Irreducible factorization of quadratic function with complex roots?

Question

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 $f(x)=8\left(x-\frac{i-1}{4}\right)\left(x+\frac{i+1}{4}\right)$?

Answer 1

$$ 1 + 4x + 8 x^2 = (1 + 2x)^2 + (2x)^2 = (1 + 2x)^2 - (2ix)^2 = $$ $$ (1 + 2x)^2 - (2ix)^2 = \left( 1 + 2x + 2ix \right) \left( 1 + 2x - 2ix \right) $$

Answer 2

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 $$ 8\left(x + \frac{i-1}4\right)\left(x + \frac{i+1}4\right) = \left((2-2i)x + \frac{(2-2i)(i+1)}4\right) \left((2+2i)x + \frac{(2+2i)(i-1)}4\right) = \bigl((2-2i)x + 1\bigr)\bigl((2+2i)x + 1\bigr) $$