How do you divide $(n^{3} + 2 n^{2} - n - 2)$ $(n^3 + 2n^2 - n - 2)$ by $(n^{2} - 1)$ $(n^2 - 1)$ ?

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How do you divide $(n^{3} + 2 n^{2} - n - 2)$ $(n^3 + 2n^2 - n - 2)$ by $(n^{2} - 1)$ $(n^2 - 1)$ ?

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Answer 1 · 2015-09-13T14:33:14

Refer to explanation

Explanation:

We have that

$\frac{n^{3} + 2 n^{2} - n - 2}{n^{2} - 1} = \frac{n^{3} - n + 2 (n^{2} - 1)}{n^{2} - 1} = \frac{n (n^{2} - 1) + 2 (n^{2} - 1)}{n^{2} - 1} = \frac{(n^{2} - 1) \cdot (n + 2)}{n^{2} - 1} = n + 2$ $(n^3 + 2n^2 - n - 2)/(n^2 - 1)=(n^3-n+2(n^2-1))/(n^2-1)= (n(n^2-1)+2(n^2-1))/(n^2-1)=((n^2-1)*(n+2))/(n^2-1)=n+2$

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How do you divide $(n^{3} + 2 n^{2} - n - 2)$ $(n^3 + 2n^2 - n - 2)$ by $(n^{2} - 1)$ $(n^2 - 1)$ ?

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How do you divide (n3+2n2−n−2)(n^3 + 2n^2 - n - 2) by (n2−1)(n^2 - 1)?

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How do you divide $(n^{3} + 2 n^{2} - n - 2)$ $(n^3 + 2n^2 - n - 2)$ by $(n^{2} - 1)$ $(n^2 - 1)$ ?