How do you divide $\frac{2 x^{3} + 7 x^{2} - 5 x - 4}{2 x + 1}$ $(2x^3+7x^2-5x-4) / (2x+1)$ ?

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How do you divide $\frac{2 x^{3} + 7 x^{2} - 5 x - 4}{2 x + 1}$ $(2x^3+7x^2-5x-4) / (2x+1)$ ?

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Answer 1 · 2016-02-28T06:27:50

$\frac{2 x^{3} + 7 x^{2} - 5 x - 4}{2 x + 1}$ $(2x^3+7x^2-5x-4)/(2x+1)$
$= \frac{x^{2} (2 x + 1) - x^{2} + 7 x^{2} - 5 x - 4}{2 x + 1}$ $=(x^2(2x+1)-x^2+7x^2-5x-4)/(2x+1)$
$= \frac{x^{2} (2 x + 1) + 6 x^{2} - 5 x - 4}{2 x + 1}$ $=(x^2(2x+1)+6x^2-5x-4)/(2x+1)$
$= \frac{x^{2} (2 x + 1) + 3 x (2 x + 1) - 3 x - 5 x - 4}{2 x + 1}$ $=(x^2(2x+1)+3x(2x+1)-3x-5x-4)/(2x+1)$
$= \frac{x^{2} (2 x + 1) + 3 x (2 x + 1) - 8 x - 4}{2 x + 1}$ $=(x^2(2x+1)+3x(2x+1)-8x-4)/(2x+1)$
$= \frac{x^{2} (2 x + 1) + 3 x (2 x + 1) - 4 (2 x + 1)}{2 x + 1}$ $=(x^2(2x+1)+3x(2x+1)-4(2x+1))/(2x+1)$

$= x^{2} + 3 x - 4$ $=x^2+3x-4$

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How do you divide $\frac{2 x^{3} + 7 x^{2} - 5 x - 4}{2 x + 1}$ $(2x^3+7x^2-5x-4) / (2x+1)$ ?

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How do you divide $\frac{2 x^{3} + 7 x^{2} - 5 x - 4}{2 x + 1}$ $(2x^3+7x^2-5x-4) / (2x+1)$ ?