Why is x^2 + 1 an Irreducible Polynomial?

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SUMMARY

The polynomial x^2 + 1 is irreducible over the real numbers but reducible over the complex numbers. This conclusion is drawn from the fact that its discriminant, calculated as b^2 - 4ac, yields a negative value (-3), indicating no real roots. In contrast, x^2 + x + 1 also has a negative discriminant, confirming it has no real solutions. Therefore, both polynomials exhibit irreducibility in the context of real number factorization.

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Why is x^2 + 1 irreducible?
 
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RTCNTC said:
Why is x^2 + 1 irreducible?

What do you think? If you factor this expression, what do you get? Moreover, what could be said about those factors in relation to the real numbers?
 
Complex numbers?
 
Indeed!

In this case we could say that the expression is irreducible over the reals, but reducible over the complex numbers. But I might let someone else chime into give a more formal approach. :)
 
Cool.
 
What about x^2 + x + 1?
 
RTCNTC said:
What about x^2 + x + 1?

What does the discriminant tell you about the roots of this quadratic polynomial?
 
b^2 - 4ac

1^2 - 4(1)(1)

1 - 4

-3

b^2 - 4ac < 0

This means no real solutions.
 

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