Can p^4 + 4 be factored? What about x^2 + 1?

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Discussion Overview

The discussion revolves around the factorability of the polynomial expressions \(p^4 + 4\) and \(x^2 + 1\). Participants explore whether these expressions can be factored over the reals and complex numbers, examining concepts of irreducibility and the nature of polynomial factors.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that \(p^4 + 4\) cannot be factored over the reals, seeking explanations for this assertion.
  • Others propose that \(p^4 + 4\) can be factored into complex factors as \((p^2 + 2i)(p^2 - 2i)\), but note that it is not factorable over the reals due to the nature of sums of squares.
  • A participant presents a factorization of \(p^4 + 4\) as \((p^2 + 2p + 2)(p^2 - 2p + 2)\), suggesting it is reducible.
  • There is a question raised about whether \(p^4 + 4\) can be considered irreducible, with a subsequent response indicating it is reducible based on the identified factors.
  • Another participant inquires about the irreducibility of \(x^2 + 1\), prompting further exploration of this concept.

Areas of Agreement / Disagreement

Participants express disagreement regarding the factorability of \(p^4 + 4\), with some asserting it is irreducible while others provide factorizations that suggest it is reducible. The status of \(x^2 + 1\) as irreducible remains unresolved, with differing opinions anticipated.

Contextual Notes

The discussion highlights the dependence on the definitions of factorability and irreducibility, particularly in relation to the real and complex number systems. The arguments presented rely on various interpretations of polynomial factorization.

mathdad
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Factor p^4 + 4.

I know that this cannot be factored but don't know the reason the expression cannot be factored. Can someone explain why it cannot be factored?
 
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If we allow for complex factors, then we can state:

$$p^4+4=\left(p^2+2i\right)\left(p^2-2i\right)$$

However, over the reals, the sum of two squares is not factorable. To see why, consider:

$$a^2+b^2=0$$ where $a$ and $b$ are non-zero real numbers.

If we subtract through by $b^2$, we have:

$$a^2=-b^2$$

The square of a real number can never be negative, so there is no factorization of the sum of two real squares over the reals and thus there are no real solutions here...what we find instead is:

$$a=\pm bi$$
 
RTCNTC said:
Factor p^4 + 4.

I know that this cannot be factored but don't know the reason the expression cannot be factored. Can someone explain why it cannot be factored?

No it can be factored

$p^4+4 = p^4 + 4p^2 + 4 - 4p^2 = (p^2+2)^2 - (2p)^2 = (p^2 + 2p +2)(p^2 -2 p +2)$
 
Can we say that p^4 + 4 is irreducible?
 
RTCNTC said:
Can we say that p^4 + 4 is irreducible?

No. It is reducible, since $p^2 - 2p + 2$ and $p^2 + 2p + 2$ are factors of $p^4 + 4$, as kaliprasad showed.

More generally, a polynomial with real coefficients is called reducible if it can be written as a product of two real polynomials of lesser degree.
 
How about x^2 + 1? Is this irreducible? If so, why?
 

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