I want to apply the method in this example here:
How could you do this for the case where you end with $ 2x^2 $ in the final equation? For example, the equation:
$$ 2x^2 - 6x + 5 = 0 $$
Using the quadratic formula, we can find that its complex roots are $ 1.5 + .5i $ and $ 1.5 - .5i $.
So how could we get to the function with $ 2x^2 $ at the start, starting with those roots?
Thanks to @Prometheus that it's as simple as multiplying by 2 to transform equations like these that equal 0. Here are my steps:
$$ \begin{align*} &(x - (1.5 + 0.5i))(x - (1.5 -0.5i)) = 0 \\ &(2)(x - (1.5 + .5i))(x - (1.5 -.5i)) = 0.\\ &\quad\text{Multiplying both sides by 2,} \\ &(2x - (3 + i))(x - (1.5 -0.5i)) = 0 \\ &2x^2 - 3x + ix - 3x + 4.5 - 1.5i -ix + 1.5i + 0.5 = 0\\ &2x^2 - 6x + 5 = 0 \end{align*} $$
I'm failing to see how this method doesn't work generally, per @dvix comment. Here is when I try it with $ 1±0.5 $
$$ \begin{align*} &(x - (1 + 0.5i))(x - (1 - 0.5i)) = 0 \\ &(2)(x - (1 + .5i))(x - (1 - .5i)) = 0.\\ &\quad\text{Multiplying both sides by 2,} \\ &(2x - (2 + i))(x - (1 - 0.5i)) = 0 \\ &2x^2 - 2x + ix - 2x + 2 - i -ix + i - 0.5i^2 = 0\\ &2x^2 - 4x + 2.5 = 0 \\ \end{align*} $$
Note that this equation is a multiple of two of the monic version of the quadratic: $ (2)(x^2 - 2x + 1.25) = 2x^2 - 4x + 2.5 = 0 $