basil said:
Hi,
I have a problem with splitting x4 + 1 into real quadratic factors. How can this be done?
Cheers.
What do you mean by this? Do you want to factor it into:
x
4 + 1 = (x
2 + ax + b)(x
2 + cx + d) where a,b,c,d are real?
If this is the case, some functions might not have
real "quadratic factors". The only way to know for sure is to look at the complex roots of
x
4 + 1 = 0
and multiply. For instance, the only roots of this are the four roots of unity:
r = e^{\frac{i \pi}{4}} , e^{\frac{i 3 \pi}{4}} , e^{\frac{i 5 \pi}{4}} , e^{\frac{i 7 \pi}{4}}
So we can factor into:
x
4 + 1 = (x - e^{\frac{i \pi}{4}})(x - e^{\frac{i 3 \pi}{4}})(x - e^{\frac{i 5 \pi}{4}})(x - e^{\frac{i 7 \pi}{4}})
Now multiply any two arbitary factors together and if you get all real numbers in the quadratic, you have a winner.
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Side note: If you haven't learned about complex numbers yet, I can't think of a better way of doing it than this.
