TITLE: On the number of roots of the polynomial $x^3+Ax^2+1=0$
QUESTION [6 upvotes]: I have the following cubic equation
$$x^{3}+Ax^{2}+1=0$$
where $A$ is an arbitrary (real) number.
I know that either:

The 3 roots will be real.
One root will be real and the other two will
be complex conjugates of each other.

I would like to find out

For what value/values of A the roots change from 3 real roots to one
real and two complex roots. 
The signs of each of the real roots (both when they are all real and when there is only one real root)

Is there an analytical way of finding this as a function of $A$ or the only option is to solve the cubic numerically?

REPLY [3 votes]: Too late at the party. But it looks good, so let's enter anyway.

TL;DR
$\quad f(x)=x^3+Ax^2+1\,$ always has a simple zero on the negative axis.
The other two zeros are

real, positive and distinct, if $\,A<-1.88988\:=\:-\dfrac{3}{\sqrt[3]{4}}\:=\:A_1$
merging into a double zero at $x=\sqrt[3]{2}$, if $\,A=A_1$
non-real (thus complex-conjugate), if $\,A>A_1$


The party-giver also asked if there is an analytical way of finding this as a function of $A$?
Let's take a step in this direction, concentrating on the case of "one real zero", and express the negative zero as a function of $A$:
Let $\,d=\big(\frac A3\big)^3 + \big(\frac 12\big)^2$, and assume $d>0$ which corresponds to $\,A>A_1$. Define
$$r \:=\:\left(-\frac 12 -\sqrt d\right)^\frac 13
 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\,,$$
$r\,$ is observed to be strictly negative. Then
$$\begin{align}
r^2\:= & \;\left(-\frac 12 -\sqrt d\right)^\frac 23
 +\, \left(-\frac 12 +\sqrt d\right)^\frac 23 -\frac23 A \\[2ex]
r^3\:= & \; -\frac23 Ar + \left(-\frac 12 -\sqrt d\right)
 +\, \left(-\frac 12 +\sqrt d\right) \\
 & \quad -\frac13 A \left\{ \left(-\frac 12 -\sqrt d\right)^\frac 13
 +\, \left(-\frac 12 +\sqrt d\right)^\frac 13\right\} \\
 = & -Ar - 1
\end{align}$$
Whence
$$1+Ar+r^3=0\quad\Longleftrightarrow\quad\frac1{r^3}+A\frac1{r^2}+1 =0 \\[2ex]
\Longrightarrow\quad x \:=\: \frac1r \:=\: -r^2-A \:=\: -\frac13 A
 -\left(\sqrt d + \frac 12\right)^\frac 23
 -\, \left(\sqrt d -\frac12\right)^\frac 23 $$
satisfies $x^3+Ax^2+1=0$.
Added in Edit: Driven by user2175783's comment as of below, I found Lecture 4 in Mathematical Omnibus: 30 Lectures on Classic Maths by D. Fuchs + S. Tabachnikov: It gives an extensive and readable account of how to explicitly solve the cubic and quartic equations.

Visual TL;DR