For some real numbers $a$ and $b$, the [equation](https://artofproblemsolving.com/wiki/index.php/Equation)
\[8x^3 + 4ax^2 + 2bx + a = 0\]
has three distinct positive roots. If the sum of the base-$2$ [logarithms](https://artofproblemsolving.com/wiki/index.php/Logarithm) of the roots is $5$, what is the value of $a$? 
$\mathrm{(A)}\ -256 \qquad\mathrm{(B)}\ -64  \qquad\mathrm{(C)}\ -8 \qquad\mathrm{(D)}\ 64  \qquad\mathrm{(E)}\ 256$
Let the three roots be $x_1,x_2,x_3$. 
\[\log_2 x_1 + \log_2 x_2 + \log_2 x_3 = \log_2 x_1x_2x_3= 5 \Longrightarrow x_1x_2x_3 = 32\]
By [Vieta’s formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%E2%80%99s_formulas),
\[8(x-x_1)(x-x_2)(x-x_3) = 8x^3 + 4ax^2 + 2bx + a\]
gives us that $a = -8x_1x_2x_3 = -256 \Rightarrow \mathrm{(A)}$.