The only irreducible polynomials over the reals are linear polynomials and quadratic polynomials that have no real roots. This conclusion is derived from the fundamental theorem of algebra, which states that any real polynomial can be factored into linear and quadratic terms. If a polynomial has complex roots, those roots must occur in conjugate pairs, leading to the formation of quadratic factors. Therefore, any real polynomial can be expressed as a product of linear and irreducible quadratic factors.
PREREQUISITESMathematics students, educators, and anyone studying algebraic structures, particularly those interested in polynomial theory and irreducibility concepts.
micromass said:I wouldn't immediately say that it's obvious, but it's certainly something that you can prove:
Let P(X) be a real polynomial. The fundamental theorem of algebra states that there exists complex number a1,...an such that P(X)=(X-a1)...(X-an). Thus we have factorized the polynomial P over the field C. However, we must factorize it over R.
For this: notice the following fact (try to prove this!): if z is a complex solution of P(X) (thus is P(z)=0), then the complex conjugate \overline{z} is a solution of P(X).
Thus, in our factorization P(X)=(X-a1)...(X-an), if ai is non-real, then one of the a1,...,an must be \overline{a_i}. And if we multiply (X-a_i)(X-\overline{a_i}), then we get a REAL quadratic polynomial!
Thus P(X) can always be written as a product of linear terms and quadratic terms. This implies that the only possible irreducible polynomials are linear or quadratic.