The roots of $x^3+mx+n$ where $n<0$ This question is from a high school book under the "applications of derivates" section.
The question says, 

Given $x^3+mx+n$ where $n<0$ and there are three roots $r_1<r_2<r_3$,
  what can be surely said about the roots?

Clearly, the product of the roots is a positive number. So, either all roots are positive, or only one of them is positive. Also, since there is no term with $x^2$, the sum of the roots is zero, so, they can't be all positive. 
Hence, the answer is $r_1<r_2<0<r_3$.
However, I fail to see how this result can be derived using derivatives.
 A: You made a mistake when you stated that the product of the three roots is negative. Indeed, by Vieta's formulas, $n = -r_1 r_2 r_3$, so the product $r_1 r_2 r_3$ must be positive.
Anyway, let us prove the result using derivatives. Let $f(x) = x^3 + m x + n$. Then $f'(x) = 3x^2 + m$.
You can prove that $m < 0$ (otherwise we get a contradiction). Then $f'(x) > 0$ if and only if $x < -k \lor x > k$, where $k = \sqrt{-m/3}$. Therefore $f$ is increasing in $]-\infty, -k[$, decreasing in $]-k, k[$, and again increasing in $]k, +\infty[$.
Since $f(0) = n < 0$ and $f$ is first decreasing and then increasing in $]0, +\infty[$, we see that there is exactly one positive root, which must be $r_3$. Then the other roots $r_1$ and $r_2$ are negative, and so $r_1 < r_2 < 0 < r_3$.
A: let $$f(x)=x^3+mx+n$$ where $$n<0$$ then we get
$$f'(x)=3x^2+m$$
and we have in the first case where $$m\geq 0$$ then our $$f(x)$$ is monotonously increasing and since we have $$n<0$$ exactly one root
in the case of $$m<0$$ we get
$$x_{1,2}=\pm\sqrt{\frac{-m}{3}}$$ then we get since $$f''(x)=6x$$ :
$$f''(x_1)>0$$ so we have a minimium point at $$(x_1;f(x_1)$$ in the other case a maximum point at $$(x_2,f(x_2))$$
can you proceed?
