Show that some of the root of the polynomial is not real.

Question

\begin{equation*} p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_3x^3+x^2+x+1. \end{equation*}

All the coefficients are real. Show that some of the roots are not real.

I don't have any idea how to do this, I only have an intuition that something is related to the coefficients of $x^2$ and $x$ and the constant 1.

Kindly help me!!

Answer 1

Adapting the argument from this answer by Noam Elkies.

If $p(x)$ has only real zeros, then so does its reciprocal $$ x^np(\frac1x)=x^n+x^{n-1}+x^{n-2}+\cdots+a_{n-1}x+a_n. $$ But by Vieta relations the sum of the roots of this polynomial is $-1$ and the sum of their pairwise products is $1$. Therefore the sum of the squares of the roots is $$(-1)^2-2\cdot1=-1<0. $$ Consequently some of the roots must be non-real.

Answer 2

If a polynomial has only real roots, its coefficients have to satisfy Newton's inequalities.

Since by looking at the coefficients of $x^0,x^1,x^2$ we can see that this is not the case, our polynomial must have some non-real root (and obviously, non-real roots come in conjugated pairs).

Answer 3

Let $\lambda_1,\ldots,\lambda_n$ be the roots of $p(x)$. We have

• coefficient of $x^0 = 1 \implies \prod_{i=1}^n \lambda_i = (-1)^n$
• coefficient of $x^1 = 1 \implies \sum_{j=1}^n \prod_{i=1,\ne j}^n \lambda_i = (-1)^{n-1}$
• coefficient of $x^2 = 1 \implies \sum_{1\le j < k \le n}\prod_{i=1,\ne j, k}^n \lambda_i = (-1)^{n-2}$

Combine these, we have

$$\sum_{i=1}^n \frac{1}{\lambda_i} = -1 \quad\text{ and }\quad \sum_{1\le j < k \le n} \frac{1}{\lambda_j\lambda_k} = 1 $$ These together imply

$$\sum_{i=1}^n \frac{1}{\lambda_i^2} = \left(\sum_{i=1}^n \frac{1}{\lambda_i}\right)^2 - 2 \left(\sum_{1\le j < k \le n} \frac{1}{\lambda_j\lambda_k}\right) = (-1)^2 - 2(1) = -1$$ If all $\lambda_i$ are real, the LHS has to be positive. Since this isn't the case, some of the roots are not real.

Yet another alternate approach

Since robjohn comes up an alternate approach using Sturm's Theorem, let's join the fun and attack this problem with another theorem. Same as everyone else, if all the roots of $p(x)$ are real, so does $$q(x) = x^n p\left(\frac1x\right) = x^n + x^{n-1} + x^{n-2} + \cdots$$ Apply Gauss-Lucas theorem $(n-2)$ times, this will imply all the roots of the polynomial

$$\frac{1}{n!}q^{(n-2)}(x) = \frac12 x^2 + \frac{1}{n}x + \frac{1}{n(n-1)} = \frac12 \left(x + \frac{1}{n}\right)^2 + \frac{n+1}{2n^2(n-1)} $$ are real. However, this contradicts with the obvious fact the polynomial on RHS doesn't have any real root at all! As a result, some of the roots of $p(x)$ are not real.

Answer 4

$p(x)$ has all roots in $\mathbb{R}$ if and only if the polynomial $q(x)=x^np(\frac1x)$ has all its roots in $\mathbb{R}$. We don't need to worry about a root of $0$ since $p(0)=1$ and $q(0)=a_n\ne0$.

The first three polynomials of the Sturm-Chain for $q(x)$ are $$ \begin{align} q(x)&=x^n+x^{n-1}+x^{n-2}+\dots\\ q'(x)&=nx^{n-1}+(n-1)x^{n-2}+(n-2)x^{n-3}+\dots\\ q'(x)\left(\tfrac1nx+\tfrac1{n^2}\right)-q(x)&=\left(-\tfrac1n-\tfrac1{n^2}\right)x^{n-2}+\dots \end{align} $$ To have $n$ real roots, the lead coefficients of the Sturm-Chain must all have the same sign, so that there are $0$ sign changes at $+\infty$ and $n$ sign changes at $-\infty$. However, the lead coefficient of the third polynomial is different from the first two. Thus, $q(x)$ cannot have $n$ real roots and, likewise, $p(x)$ cannot have $n$ real roots.

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Note that if $q$ has $d$ repeated roots then $\deg(\gcd(q,q'))=d$. Thus, the Sturm-Chain ends with a degree $d$ polynomial rather than a constant. This means we need $n-d$ sign changes to get the $n-d$ distinct real roots. This still means we need all the lead coefficients of the Sturm-Chain to have the same sign.

Answer 5

Counterexample: If $n=0$, then $p$ has no roots.