<s>[INST] Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots. 
Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.

For an equation to have nonnegative roots, it mustn't only have negative roots.
Let $y = x^2 - x + 4$ $(y > 0)$, we have that $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = mx^2 + (2m - 1)xy + my^2$$
For the equation $$mx^2 + (2m - 1)xy + my^2 = 0$$ to have only negative roots, it must be satisfied that $\left\{ \begin{align} [(2m - 1)y]^2 - 4m^2y^2 \ge 0\\ (1 - 2m)y < 0\\ my^2 > 0 \end{align} \right.$ $\implies \left\{ \begin{align} (1 - 4m)y^2 \ge 0\\ 1 - 2m < 0\\ m > 0 \end{align} \right.$ $\implies \left\{ \begin{align} 1 - 4m \ge 0\\ m > \dfrac{1}{2}\\ m > 0 \end{align} \right.$ $\iff \left\{ \begin{align} m \le \frac{1}{4}\\ m > \dfrac{1}{2} \end{align} \right. \implies m \in \varnothing$.
Thus for $\forall m \in \mathbb R$, the equation $$mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
Is the above solution correct? And should my attempt was inaccurate and yours is (hopefully) helpful, please post an answer.
 [/INST] Express $m$ as a function of $x\ge 0$,
$$m(x) = - \frac{x(x^2-x+4)}{(x^2+4)^2}\le0\tag 1$$
It is equivalent to find the lower bound of $m(x)$ for $x>0$. Set $m'(x) = 0$ to get
$$x^4-2x^3+8x-16=(x-2)(x+2)(x^2-2x+4) = 0$$
which shows that the minimum is at $x=2$. Plug it into $(1)$ to get $m(2) = -\dfrac3{16}$. Thus, below are all values of $m$ for non-negative roots of $x$, 
$$-\frac3{16} \le m \le 0$$
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