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

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Answer 1 · 2017-12-26T01:05:24

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

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

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