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

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

1 Answer

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Answer 1 · 2015-07-17T15:42:55

Long divide to find:

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

Explanation:

Write the numerator and denominator in standard form, with descending powers of $x$ $x$ and $0$ $0$ coefficients for otherwise missing powers. Then long divide:

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So:

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

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

1 Answer

Explanation:

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