Proof that every polynomial of odd degree has one real root

Question

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.

Answer 1

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 $P(y)=0$.

Answer 2

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.

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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 $(-R,R)$.

Answer 3

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.

Answer 4

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?

Answer 5

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.