$$x^6\pm x^5\pm x^4 \pm x^3 \pm x^2 \pm x \pm 1 = p(x)$$
What is the set $A$ of possible polynomials from the class of polynomials $p(x)$ such that the polynomial only has real roots.
I am confused over how to approach the problem. Should I use Descarte's rule?
Descartes' rule just gives informations on the number of positive/negative real roots.
Newton's inequalities are the key, here. Assuming that such a polynomial completely splits over $\mathbb{R}$, we must have: $$ \left(\frac{1}{6}\right)^2 \geq \pm \frac{1}{\binom{6}{0}}\cdot\frac{1}{\binom{6}{2}} = \pm\frac{1}{15}$$ but that may happen only if the sign is negative. So we have that the coefficient of $x^6$ and the coefficient of $x^4$ have to be opposite. Can you fill the missing details?
