is n^2-n already factored? If so, how to solve n^2-n=0?

mizkiki55

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is n^2-n already factored if not how do you factor it. if it is how do you solve n^2-n = 0
 
mizkiki55 said:
is n^2-n already factored
Is 3x - 3 fully factored? (Hint: No.) :oops:

mizkiki55 said:
if not how do you factor it.
Hint: You would factor n[sup:3mtuolkp]2[/sup:3mtuolkp] - n in the exact same way that you'd factor 3x - 3. :wink:

Eliz.
 
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