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

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Answer 1 · 2015-07-14T15:49:58

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

You use the process of long division.

Long Division

So,

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

Check:

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

How do you divide 2n^2 - 3n - 12n^2 - 3n - 1 by 2n - 52n - 5?