This result is believed to have been first described by Réné Descartes in his 1637 work La Géométrie. In 1828, Carl Friedrich Gauss improved the rule by proving that when there are fewer roots of polynomials than there are variations of sign, the parity of the difference between the two is even.

Let $p(x)=x^{3}+3x^{2}-x-3$. Looking at the list of coefficients, we have $1,3,-1,-3$, so there is only one variation in sign (from $3$ to $-1$).

Thus $v=1$. Since $0\leq N\leq\frac{1}{2}$ then we must have $N=0$. Thus $v-2N=1$ and so there is exactly one positive real root of $p(x)$.

To find the negative roots, we examine $p(-x)=-x^{3}+3x^{2}+x-3$. The coefficient list is $-1,3,1,-3$, so here are are two variations in sign (from $-1$ to $3$ and $1$ to $-3$. Thus $v=2$ and so $0\leq N\leq\frac{2}{2}=1$.

Thus we have two possible solutions, $N=0$ and $N=1$, and two possible values of $v-2N$. Therefore there are either two negative real roots or none at all.

We note that $p(-1)=(-1)^{3}+3\cdot(-1)^{2}-(-1)-3=0$, hence there is at least one negative root. Therefore there must be exactly two.