How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 9 x + 3}{x^{2} - 3}$ $(x^4-2x^3-2x^2+9x+3)/(x^2-3)$ ?

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How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 9 x + 3}{x^{2} - 3}$ $(x^4-2x^3-2x^2+9x+3)/(x^2-3)$ ?

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Answer 1 · 2016-05-02T16:11:42

$\frac{x^{4} - 2 x^{3} - 2 x^{2} + 9 x + 3}{x^{2} - 3} = (x^{2} - 2 x + 1) + \frac{3 x + 6}{x^{2} - 3}$ $(x^4-2x^3-2x^2+9x+3)/(x^2-3) = (x^2-2x+1) + (3x+6)/(x^2-3)$

Explanation:

I like to long divide the coefficients, not forgetting to include $0$ $0$ 's for any missing powers of $x$ $x$ . In our example, that means the missing $x$ $x$ term in the divisor, which is therefore repesented as $1, 0, - 3$ $1, 0, -3$ .

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The process is similar to long division of numbers.

We find a quotient $1, - 2, 1$ $1, -2, 1$ , meaning $x^{2} - 2 x + 1$ $x^2-2x+1$ and remainder $3, 6$ $3, 6$ , meaning $3 x + 6$ $3x+6$

So: $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 9 x + 3}{x^{2} - 3} = (x^{2} - 2 x + 1) + \frac{3 x + 6}{x^{2} - 3}$ $(x^4-2x^3-2x^2+9x+3)/(x^2-3) = (x^2-2x+1) + (3x+6)/(x^2-3)$

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How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 9 x + 3}{x^{2} - 3}$ $(x^4-2x^3-2x^2+9x+3)/(x^2-3)$ ?

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Explanation:

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How do you divide $\frac{x^{4} - 2 x^{3} - 2 x^{2} + 9 x + 3}{x^{2} - 3}$ $(x^4-2x^3-2x^2+9x+3)/(x^2-3)$ ?