How to Find the Real and Imaginary Solutions of a Polynomial

Introduction

It has been of 4 decades since imaginary numbers. Like it wasn't hard enough with real ones. I think the basic theory is that you make everything a square too of negative 1 or $\sqrt{-1}$

So if you see $\sqrt{-2}$ This is actually +/- 2$\sqrt{-1}$ or 2i where i is an imaginary number and 2 is real

Example 1

So this was a little dull to say the least because the only thing you need to be aware of is $\sqrt{-1}$. So here is an example. I guess noting the ratios of the left and rights being the same is worthwhile.

1x³ - 3x² + 9x - 27 = 0

So the coefficients on the left are 1,-3, and on the right 9,-27 so this means you can factor by grouping

x²(x -3) +9(x-3) = 0
(x-3)(x² +9) = 0

So setting each to 0 we get

 x = 3

For the second we get

 x² = -9 
 x  = +/- $\sqrt{-9}$
 x  = +/- $\sqrt{9}$ * $\sqrt{-1}$
 x  = +/- 9i

And that's it.