TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-27

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Every odd-degree real polynomial has a real root

Statement

Let $f (x) = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} \in R [x]$ have odd degree $n \geq 1$ . Then there exists $c \in R$ with $f (c) = 0$ .

Facts & Assumptions

Given: A real polynomial $f (x) = a_{n} x^{n} + \dots + a_{0}$ of odd degree $n \geq 1$ .

[F1]

The degree hypothesis means $a_{n} \neq 0$ and $a_{i} = 0$ for every $i > n$ (Degree, leading coefficient and monic polynomial, with the zero polynomial having no degree).

[L1]

Every real polynomial function is continuous on $R$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function).

[L2]

A continuous function on a closed interval takes every intermediate value between its endpoint values (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ).

[F2]

The real numbers form an ordered field, so absolute values and the order laws behave as usual (The reals form a totally ordered field).

Proof

technique · direct

1.1F1F2algebra

Put $S : = \sum_{i < n} ∣ a_{i} ∣$ and $R : = 1 + \frac{S}{∣ a_{n} ∣} .$ Then $R > 1$ , and therefore $\sum_{i < n} ∣ a_{i} ∣ R^{i} \leq S R^{n - 1} < ∣ a_{n} ∣ R^{n} .$

2.1step 1.1F2algebra

The estimate in step 1.1 gives $∣ \sum_{i < n} a_{i} R^{i} ∣ \leq \sum_{i < n} ∣ a_{i} ∣ R^{i} < ∣ a_{n} ∣ R^{n} .$ Hence $f (R) = a_{n} R^{n} + \sum_{i < n} a_{i} R^{i}$ has the same sign as $a_{n}$ .

2.2step 1.1F1F2algebra

Because $n$ is odd, $(- R)^{n} = - R^{n}$ . Also $∣ \sum_{i < n} a_{i} (- R)^{i} ∣ \leq \sum_{i < n} ∣ a_{i} ∣ R^{i} < ∣ a_{n} ∣ R^{n},$ so $f (- R) = a_{n} (- R)^{n} + \sum_{i < n} a_{i} (- R)^{i}$ has the opposite sign from $a_{n}$ . Thus $f (- R)$ and $f (R)$ have opposite signs.

3.1L1L2step 2.2∎

By [L1], the polynomial function $f$ is continuous on $[- R, R]$ . Since step 2.2 shows that $0$ lies between $f (- R)$ and $f (R)$ , [L2] gives a point $c \in [- R, R]$ with $f (c) = 0$ .

Depends on

Used by

An irreducible polynomial over ℝ has degree 1 or an even degree Corollary
x³-2 over ℝ and over ℂ Example
x⁵-x-1 has a real root Example

Dependency tree · two levels

35 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Keith Conrad, Applications of Galois Theory, Theorem 2.1 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, v5.10, Chapter 5 (standard reference, not scraped)