How do you simplify #r^2 - t^0?

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Answer 1 · 2015-06-02T23:58:59

#r^2 - t^0 = ?#

Anything raised to the 0 is 1. Here is a quick example of why:

#(2^4)/(2^1) = 16/2 = 2^(4-1) = 2^(3) = 8#

Thus:

#(2^4)/(2^4) = 16/16 = 2^(4-4) = 2^0 = 1#

So we just get:

#r^2 - cancel(t^0) = r^2 - 1#

And you can, yes, factor it out to be #(r-1)(r+1)#.