  By Descartes' rule of signs, a real degree $d$ polynomial $P$ with all
nonvanishing coefficients, with $c$ sign changes and $p$ sign preservations in
the sequence of its coefficients ($c+p=d$) has $pos\leq c$ positive and
$neg\leq p$ negative roots, where $pos\equiv c($\, mod $2)$ and $neg\equiv
p($\, mod $2)$. For $1\leq d\leq 3$, for every possible choice of the sequence
of signs of coefficients of $P$ (called sign pattern) and for every pair $(pos,
neg)$ satisfying these conditions there exists a polynomial $P$ with exactly
$pos$ positive and exactly $neg$ negative roots (all of them simple). For
$d\geq 4$ this is not so. It was observed that for $4\leq d\leq 10$, in all
nonrealizable cases either $pos=0$ or $neg=0$. It was conjectured that this is
the case for any $d\geq 4$. We show a counterexample to this conjecture for
$d=11$. Namely, we prove that for the sign pattern $(+,-,-,-,-,-,+,+,+,+,+,-)$
and the pair $(1,8)$ there exists no polynomial with $1$ positive, $8$ negative
simple roots and a complex conjugate pair.
