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Can all the roots of the polynomial equation (with real co-efficients) $a_nx^n+…+a_3x^3+x^2+x+1=0$ be real ? I tried using Vieta’s formulae $\prod \alpha=\dfrac {(-1)^n}{a_n}$ , $(\prod \alpha )(\sum \dfrac 1 \alpha)=(-1)^{n-1}\dfrac 1 a_n$ , so $\sum \dfrac1\alpha = -1$ but this is not going anywhere , please help .

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The answer is NO.

Suppose $P=1+x+x^2+\sum_{k=3}^{n} a_kx^k$ has $n$ real roots.

Obviously zero is not a root of $P$,

so $Q(x)=x^nP(\frac{1}{x})$ also has $n$ real roots. Let us denote them

by $t_1,t_2,\ldots,t_n$ (the $t_i$ are not necessarily distinct).

Since the expansion of $Q(x)$ starts with $x^n+x^{n-1}+x^{n-2}+\ldots$, we see

that

$$

\sum_{i=1}^{n} t_i=(-1), \ \sum_{1\leq i < j \leq n}t_it_j=1 \tag{1}

$$

But

$$

\begin{array}{lcl}

\sum_{1\leq i < j \leq n}(t_i-t_j)^2 &=&

(n-1)\sum_{i=1}^n t_i^2-2\sum_{1\leq i < j \leq n}t_it_j \\

&=& (n-1)\left(\sum_{i=1}^n t_i\right)^2-

2n\sum_{1\leq i < j \leq n}t_it_j \\

&=& (n-1)-(2n) \\

&=& -(n+1)

\end{array} \tag{2}

$$

which is impossible. This concludes the proof.

Here is another approach. Rolle’s Theorem (a special case of the Mean Value Theorem) states that for a function $f$ continuous on $[a,b]$ and differentiable on $(a,b)$, $f(a) = f(b)$ implies the existence of a $c \in (a,b)$ such that $f(c) = 0$.

We can then say that if a polynomial has $k$ real roots, its derivative must have $k-1$ real roots where each root of the derivative lies between the roots of the original polynomial. Now every root $r$ of a polynomial $P(x)$ of degree $n$ corresponds to a root $r’ = \frac{1}r$ of $x^nP\left( \frac{1}x \right)$. Thus, $$P(x) = \sum_{k = 3}^n a_kx^k+x^2+x+1$$ has all real roots iff $$G(x) = x^n+x^{n-1}+x^{n-2} + \sum_{k = 3}^na_kx^{n-k}$$ has all real roots. Taking the $n-2$th derivative of $G$ gives us that $$\frac{n!}{2}x^2+(n-1)!x+(n-2)!1$$ has $2$ real roots or $n(n-1)x^2+(n-1)x+1$ has real roots. However, $$G_{\delta} = (n-1)^2-4n(n-1) < 0$$ for all $n > 1$. Thus the result follows.

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