On the number of roots of the polynomial $x^3+Ax^2+1=0$

If we can show that the value of $A$ changes the amount of maxima or minima from 1 to 2 such that one extreme is at a positive $y$-location and the other at a zero or negative $y$-location, then we know that this value of $A$ is the border for all real solutions versus 1 real and 2 complex solutions.

Denote $f(x) = x^3 + Ax^2 + 1.$

$$f'(x) = 3x^2+2Ax=x(3x+2A) = 0.$$

We see that $f(0)$ is always either a local maximum or minimum and $f(0) > 0.$ The other zero is given by $x = -2A/3 = \xi.$

We need $f(\xi) \leq 0.$

$$f(\xi) = \frac{4}{27}A^3 + 1 \leq 0$$

solves for the values of $A$ where 3 real solutions are guaranteed.

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If there are 3 real roots, then since $x = 0$ is always a local extreme, at least one has to be positive and another has to be negative. Also, if $A \leq -3/\sqrt[3]{4}$ and the location of the moving extreme is $x=-2A/3 > 0,$ then the remaining real root has to be positive.

If there's only 1 real root, then $A > -3/\sqrt[3]{4}$ and the moving extreme is located at $x < \sqrt[3]{2}.$ Thus the only real root must be negative.

hint

Put $$A=-\frac 32B$$ and $$f(x)=x^3-\frac 32Bx^2+1.$$

$$f'(x)=3x(x-B).$$

$$f(0)=1$$ to have three real roots, we need

$$B>0 \text{ and } f(B)<0.$$

The stationary points of $f(x)$ lie at $x=0$ and at $x=-\frac{2A}{3}$ and the sign of $$ f(0)f\left(-\tfrac{2A}{3}\right) = 1+\tfrac{4}{27}A^3$$ decides if there are three real roots (<0), a simple real root (>0) or a double real root and a simple real root (=0). This happens since the sign of $ f(0)f\left(-\tfrac{2A}{3}\right)$ only depends on the stationary values having the same sign or not.