How to factor a polynomial with complex roots of the form $a+bi$, where $a\neq 0$? (Not just find the root)

Question

For example, the quadratic formula reveals that the roots of $x^2 - 4x + 5$ are $x = 2\pm 2i $

But how do we use these roots to actually factor $x^2 - 4x + 5$?

My best guess was that, since $x = 2\pm 2i $, we would have $x^2 - 4x + 5 = (x-2-2i)(x-2+2i)$

But that is incorrect because $(x-2-2i)(x-2+2i) = x^2 - 8i \neq x^2 - 4x + 5 $

So how do we factor $x^2 - 4x + 5$ ?

Answer 1

You got the wrong roots in the first place. The roots are $2 \pm i$, please check your work (application of the quadratic formula).

Once you correct that, your factorisation will be correct. But you made another mistake in expanding the product of the factors.

$(x-2-i)(x-2+i) = (x-2)^2 - i(x-2)+i(x-2) +(i)(-i) = (x-2)^2 + 1 = x^2 - 4x +4 + 1 = x^2 -4x + 5$ as expected.

There is no factorisation possible purely in the reals.