MHB Why is x^2 + 1 an Irreducible Polynomial?

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The polynomial x^2 + 1 is irreducible over the real numbers because it has no real roots, as indicated by its positive discriminant. However, it is reducible over the complex numbers, where it can be factored into (x - i)(x + i). Similarly, the polynomial x^2 + x + 1 is also irreducible over the reals, as its discriminant is negative (-3), indicating no real solutions. This discussion highlights the importance of analyzing discriminants to determine the factorability of polynomials in different number systems. Understanding these concepts is crucial for polynomial factorization in algebra.
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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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