SOLUTION: I don't understand how (n-2)/(n^3-8) becomes 1/(n^2+2n+4). I get how you take out the common factor of (n-2), but why does the denominator become n^2+2n+4 instead of n^2+4?

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Question 946254: I don't understand how (n-2)/(n^3-8) becomes 1/(n^2+2n+4).
I get how you take out the common factor of (n-2), but why does the denominator become n^2+2n+4 instead of n^2+4?
Found 2 solutions by Alan3354, MathTherapy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
I don't understand how (n-2)/(n^3-8) becomes 1/(n^2+2n+4).
I get how you take out the common factor of (n-2), but why does the denominator become n^2+2n+4 instead of n^2+4?
===================================
(n-2)*(n^2+4) = n^3 - 2n^2 + 4n - 8
----
That's <> n^ - 8
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
I don't understand how (n-2)/(n^3-8) becomes 1/(n^2+2n+4).
I get how you take out the common factor of (n-2), but why does the denominator become n^2+2n+4 instead of n^2+4?




 ----- Factoring denominator

 ---- Cancelling n - 2 in numerator and denominator

 

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