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

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

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Answer 1 · 2018-07-21T16:57:36

$x^{2} - 3 x + 1 + \frac{16 x - 10}{x^{2} + 3}$ $x^2-3 x+1+(16 x-10)/(x^2+3)$

Explanation:

$\frac{- x^{4} - 3 x^{3} - 2 x^{2} + 7 x - 7}{x^{2} + 3}$ $(-x^4-3 x^3-2 x^2+7 x-7)/(x^2+3)$

$color(white)(......)color(white)(.....)-x^2-3x+1$
$x^2+3|overline(-x^4-3x^3-2x^2+7x-7)$
$color(white)(.............)ul(-x^4+"0"-3x^2)$
$color(white)(.....................)-3x^3+x^2+7x$
$color(white)(......................)ul(-3x^3+""0-9x)$
$color(white)(...................................)x^2+16x-7$
$color(white)(.....................................)ul(x^2+""0+3)$
$color(white)(.............................................)16x-10$

$(-x^4-3 x^3-2 x^2+7 x-7)/(x^2+3) = -x^2-3x+1+(16x-10)/(x^2+3)$

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

1 Answer

Explanation:

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Explanation:

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