Solution:
Assume that $P(x)$ has 4 positive real roots $x_{1}$, $x_{2}$, $x_{3}$ and $x_{4}$. Using Viete's theorem, one can obtain the following equations

$$\begin{array}{c} x_{1} + x_{2} + x_{3} + x_{4} = \frac{8a}{a} = 8 \\ x_{1}x_{2} + x_{1}x_{3} + x_{1}x_{4} + x_{2}x_{3} + x_{2}x_{4} + x_{3}x_{4} = \frac{b}{a} \\ x_{1}x_{2}x_{3} + x_{1}x_{2}x_{4} + x_{1}x_{3}x_{4} + x_{2}x_{3}x_{4} = \frac{32c}{a} \\ x_{1}x_{2}x_{3}x_{4} = \frac{16c}{a} \end{array} \quad (5)$$
From 4 and 5, we have

$$\frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} + \frac{1}{x_{4}} = \frac{x_{1}x_{2}x_{3} + x_{1}x_{2}x_{4} + x_{1}x_{3}x_{4} + x_{2}x_{3}x_{4}}{x_{1}x_{2}x_{3}x_{4}} = \frac{\frac{32c}{a}}{\frac{16c}{a}} = 2 \quad (6)$$
Using AM-GM (arithmetic mean geometric mean inequality) on the sums in equation (2) and (6) we get

$$\frac{8}{4} = \frac{x_{1} + x_{2} + x_{3} + x_{4}}{4} \geqslant \sqrt[4]{x_{1} \cdot x_{2} \cdot x_{3} \cdot x_{4}}$$

$$\frac{2}{4} = \frac{\frac{1}{x_{1}} + \frac{1}{x_{2}} + \frac{1}{x_{3}} + \frac{1}{x_{4}}}{4} \geqslant \sqrt[4]{\frac{1}{x_{1}} \cdot \frac{1}{x_{2}} \cdot \frac{1}{x_{3}} \cdot \frac{1}{x_{4}}}$$
Multiplying these together yields $1 \geqslant 1$ and so equality must occur in both AM-GM's. Therefore $x_{1} = x_{2} = x_{3} = x_{4}$. Furthermore for Equation (2) we get $x_{1} = x_{2} = x_{3} = x_{4} = 2$. Substituting this into equation (3) gives us $b = 24a$. Substituting this into equation (5) gives us $c = a$. So $(a, b, c) = (a, 24a, a)$ where $a$ is any non-zero real number, and the polynomial $P(x)$ is

$$P(x) = a(x - x_{1})(x - x_{2})(x - x_{3})(x - x_{4}) = a(x - 2)^{4}.$$