<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] The factorization you got also is an irreducible factorization. To get to the other one, note that 
$$ 8 = 2^2 + 2^2 = (2+2i)(2-2i), $$
hence 
$$ 8\left(x + \frac{i-1}4\right)\left(x + \frac{i+1}4\right)
 = \left((2-2i)x + \frac{(2-2i)(i+1)}4\right) \left((2+2i)x + \frac{(2+2i)(i-1)}4\right) = \bigl((2-2i)x + 1\bigr)\bigl((2+2i)x + 1\bigr) $$
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