<s>[INST] Irreducible factorization of quadratic function with complex roots? How can we get the irreducible factorization of function
$$f(x)=1+4x+8x^2$$
which is $$f(x)=(1+(2+2i)x)(1+(2-2i)x)?$$
If we find the discriminant, which is $\Delta=-16$, we get
$$\sqrt{\Delta}=4i,x_1=\frac{i-1}{4},x_2=\frac{-1-i}{4}$$
Now we have $$f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$$
My question is how can we get $f(x)=(1+(2+2i)x)(1+(2-2i)x)$ from $f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$?
 [/INST] $$ 1 + 4x + 8 x^2 = (1 + 2x)^2 + (2x)^2 = (1 + 2x)^2 - (2ix)^2 =   $$
$$  (1 + 2x)^2 - (2ix)^2 = \left( 1 + 2x + 2ix \right) \left( 1 + 2x - 2ix \right)   $$
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