roots of a cubic polynomial Consider a cubic polynomial of the form
$$f(x)=a_3x^3+a_2x^2+a_1x+a_0$$
where the coefficients are non-zero reals. Conditions for which this equation has three real simple roots are well-known. What conditions would guarantee that none of these roots is positive? In other words, what constraints on the parameters would guarantee that the polynomial has no positive roots? Please provide references also, if possible.
 A: Using Routh–Hurwitz: 


*

*if $a_0, a_1, a_2, a_3 > 0$, you arrive at the condition $a_2 a_1 >
   a_3 a_0$.

*if $a_0, a_1, a_2, a_3 < 0$, you also arrive at the condition $a_2 a_1 >
   a_3 a_0$.

*if $a_3$ and $a_2$ have different signs, then the polynomial has positive roots.

A: You can do it like this. 
If $a_3>0$ then $f(x)$ is increasing except for the region between its two turning points. To have none of the three zeros positive requires that $f(0)\geq 0$ i.e. $a_0 \geq 0$. We also need that $f(x)$ is increasing at $x=0$.
The turning points come between zeros of $f(x)$ so the other condition is that the zeros of the derivative $f'(x)$ are both negative (this deals with increasing at $x=0$ and the values can easily be expressed in terms of the coefficients). [if we did not have the condition that there were three distinct roots already, it would be possible for the derivative to have no real roots, and the condition on $a_0$ would then ensure that the one real root was not positive].
If $a_3<0$ apply the same criteria to $-f(x)$.
With $a_3>0$ the condition for three distinct real roots is equivalent to the local maximum (existing and) being greater than zero, and the local minimum being less than zero.
