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simplify the expression n^2-2n/n^2-3n
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Assuming your problem looks like \[\frac{n^2-2n}{n^2-3n}\] You can first factor an n out of both the top and bottom \[\frac{(n)n-2}{(n)n-3}\] And then cancel them, leaving you with a nice, neat \[\frac{n-2}{n-3}\] It's worth noting that the original equation is undefined at n=0, and both forms are undefined at n=3. It has a single root at x=2.
so what would be the final form
?
That would be the final form. There are a number of others you could get to, like \[\frac{1}{n-3}+1\]but there's little reason to do so.
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