+0

complex numbers

0
36
1
+8

does anyone know how to solve these

nellycrane  Sep 29, 2018
#1
+2248
+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