How do you divide $\frac{x^{2} - 3}{- 6 x^{4} + 2 x^{2} - 8 x + 1}$ $(x^2-3) / (-6x^4+2x^2-8x+1)$ ?

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How do you divide $\frac{x^{2} - 3}{- 6 x^{4} + 2 x^{2} - 8 x + 1}$ $(x^2-3) / (-6x^4+2x^2-8x+1)$ ?

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Answer 1 · 2017-12-26T01:05:24

$- 2 (3 x^{2} + 8) - \frac{8 x + 47}{- 6 x^{4} + 2 x^{2} - 8 x + 1}$ $-2(3x^2+8)-{8x+47}/{-6x^4+2x^2-8x+1}$

Explanation:

$- 6 x^{2} - 16$ $-6x^2-16$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ - 6 x^{4} + 2 x^{2} - 8 x + 1 ∣ x^{2} - 3$ $bar(-6x^4+2x^2-8x+1)|x^2-3$
$-$ $-$
$- 6 x^{4} - 18 x^{2}$ $-6x^4-18x^2$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0 x^{4} - 16 x^{2} - 8 x + 1$ $bar(0x^4-16x^2-8x+1)$
$-$ $-$
$0 x^{4} - 16 x^{2} + 0 x + 48$ $0x^4-16x^2+0x+48$
$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 0 x^{4} - 0 x^{2} - 8 x - 47$ $bar(0x^4-0x^2-8x-47)$

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How do you divide $\frac{x^{2} - 3}{- 6 x^{4} + 2 x^{2} - 8 x + 1}$ $(x^2-3) / (-6x^4+2x^2-8x+1)$ ?

1 Answer

Explanation:

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How do you divide (x^2-3) / (-6x^4+2x^2-8x+1)x2−3−6x4+2x2−8x+1(x^2-3) / (-6x^4+2x^2-8x+1)?

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Explanation:

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How do you divide $\frac{x^{2} - 3}{- 6 x^{4} + 2 x^{2} - 8 x + 1}$ $(x^2-3) / (-6x^4+2x^2-8x+1)$ ?