Factoring $x^{4} +1$, using real factoring to the second degree Factoring to the second degree using real numbers
$$x^{4} +1$$
I know that $ x^{4} +1=(x^{2} + i)(x^{2}-i).\;$ But these are complex but I thought using these in some kind of way? Got no where! 
And then I tried to guess, two solutions are $\pm (-1)^{1/4},\, $
which gave me $(x-\sqrt{i})(x+\sqrt{i}).\;$ But I have a feeling that I am really off here...
Hmm... Thinking! 
What kind of techniques do you use?  

Your approach starts off perfectly fine: you do have
$$ x^4 + 1 = (x + \sqrt{\mathbf{i}})(x - \sqrt{\mathbf{i}})(x + \sqrt{-\mathbf{i}})(x - \sqrt{-\mathbf{i}}) $$
You just need to group them into conjugate pairs: (I assume you take the principal square roots)
$$ x^4 + 1 = \left( (x + \sqrt{\mathbf{i}})(x + \sqrt{\mathbf{-i}}) \right)\left( (x - \sqrt{\mathbf{i}})(x - \sqrt{\mathbf{-i}}) \right) $$
$$ x^4 + 1 = \left( x^2 + (\sqrt{\mathbf{i}} + \sqrt{-\mathbf{i}}) x + 1 \right) 
\left( x^2 - (\sqrt{\mathbf{i}} + \sqrt{-\mathbf{i}}) x + 1 \right) $$
The only simplification that remains is that you can compute
$$ \sqrt{\mathbf{i}} + \sqrt{-\mathbf{i}} = \sqrt{2}$$
