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

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Answer 1 · 2015-09-13T14:33:14

Refer to explanation

We have that

$(n^3 + 2n^2 - n - 2)/(n^2 - 1)=(n^3-n+2(n^2-1))/(n^2-1)= (n(n^2-1)+2(n^2-1))/(n^2-1)=((n^2-1)*(n+2))/(n^2-1)=n+2$

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