Hint: This can give some information about the possible location of roots, to help eliminate what you actually have to test. (Note: everything here refers to real roots and real zeroes.)

Write your polynomial as the function $$p(x) = x^3-6x^2-2x+40$$ and note that its derivative $$p'(x) = 3x^2 -12x-2$$
has exactly two distinct real zeroes $d_0&lt;d_1$ given by $2\pm\frac16\sqrt{42}$. So, moving from left to right, the graph of $p$ rises to the left of $d_0$, falls between $d_0$ and $d_1$, and rises again to the right of $d_1$.

Then the possibilities depend on the values of $y_j \equiv p(d_j)$:


if $y_0&lt;0$, $p$ has exactly one zero (it is to the right of $d_1$);
if $y_0=0$, $p$ has exactly two zeroes ($d_0$, and another to the right of $d_1$);


otherwise $y_0&gt;0$, and we have that


if $y_1&lt;0$, $p$ has three zeroes (one to the left of $d_0$, one between $d_0$ and $d_1$, and one to the right of $d_1$)
if $y_1=0$, $p$ has two zeroes (one to the left of $d_0$, and one at $d_1$;
otherwise $y_1&gt;0$, and $p$ has one zero (it is to the left of $d_0$)


Roots of this third degree polynomial

  The eagle flies at midnight.


What's the origin and meaning of this idiom?
