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Given a cubic polynomial $f(x) = ax^{3} + bx^{2} + cx +d$ with arbitrary real coefficients and $a\neq 0$. Is there an easy test to determine when all the real roots of $f$ are negative?

The Routh-Hurwitz Criterion gives a condition for roots lying in the open left half-plane for an arbitrary polynomial with complex coefficients which helps a little, but this criterion doesn't help me when the complex roots lie in the right half plane.

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    $\begingroup$ I think Descarte's Rule of signs may be useful here. purplemath.com/modules/drofsign.htm $\endgroup$ – Kartik Jul 16 '15 at 11:08
  • $\begingroup$ @Kartik I've never seen this before, I'll take a look. Thanks $\endgroup$ – JessicaK Jul 16 '15 at 11:17
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    $\begingroup$ Without loss of generality, suppose $a = 1$. You want to check whether the polynomial has a zero in $[0,+\infty)$. If $d = 0$, you have $0$ as a zero (trivial). If $d < 0$, the intermediate value theorem tells you it has a zero in $(0,+\infty)$. If $d > 0$, check the derivative. If $f'$ has no positive real zero or a double zero, $f$ is monotnoically increasing on $[0,+\infty)$, so it has no non-negative real zero. If $f'$ has two distinct zeros, $z_1 < z_2$ with $z_2 > 0$, check $f(z_2)$ if $f(z_2) > 0$, $f$ has no zero in $[0,+\infty)$ otherwise it does. $\endgroup$ – Daniel Fischer Jul 16 '15 at 16:09
  • $\begingroup$ @DanielFischer If you would like to put that as an answer I will be happy to accept it. $\endgroup$ – JessicaK Jul 20 '15 at 13:58
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We can assume that $a = 1$, since dividing by $a$ doesn't change the zeros. Then we know that $\lim\limits_{x\to +\infty} f(x) = +\infty$. We want to check whether $f$ has a zero in $[0,+\infty)$.

If $d = 0$, we have $f(0) = 0$. There can be circumstances when that should count as negative. Then we are reduced to checking a quadratic polynomial, whose zeros we know to find.

If $f(0) = d < 0$, then $f$ has a zero in $(0,+\infty)$ by the intermediate value theorem.

If $d > 0$, we check the derivative,

$$f'(x) = 3x^2 + 2bx + c = 3\biggl(x+\frac{b}{3}\biggr)^2 - \frac{b^2-3c}{3}.$$

If $f'$ has no positive zero, or a double zero, then $f$ is strictly increasing on $[0,+\infty)$, so then $f$ has no non-negative real zero. If $b^2 < 3c$, then $f'$ has two non-real zeros, and if $b^2 = 3c$, $f'$ has a double zero at $-\frac{b}{3}$. If $b^2 > 3c$, then $f'$ has two distinct real zeros. The larger of these is

$$\zeta = \frac{\sqrt{b^2-3c}-b}{3}.$$

We have $\zeta > 0$ if and only if $b \leqslant 0$ or $b > 0$ and $c < 0$. Then $f$ has no positive zero if and only if $f(\zeta) > 0$.

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If you are interested in only the roots you can normalize and take $a=1$. Then a necessary condition is that $b,c,d>0$. As it has 3 negative roots its two turning points should be negative too. That is $f'(x)= 3x^2-2bx+c$ should have real negative roots, which can be easily translated to a condition on the discriminant $b^2-3c\ge0$.

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  • $\begingroup$ I'm not sure I understand. If you consider $x^{3}-x^{2}+x+1$, this has a negative real root and two complex roots but your necessary condition doesn't hold. $\endgroup$ – JessicaK Jul 16 '15 at 11:16
  • $\begingroup$ I said $b,c,d>0$. You are bringing $-1$ as coefficient of $x^2$. Two conditions are needed. $\endgroup$ – P Vanchinathan Jul 16 '15 at 15:29
  • $\begingroup$ The question is not when all three roots are real and negative, but when the real roots the polynomial has are all negative. For example when it has one negative real root and two non-real roots. $\endgroup$ – Daniel Fischer Jul 16 '15 at 16:01

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