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

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

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Answer 1 · 2018-07-10T06:53:37

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

Explanation:

Perform a long division

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

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

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

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

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

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

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

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

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

Therefore,

$\frac{- x^{5} + 0 x^{4} - 4 x^{3} + 0 x^{2} + x - 12}{x^{2} - x + 3}$ $(-x^5+0x^4-4x^3+0x^2+x-12)/(x^2-x+3)$

$= (- x^{3} - x^{2} - 2 x + 1) + \frac{8 x - 15}{x^{2} - x + 3}$ $=(-x^3-x^2-2x+1)+(8x-15)/(x^2-x+3)$

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

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

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Explanation:

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