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How do you divide #2n^2 - 3n - 1# by #2n - 5#?

1 Answer

Jul 14, 2015

#(2n^2 -3n -1)/(2n-5) = n + 1 - 4/(2n-5)#

Explanation:

You use the process of long division.

Long Division

So,

#(2n^2 -3n -1)/(2n-5) = n + 1 - 4/(2n-5)#

Check:

#(2n-5)(n+1+4/(2n-5)) = (2n-5)(n+1) + 4 = 2n^2 -3n -5 +4 = 2n^2 - 3n - 1#

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