Math 2.0

statement proposition

Polynomials of degree two or three are irreducible iff they have no roots

Statement

Let KK be a field and let f∈K[x]f \in K[x] with deg⁡f=2\deg f = 2 or deg⁡f=3\deg f = 3. Then ff is irreducible over KK if and only if ff has no root in KK.

Remarks

The proof: if f=ghf = g h with deg⁡g,deg⁡h≥1\deg g, \deg h \ge 1, then deg⁡g+deg⁡h≤3\deg g + \deg h \le 3, so one factor has degree 11. A factor ax+ba x + b of degree 11 gives the root −b/a-b/a. The factor theorem gives the converse. For example, x3−2x^3 - 2 has no root in Q\mathbb{Q}, so it is irreducible over Q\mathbb{Q}. The statement fails in degree 44: (x2+1)2(x^2 + 1)^2 has no real root, but it is not irreducible over R\mathbb{R}.

Explanations

L1 human

A factorisation of a polynomial of degree 22 or 33 must have a factor of degree 11. A factor of degree 11 is the same as a root. So to test irreducibility in low degree, you only look for roots.

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  1. Group concept
  2. Ring definition
  3. Commutative ring definition
  4. Polynomial ring definition
  5. Unit of a ring definition
  6. Zero divisor definition
  7. Degree of a polynomial definition
  8. Polynomial evaluation definition
  9. Integral domain definition
  10. Leading coefficient definition
  11. Root of a polynomial definition
  12. Irreducible element definition
  13. Monic polynomial definition
  14. Degree of a product statement
  15. Irreducible polynomial definition
  16. Division algorithm for polynomials statement
  17. Remainder theorem statement
  18. Factor theorem statement
  19. Polynomials of degree two or three are irreducible iff they have no roots statement

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Sources

  1. Web https://en.wikipedia.org/wiki/Irreducible_polynomial, Simple examples