Unlike the situation in \(\mathbb{Q}[x]\), it is possible to give an explicit description of all the irreducible polynomials in \(\mathbb{R}[x]\) and \(\mathbb{C}[x]\). Consequently, you can immediately tell if a polynomial in \(\mathbb{R}[x]\) or \(\mathbb{C}[x]\) is irreducible without any elaborate tests or criteria. These facts are a consequence of the following theorem, which was first proved by Gauss in 1799:

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Theorem 4.26: The Fundamental Theorem of Algebra

Every nonconstant polynomial in \(\mathbb{C}[x]\) has a root in \(\mathbb{C}\).

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This theorem is sometimes expressed in other terminology by saying that the field \(\mathbb{C}\) is algebraically closed. Every known proof of the theorem depends significantly on facts from analysis and/or the theory of functions of a complex variable. For this reason, we shall consider only some of the implications of the Fundamental Theorem on irreducibility in \(\mathbb{C}[x]\) and \(\mathbb{R}[x]\). For a proof, see Hungerford [5].

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Corollary 4.27

A polynomial is irreducible in \(\mathbb{C}[x]\) if and only if it has degree 1.

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Corollary 4.28

Every nonconstant polynomial \(f(x)\) of degree \(n\) in \(\mathbb{C}[x]\) can be written in the form \((x - a_1)(x - a_2) \cdots (x - a_n)\) for some \(c, a_1, a_2, \dots, a_n \in \mathbb{C}\). This factorization is unique except for the order of the factors.

Proof

By Theorem 4.14, \(f(x)\) is a product of irreducible polynomials in \(\mathbb{C}[x]\). Each of them has degree 1 by Corollary 4.27, and there are exactly \(n\) of them by Theorem 4.2. Therefore,

\[ f(x) = (r_1x + s_1)(r_2x + s_2) \cdots (r_nx + s_n) \] \[ = r_1(x - (-r_1^{-1}s_1))r_2(x - (-r_2^{-1}s_2)) \cdots r_n(x - (-r_n^{-1}s_n)) \] \[ = c(x - a_1)(x - a_2) \cdots (x - a_n), \]

where \(c = r_1r_2 \cdots r_n\) and \(a_i = -r_i^{-1}s_i\). Uniqueness follows from Theorem 4.14. See Exercise 25 in Section 4.3. \(\blacksquare\)

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Theorem 4.30

A polynomial \(f(x)\) is irreducible in \(\mathbb{R}[x]\) if and only if \(f(x)\) is a first-degree polynomial or

\[ f(x) = ax^2 + bx + c \]

with \(b^2 - 4ac < 0\).

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Proof

The proof that the two kinds of polynomials mentioned in the theorem are, in fact, irreducible is left to the reader (Exercise 7). Conversely, suppose \(f(x)\) has degree \(\geq 2\) and is irreducible in \(\mathbb{R}[x]\). Then \(f(x)\) has a root \(\omega\) in \(\mathbb{C}\) by Theorem 4.26. Lemma 4.29 shows that \(\bar{\omega}\) is also a root of \(f(x)\). Furthermore, \(\omega \neq \bar{\omega}\) (otherwise \(\omega\) would be a real root of \(f(x)\), contradicting the irreducibility of \(f(x)\). Consequently, by the Factor Theorem, \(x - \omega\) and \(x - \bar{\omega}\) are factors of \(f(x)\) in \(\mathbb{C}[x]\); that is, \(f(x) = (x - \omega)(x - \bar{\omega})h(x)\) for some \(h(x)\) in \(\mathbb{C}[x]\). Let \(g(x) = (x - \omega)(x - \bar{\omega})\). Then \(f(x) = g(x)h(x)\) in \(\mathbb{R}[x]\). Furthermore, if \(\omega = r + si\) (with \(r, s \in \mathbb{R}\)), then

\[ g(x) = (x - \omega)(x - \bar{\omega}) = (x - (r + si))(x - (r - si)) = x^2 - 2rx + r^2 + s^2. \]

Hence, the coefficients of \(g(x)\) are real numbers.

We now show that \(h(x)\) also has real coefficients. The Division Algorithm in \(\mathbb{C}[x]\) shows that if \(f(x)\) and \(g(x)\) are polynomials in \(\mathbb{R}[x]\), then \(h(x)\) is in \(\mathbb{R}[x]\). Since \(f(x) = g(x)h(x)\) and \(f(x)\) is irreducible in \(\mathbb{R}[x]\), \(h(x)\) must be a constant, and \(f(x)\) is irreducible in \(\mathbb{R}[x]\) and is a quadratic polynomial in \(\mathbb{R}[x]\) and hence has the form \(ax^2 + bx + c\) for some \(a, b, c \in \mathbb{R}\). Since \(f(x)\) has no roots in \(\mathbb{R}\), the quadratic formula (Exercise 6) shows that \(b^2 - 4ac < 0\). \(\blacksquare\)

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Corollary 4.31

Every polynomial \(f(x)\) of odd degree in \(\mathbb{R}[x]\) has a root in \(\mathbb{R}\).

Proof

By Theorem 4.14, \(f(x) = p_1(x)p_2(x) \cdots p_k(x)\) with each \(p_i(x)\) irreducible in \(\mathbb{R}[x]\). Each \(p_i(x)\) has degree 1 or 2 by Theorem 4.30. Theorem 4.2 shows that

\[ \deg(f(x)) = \deg(p_1(x)) + \deg(p_2(x)) + \cdots + \deg(p_k(x)). \]

Since \(f(x)\) has odd degree, at least one of the \(p_i(x)\) must have degree 1. Therefore, \(f(x)\) has a first-degree factor in \(\mathbb{R}[x]\) and, hence, a root in \(\mathbb{R\}). \blacksquare

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It may seem that the Fundamental Theorem and its corollaries settle all the basic questions about polynomial equations. Unfortunately, things aren’t quite that simple. None of the known proofs of the Fundamental Theorem provides a constructive way to find the roots of a specific polynomial.* Therefore, even though we know that every polynomial equation has a solution in \(\mathbb{C}\), we may not be able to solve a particular equation.

Polynomial equations of degree less than 5 are no problem. The quadratic formula shows that the solutions of any second-degree polynomial equation can be obtained from the coefficients of the polynomials by taking sums, differences, products, quotients, and square roots. There are analogous, but more complicated, formulas involving cube and fourth roots for third- and fourth-degree polynomial equations (see page 423 for one version of the cubic formula). However, there are no such formulas for finding the roots of all fifth-degree or higher-degree polynomials. This remarkable fact, which was proved nearly two centuries ago, is discussed in Section 12.3.