Let \(p\in\RR[x]\) be a nonzero polynomial, and let \(V(p)\) be the number of sign changes in the sequence of its nonzero coefficients, listed in order of degree. The number \(r^+\) of positive real roots of \(p\), counted with multiplicity, satisfies \[ r^+\le V(p)\qquad\text{and}\qquad V(p)-r^+\in2\ZZ. \] The negative real roots of \(p\) are the negatives of the positive real roots of \(p(-x)\), so their number \(r^-\), counted with multiplicity, satisfies \(r^-\le V(p(-x))\) and \(V(p(-x))-r^-\in2\ZZ\).
Consequently, if \(p(0)\ne0\) and \(\deg p=d\), then \(p\) has at least \(d-V(p)-V(p(-x))\) nonreal complex roots, counted with multiplicity.
Let \(f(x)=x^3+x^2-x-1\). Its coefficient signs are \(+,+,-,-\), so \(V(f)=1\) and \(f\) has exactly one positive root. The polynomial \(f(-x)=-x^3+x^2+x-1\) has signs \(-,+,+,-\), so \(V(f(-x))=2\) and \(f\) has \(2\) or \(0\) negative roots; in fact \(f=(x-1)(x+1)^2\) has the negative root \(-1\) with multiplicity \(2\).
Let \(g(x)=x^3-1\). Then \(V(g)=1\), so \(g\) has exactly one positive root, and \(g(-x)=-x^3-1\) has \(V(g(-x))=0\), so \(g\) has no negative root. Hence \(g\) has at least \(3-1-0=2\) nonreal roots, so exactly \(2\).