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2 July, 11:31

Factor the expression over the complex number

x^2+25

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  1. 2 July, 11:44
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    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)
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