



 
Well, you can, but only by using imaginary numbers: \[\left[ x + \sqrt{5} \right] \left[ x  \sqrt{5} \right] = x^2  \left(\sqrt{5}\right)^2 = x^2  (5) = x^2 + 5.\] The simplest way to show it can't factor over the reals is to note that it's a quadratic, so any nontrivial factor has to be linear; and zeros of polynomials correspond to linear factors and vice versa. A zero of a polynomial is a value $x$ such that $P(x) = 0.$ But $x^2 + 5$ is always greater than or equal to 5, hence strictly greater than 0. Thus it has no zeros and does not factor. Good Hunting!  


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