#1**+1 **

Hi, nellycrane!

I think you will be better off if I give you an easier problem like \(x^2-4 \) . You are probably well aware that this is a difference of squares. \((x+2)(x-2)\) would be the corresponding factorization.

Now, let's say that you want to factor \(x^2-30\) . This is not possible if you restrict yourself to the rational number set. However, it can be factored as \((x+\sqrt{30})(x-\sqrt{30})\).

In general, \(a^2-b^2=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) . You can apply this knowledge to the problems at hand.

\(x^2+50\\ a=x^2,b=-50;\\ (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})\\ (\sqrt{x^2}+\sqrt{-50})(\sqrt{x^2}-\sqrt{-50}) \)

You can simplify this to get the factorization amongst the complex numbers. Good luck!

TheXSquaredFactor Sep 30, 2018