There's infinitely many such polynomials!

One example of a quartic polynomial with the given conditions is as follows:

The above graph can be shifted horizontally and vertically without changing the area of the triangle, so for convenient calculations, let's shift it so that the middle turning point is at the origin.

Then the quartic will have a double root at $0$ and (since it must be symmetrical) two other roots at $x = \pm p$ , therefore giving it an equation $y = x^2(x - p)(x + p) = x^4 - p^2x^2$ for some real value of $p$ .

The turning points are when its derivative is $0$ , so $4x^3 - 2p^2x = 0$ , which solves to $x = 0$ and $x = \pm \frac{\sqrt{2}}{2}p$ , and gives the turning points $(0, 0)$ and $(\pm \frac{\sqrt{2}}{2}p, -\frac{1}{4}p^4)$ .

Since the ratio of the height of an equilateral triangle to half its side is $\sqrt{3}$ , using the coordinates of the right turning point we have $\sqrt{3} = \frac{\frac{1}{4}p^4}{\frac{\sqrt{2}}{2}p}$ , which solves to $p = \sqrt[6]{24}$ .

Then the bottom side of the equilateral triangle is $s = 2\frac{\sqrt{2}}{2}p = \sqrt{2}\sqrt[6]{24}$ , so its area $K = \frac{\sqrt{3}}{4}s^2 = \frac{\sqrt{3}}{4}(\sqrt{2}\sqrt[6]{24})^2 = \sqrt[6]{243}$ . Therefore, $A = \boxed{243}$ .

Consider, WLOG*, the monic quartic $y=(x^2-a^2)^2$ .
This quartic has turning points $( \pm a, 0), (0, a^4)$ .
For the points to form an equilateral triangle, $a^4 = a \sqrt 3$ , i.e. $a^3 = \sqrt{3}$ .
Area of triangle is given by $\frac 12 \cdot 2a\cdot a^4 = a^5 = \sqrt[6]{A}$ (given).
Hence $A=a^{30}=(\sqrt{3})^{10}=3^5 = \color{#D61F06}243$ .

* This quartic has two real coincident roots at each of $(\pm a, 0)$ but it can easily be converted to a monic quartic with four real distinct real roots with a simple vertical translation, and the analysis here would still hold.