Power 4 equation to Quadratic factors

[basil]

Hi,

I have a problem with splitting x⁴ + 1 into real quadratic factors. How can this be done?

Cheers.

 

[micromass]

Hi basil!

Say we split it as

x^4+1=(x^2+ax+b)(x^2+cx+d)

Try to factor the left side. You'll obtain a system of equations that you need to solve...

 

[gb7nash]

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:

x⁴ + 1 = (x² + ax + b)(x² + 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

x⁴ + 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:

x⁴ + 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.

 

[gb7nash]

Hi basil!

Say we split it as

x^4+1=(x^2+ax+b)(x^2+cx+d)

Try to factor the left side. You'll obtain a system of equations that you need to solve...

This might be take less time for this problem. I'd race you but I have no paper.

 

[micromass]

If this is the case, some functions might not have real "quadratic factors".

All real polynomials can be split into real quadratic and linear factors!

 

[BloodyFrozen]

Just for reference, and function with a power of 4 is a quartic.