Question

What techniques are there to factor quadratic equations that have complex conjugate roots? I do not always want to go through quadratic formula to be able to find these out...

Answer 1

use the quadratic formula..

Answer 2

if the quadratic has complex solutions, then by definition it can't be factored. But if you don't want to use quadratic formula, you can notice the following:
Let \[x = a \pm bi\]
then
\[x-a = \pm bi\]
\[(x-a)^{2} = -b^{2}\]
\[(x-a)^{2} +b^{2} = 0\]
if you can put the quadratic in this form , you will know the complex solutions
Note: if b^2 is negative then solutions are real