How do you divide $(-x^5-4x^3+x-12)/(x^2-x+3)$ ?

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Answer 1 · 2018-07-10T06:53:37

The remainder is $=(8x-15)$ and the quotient is $=(-x^3-x^2-2x+1)$

Perform a long division

$color(white)(aa)$ $-x^5+0x^4-4x^3+0x^2+x-12$ $color(white)(aa)$ $|$ $x^2-x+3$

$color(white)(aa)$ $-x^5+x^4-3x^3$ $color(white)(aaaaaaaaaaaaaaaa)$ $|$ $-x^3-x^2-2x+1$

$color(white)(aaaa)$ $0-1x^4-1x^3+0x^2$

$color(white)(aaaaaa)$ $-1x^4+1x^3-3x^2$

$color(white)(aaaaaaaaa)$ $0-2x^3+3x^2+x$

$color(white)(aaaaaaaaaaa)$ $-2x^3+2x^2-6x$

$color(white)(aaaaaaaaaaaaaa)$ $0+x^2+7x-12$

$color(white)(aaaaaaaaaaaaaaaa)$ $+x^2-1x+3$

$color(white)(aaaaaaaaaaaaaaaaa)$ $-0+8x-15$

Therefore,

$(-x^5+0x^4-4x^3+0x^2+x-12)/(x^2-x+3)$

$=(-x^3-x^2-2x+1)+(8x-15)/(x^2-x+3)$

The remainder is $=(8x-15)$ and the quotient is $=(-x^3-x^2-2x+1)$

How do you divide (-x^5-4x^3+x-12)/(x^2-x+3)(-x^5-4x^3+x-12)/(x^2-x+3)?