 24}{2 x + 2} = \frac{2 (3 x^{2} - 2 x - 12)}{2 (x + 1)}$ $(6x^2-4x-24)/(2x+2)=(2(3x^2-2x-12))/(2(x+1))$
$= \frac{3 x^{2} - 2 x - 12}{x + 1}$ $=(3x^2-2x-12)/(x+1)$

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How do you divide $\frac{6 x^{2} - 4 x - 24}{2 x + 2}$ $( 6x^2 - 4x - 24 )/(2x+2)$ ?

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How do you divide $\frac{6 x^{2} - 4 x - 24}{2 x + 2}$ $( 6x^2 - 4x - 24 )/(2x+2)$ ?