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

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

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Answer 1 · 2016-08-29T16:09:50

$\frac{x^{4} - 2 x^{3} - 2 x^{2} + 5 x + 3}{x^{2} - 3} = x^{2} - 2 x + 1 - \frac{x + 6}{x^{2} - 3}$ $(x^4-2x^3-2x^2+5x+3)/(x^2-3) = x^2-2x+1-(x+6)/(x^2-3)$

Explanation:

One way is to long divide the coefficients, not forgetting to include a $0$ $0$ for the coefficient of the missing $x$ $x$ term in the divisor...

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We find:

$\frac{x^{4} - 2 x^{3} - 2 x^{2} + 5 x + 3}{x^{2} - 3} = x^{2} - 2 x + 1 - \frac{x + 6}{x^{2} - 3}$ $(x^4-2x^3-2x^2+5x+3)/(x^2-3) = x^2-2x+1-(x+6)/(x^2-3)$

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

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