8. Problem: Find all non-zero real numbers $a, b, c$ such that the following polynomial has four (not necessarily distinct) positive real roots.
$$
P(x)=a x^{4}-8 a x^{3}+b x^{2}-32 c x+16 c
$$
Solution: (Viet Hoang)
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
$$
\begin{aligned}
x_{1}+x_{2}+x_{3}+x_{4} & =\frac{8 a}{a}=8 \\
x_{1} x_{2}+x_{1} x_{3}+x_{1} x_{4}+x_{2} x_{3}+x_{2} x_{4}+x_{3} x_{4} & =\frac{b}{a} \\
x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4} & =\frac{32 c}{a} \\
x_{1} x_{2} x_{3} x_{4} & =\frac{16 c}{a}
\end{aligned}
$$

From 4 and 5, we have
$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=\frac{x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}}{x_{1} x_{2} x_{3} x_{4} i}=\frac{\frac{32 c}{a}}{\frac{16 c}{a}}=2
$$

Using AM-GM (arithmetic mean geometric mean inequality) on the sums in equation (2) and (6) we get
$$
\begin{array}{c}
\frac{8}{4}=\frac{x_{1}+x_{2}+x_{3}+x_{4}}{4} \geqslant \sqrt[4]{x_{1} \cdot x_{2} \cdot x_{3} \cdot x_{4}} \\
\frac{2}{4}=\frac{\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}}{4} \geqslant \sqrt[4]{\frac{1}{x_{1}} \cdot \frac{1}{x_{2}} \cdot \frac{1}{x_{3}} \cdot \frac{1}{x_{4}}}
\end{array}
$$

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 $b=24 a$. Substituting this into equation (5) gives us $c=a$. So $(a, b, c)=(a, 24 a, a)$ where $a$ is any non-zero real number, and the polynomial $P(x)$ is
$$
P(x)=a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)=a(x-2)^{4} .
$$