Not a very rigorous proof by any means, but the way I think about it is as follows:
By substituting a negative value for n into the expression we get (-n^2) - (-4n) + 21
Evaluating, we can see that this will equal n^2 + 4n + 21, which will equal a positive for all values.
That takes care of the negetive end of things, so let's look at the positive side.
Factoring the first part of the expression, we see that it can be rewritten as n(n - 4) + 21
From this, it can be noted that n will "break even" at the value n = 4 and be positive for any value greater than 4. This means that we only have to check the values 0, 1, 2, and 3.
Subsituting these values in, we get:
0 + 21 = 21;
-3 + 21 = 18;
-4 + 21 = 17
-3 + 21 = 18.
Since all three situations return a positive value, we can conclude that n^2 - 4n + 21 will always equal a positive number.
Hope this helps!
