MHB Can you factor the following two polynomials?

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The discussion centers on the factorization of the polynomials x^4 + 4 and x^4 + 3x^2y^2 + 2y^4 + 4x^2 + 5y^2 + 3. Participants note that x^4 + 4 cannot be factored over the integers or real numbers, but can be expressed using complex factors. The polynomial can be rewritten as (x^2 + 2 + 2x)(x^2 + 2 - 2x) through a specific manipulation. Additionally, the equation x^4 + 4 = 0 has four imaginary solutions that form conjugate pairs, leading to further factorization. The conversation highlights the complexity of polynomial factorization and the existence of imaginary roots.
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Can you factor the following polynomials over integers?

[math] x^4 + 4[/math]

[math] x^4 + 3 ~x^2~y^2 + 2 ~y^4 + 4 ~x^2 + 5 ~y^2 + 3[/math]

If not, you can get help from the following free math tutoring YouTube channel "Math Tutoring by Dr. Liang"

https://www.youtube.com/channel/UCWvb3TYCbleZjfzz8HEDcQQ
 
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Hint: [math]x^4 +4 = x^4 + 4x^2 - 4x^2 + 4[/math]

-Dan
 
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I don't see how that helps. $x^4+ 4$ obviously cannot be factored over the integer, or even over the real numbers.
 
HallsofIvy said:
I don't see how that helps. $x^4+ 4$ obviously cannot be factored over the integer, or even over the real numbers.
As linear factors, yes. But:
[math]x^4 + 4 = x^4 + (4 x^2 - 4 x^2) + 4 = (x^4 + 4 x^2 + 4) - 4 x^2 = (x^2 + 2)^2 - 4 x^2 = (x^2 + 2 + 2 x)(x^2 + 2 - 2 x)[/math]

-Dan
 
Just to mention an alternative approach, $$x^4+4=0$$ has 4 imaginary solutions that form conjugate pairs.
The solutions are $$x=\pm 1\pm i$$

If we then put the conjugate pairs together, we get $$(x-(1+i))(x-(1-i))=x^2-2x+2$$ and $$(x-(-1+i))(x-(-1-i))=x^2+2x+2$$ just like topsquark found.
 
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