Here's a somewhat tough factoring problem if anyone would like to have a go.
Yes, it factors, though it may take some ingenuity.
Some may not find it that bad, but what the heck, have fun if you wish.
Now we solve . Note that , let us call this number . Let . That means the solution to is .
Now we solve . The solutions are: .
That means the full factorization is:
...my factorization looks stupid and incomplete now
I like the clever approaches, but here is what I was getting at. I failed to elaborate. I apologize.
Here's what I done. I must admit, I pondered it for a while.
Now, if we divide:
So, we (as Kriz says) happily have:
There's more than one way to factorize an expression like this. I just thought it was a cool factoring problem.
So this means galactus' 8th degree factor has 4 quadratic factors as well... (Too bad I have never learned any general technique to find them!)
This my method. Consider the complex equation
Multiplying by ,
Hence the solutions to are all the 15th roots of unity minus the 5th roots of unity, i.e.
But two of the solutions (with ) are
These are roots of the equation .
"Every polynomial can be factored (over the real numbers) into a product of linear factors and irreducible quadratic factors."
Apologies and thanks for the catch!