Respuesta :
Consider the polynomial . It cannot be factored over the real numbers, since its graph has no x-intercepts. (The graph is just the standard parabola shifted up by one unit!)
How can we tell that the polynomial is irreducible, when we perform square-completion or use the quadratic formula? Let's try square-completion: Not much to complete here, transferring the constant term is all we need to do to see what the trouble is:
We can't take square roots now, since the square of every real number is non-negative!
Here is where the mathematician steps in: She (or he) imagines that there are roots of -1 (not real numbers though) and calls them i and -i. So the defining property of this imagined number i is that
Now the polynomial has suddenly become reducible, we can write
x to the power of 2 + 1 = (x - i)(x + i)