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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-> 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?
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(n-2)*(n^2+4) = n^3 - 2n^2 + 4n - 8
----
That's <> n^ - 8
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
