# Proof that every polynomial of odd degree has one real root

I want to prove that every real polynomial of odd degree has at least one real root, using the intermediate value theorem.

Let $P(x) = x^{2n+1} + a_n x^{2n} + . . . + a_0$ for each $a_i \in \mathbb{R}$ and $n \in \mathbb{N}$.

By the fundamental theorem of algebra I know that $P(x)$ has exactly $2n+1$ complex roots, so

$P(x) = (x+r_1)(x+r_2) . . . (x+r_{2n+1})$ for each $r_i \in \mathbb{C}$

I do not know how to complete this but I do know that, at some point, I probably have to show that each root with imaginary part non zero has to come in conjugate pairs, and since $2n+1$ is odd there is at least $1$ root that is imaginary part $0$ and thus real.

• We do not need the Fundamental Theorem of Algebra at all. If we decide to use it, we do not need the IVT. Feb 25, 2014 at 7:44
• $andre my book says that I must do this proof with the fundamental theorem of algebra and IVT, that is why I have included this. Feb 25, 2014 at 7:44 • Do you mean you should supply two proofs, one with the Fundamental Theorem and another with IVT? Feb 25, 2014 at 7:45 • @terribleatmath That's impossible unless you accidentally prove the result twice in the same proof. Feb 25, 2014 at 7:47 • Show that$P(x) \to + \infty$as$x \to + \infty$and$P(x) \to - \infty$as$x \to - \infty$. Feb 25, 2014 at 7:49 ## 5 Answers Method of FTA: $$P(\overline z)=\sum_{k=0}^{2n+1}a_k\overline z^k=\sum_{k=0}^{2n+1}\overline a_k\overline{z^k}=\sum_{k=0}^{2n+1}\overline{a_kz^k}=\overline{\sum_{k=0}^{2n+1}a_kz^k}=\overline{P(z)}$$ which states $$z$$ is a root for $$P(z)=0$$ iff its complex conjugate $$\bar z$$ is. According to FTA, there are odd number of roots for a polynomial of odd degree. That implies there must exist at least one root $$z$$ satisfying $$z=\bar z$$, hence the real root. Method of IVT: $$\frac{P(x)}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_k\frac{x^k}{x^{2n+1}}=1+\sum_{k=0}^{2n}a_kx^{k-(2n+1)}$$ For any $$\varepsilon>0$$, there exists $$N>0$$ such that for all $$|x|>N$$, $$\left|\sum_{k=0}^{2n}a_kx^{k-(2n+1)}\right|<\varepsilon$$. Hence for $$x>N$$, we have $$P(x)>x^{2n+1}-\varepsilon x^{2n+1}>0$$ and similarly for $$x<-N$$, we have $$P(x)<0$$. Then IVT implies there exists some $$y$$ such that $$P(y)=0$$. • Hello @Shuchang. Could you please elaborate a little more on how for any$\varepsilon>0$, there is a$N>0$such that for all$\left|x\right|>N$we have$\left| \sum_{k=0}^{2n} a_k x^{k-(2n+1)} \right|<\varepsilon$? Nov 20, 2020 at 12:55 • this means that as$x$gets bigger and bigger (in magnitude), the sum gets smaller and smaller (but remains positive) ie : the sum tends to 0 as$x\to +\infty$(definition of the limit) – T.D Feb 22, 2021 at 15:10 • How do you know$x^{2n+1}(1-\epsilon) > 0 $? What if$\epsilon > 1$? That becomes negative. Apr 10, 2021 at 23:49 Indeed it is true that all proofs of the fundamental theorem of algebra need some piece of analysis. Even the most algebraic proof of FTA (Euler, Gauß II) relies on the fact that all odd-degree real polynomials have at least one real root. First consider the case of relatively large positive$x$. Assuming$x\ge 1$as provisional lower bound, then$1\le x^k\le x^{2n}$for$0\le k\le 2n$and the value of the polynomial is bounded below by $$P(x)\ge x^{2n+1}-\sum_{k=0}^{2n}|a_k|x^k\ge x^{2n+1}-x^{2n}\sum_{k=0}^{2n}|a_k| =x^{2n}\left(x-\sum_{k=0}^{2n}|a_k|\right)$$ We can now try to push the last expression on the right into positive territory by increasing the lower bound for$x$. At the Lagrange root bound $$R=\max\left(1,\sum_{k=0}^{2n}|a_k|\right),$$ the right side for$x\ge R$gives a non-negative bound. Increasing the lower bound to$x\ge 2R$will result in $$x≥2R \implies P(x)\ge (2R)^{2n}\cdot R\ge 2^{2n}>0.$$ The same reasoning can be applied to$-P(-x)=x^{2n+1}-a_{2n}x^{2n}+a_{2n-1}x^{2n-1}\mp...-a_0$, so that $$x≤-2R \implies P(x)≤-(2R)^{2n}\cdot R≤-2^{2n}<0.$$ In total one obtains $$P(-2R)≤-(2R)^{2n}\cdot R ≤ -2^{2n}<0<2^{2n}≤(2R)^{2n}\cdot R≤P(2R)$$ which allows to apply the intermediate value theorem for$P$concluding for a real root of$P$inside$(-2R, 2R)$, but really already inside$(-R,R)$. • I'm curious on what you mean by your first assertion. I'm feel that many proofs of the FTA doesn't require the fact that all odd degree polynomials have at least one real root. For example, off the top of my head, the proof using Liouville's theorem doesn't seem to require this fact. – EuYu Feb 25, 2014 at 8:24 • This is then an almost purely analytical proof. If one tries to prove the FT of Algebra with almost exclusively algebraic means one sees that it is impossible, you need properties of real numbers, continuous functions and the intermediate value theorem in some fashion. In the maximum principle of harmonic functions these basic facts are buried some layers deep. Feb 25, 2014 at 8:49 • I split the first sentence to reflect that. All proofs need calculus, the minimal amount of calculus is required for the Euler-Gauß II proof using only the roots of odd-degree polynomials and complex square roots. Feb 25, 2014 at 8:55 • @LutzL hey i put a bounty on this question because I am desperate for an answer, I am almost close -- I will give it to you if you can answer this for me. I ended up proving that a polynomial$ax^3 + bx^2 + cx + d$always has at least one real root. Can i extend this to ALL odd degree polynomials ? Feb 28, 2014 at 1:40 • If it were that easy, then the FTA would not be such a famous theorem with 100 years of failed or incomplete proof attempts. If a root is known, then factoring by deflation a la Horner-Ruffini is easy. Proving the existence of the root is non-trivial. Feb 28, 2014 at 3:19 Let$p(x)=a_0 + a_1 x + \dots + a_n x^n$a polynomial$p: \mathbb{R} \to \mathbb{R}$with$n$odd and$a_n \neq 0$. Suppose$a_n >0$. We can write$p(x) = a_n x^n \cdot r(x)$with $$r(x) = \frac{a_0}{a_n} \cdot \frac{1}{x^n} + \frac{a_1}{a_n} \cdot \frac{1}{x^{n-1}}+ \cdots + \frac{a_{n-1}}{a_n} \cdot \frac{1}{x} + 1.$$ So, we have $$\lim_{x \to +\infty} r(x) = \lim_{x \to -\infty} = 1.$$ Thus $$\lim_{x \to +\infty} p(x) = \lim_{x \to +\infty} a_n x^n = +\infty$$ and $$\lim_{x \to -\infty} p(x) = \lim_{x \to -\infty} a_n x^n = -\infty$$ because$n$is odd. Therefore the interval$p(\mathbb{R})$is ilimited inferior and superiorly, i.e.$p(\mathbb{R}) = \mathbb{R}$. This means that$p: \mathbb{R} \to \mathbb{R}$is sujective. In particular, there is$c \in \mathbb{R}$that$p(c)=0$. For a polynomial with even degree, take$p(x)=x^2+1$. This have not real roots. • Are you allowed to use limit properties if one sequence diverges? e.g. look at the "Product Rule" here: math24.net/properties-limits. it requires that the limits exist in order to conclude something about the product. Apr 28, 2019 at 2:10 • If you have$\lim_{x\to\infty}f(x)=L>0$and$\lim_{x\to\infty}g(x)=\infty$, then it is safe to say that$\lim_{x\to\infty}f(x)g(x)=\infty$(prove this), which is what we need for$a_n>0$in the polynomial. Eventually, something similar can be done for$a_n<0\$ Dec 8, 2021 at 6:07

Let $$f(x)$$ be a polynomial of odd degree function. Then $$f(x)\to +\infty$$ as $$x\to +\infty$$ and $$f(x)\to -\infty$$ as $$x \to -\infty$$, or vice versa, depending on whether the leading coefficient is positive or negative. Hence, there are $$a, b \in \mathbb{R}$$ such that $$f(a) < 0$$ and $$f(b)> 0$$. Now IVT applies to give an $$x \in [a,b]$$ such that $$f(x) = 0$$. Do you think this is a valid prove?

Let us assume the basic facts about complex field and real number field.

Theorem: (1) The complex field $$\mathbb{C}$$ is algebraically closed; (2) $$\dim_{\mathbb{R}}\mathbb{C}=2$$.

Now let us prove the statement. We first show the following

Lemma: For any $$\alpha\in \mathbb{C}$$, there exists $$a,b,c\in \mathbb{R}$$ such that $$a\alpha^2+b\alpha+c=0.$$

Proof: Consider the set $$S=\{1,\alpha,\alpha^2\}\in \mathbb{C}$$. Since $$\dim_{\mathbb{R}}\mathbb{C}=2$$, the set $$S$$ must be linearly dependent over $$\mathbb{R}$$. The lemma follows.

Lemma: Every $$f\in \mathbb{R}[x]$$ with $$\deg(f)\ge 3$$ is reducible.

Proof: If $$f$$ has a rood in $$\mathbb{R}$$, then of course $$f$$ is reducible. Suppose that $$f$$ has no real root. Since $$\mathbb{C}$$ is algebraically closed, $$f$$ must has a complex root $$\alpha\in \mathbb{C}$$. By the above Lemma, there exists a degree 2 polynomial $$g\in \mathbb{R}[x]$$ such that $$g(\alpha)=0$$. We can moreover assume that $$g$$ is monic. By Euclidean division, we can write $$f=gd+r$$ with $$r\in \mathbb{R}[x]$$, and $$\deg(r)\le 1$$ or $$r=0$$. Moreover, $$r(\alpha)=f(\alpha)-g(\alpha)d(\alpha)=0$$. If $$\deg(r)=1$$, then it is easy to get a contradiction since $$\alpha$$ is assumed to be non-real. Thus the condition $$r(\alpha)=0$$ must imply that $$r=0$$ (as a polynomial). Thus $$f=dg$$. Since $$\deg(f)\ge 3$$ and $$\deg(g)=2$$, $$g$$ must be a proper factor of $$f$$. Thus $$f$$ is reducible.

Lemma: Every odd degree polynomial $$f\in \mathbb{R}[x]$$ must have a real root.

Proof: Consider the prime factorization of $$f=p_1^{r_1}\dots p_k^{r_k}$$ with irreducible polynomials $$p_i\in \mathbb{R}[x]$$. By the above Lemma, $$\deg(p_1)=1$$ or $$2$$. If there is one $$p_i$$ with $$\deg(p_i)=1$$, we are done. If each $$\deg(p_i)=2$$, then $$\deg(f)$$ is even, which contradicts to $$\deg(f)$$ is odd. We are done.