MHB Does the Polynomial $P(x)=x^3+mx^2+nx+k$ Have Three Distinct Real Roots?

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The polynomial \( P(x) = x^3 + mx^2 + nx + k \) is analyzed under the conditions \( n < 0 \) and \( mn = 9k \). The discussion focuses on proving that these conditions ensure the polynomial has three distinct real roots. Various approaches to the proof are shared, highlighting the importance of the relationships between the coefficients. The conversation emphasizes the significance of the discriminant and the behavior of the polynomial's derivative. Ultimately, the conditions provided are sufficient for establishing the existence of three distinct real roots.
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A polynomial $P(x)=x^3+mx^2+nx+k$ is such that $n<0$ and $mn=9k$.

Prove that the polynomial has three distinct real roots.
 
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anemone said:
A polynomial $P(x)=x^3+mx^2+nx+k$ is such that $n<0$ and $mn=9k$.

Prove that the polynomial has three distinct real roots.

If $m = 0$, then $k = 0$ and $P(x) = x^3 + nx$. So $P$ has three real distinct roots: $0, \sqrt{-n}$ and $-\sqrt{-n}$.

Now suppose $m \neq 0$. If $m > 0$, then $P(x)$ has one sign change and $P(-x)$ has two sign changes. By Descartes' rule of signs, $P$ has one positive root and zero or two negative roots. Since $P(0) = \frac{mn}{9}$ and $P(-m) = -\frac{8mn}{9}$ have opposite signs, the intermediate value theorem ensures that $P$ has a negative root. Therefore $P$ has two negative roots. By a similar argument, if $m < 0$, then $P$ has one negative root and two positive roots. Hence in both cases, $P$ has three real roots, two of which have different signs. So it suffices to show that $P$ has no double root.

Since $P$ is cubic, if it has a double root, then that root is a critical point. The critical points of $P$ are $\frac{-m+\sqrt{m^2-3n}}{3}$ and $\frac{-m-\sqrt{m^2-3n}}{3}$. Neither of these satisfy $P$. Indeed, by the second derivative test, these points give a global min and max for $P$. But $P$ is nonconstant, so they cannot be roots of $P$. So, $P$ has no double root. Therefore, $P$ has three distinct real roots, with exactly two roots of the same sign.
 
Well done, Euge! And thanks for participating!

Here is one solution that I saw online that approaches it differently than you, so I will share it with you and the rest of the members:

Consider the derivative $P'(x)=3x^2+2mx+n$. Since $n<0$, it has two real roots $x_1$ and $x_2$. Since $P(x)\rightarrow \pm \infty$ as $x\rightarrow \pm \infty$, it is sufficient to check that $P(x_1)$ and $P(x_2)$ have different signs, i.e., $P(x_1)P(x_2)<0$.

Dividing $P(x)$ by $P'(x)$ and using the equality $mn=9k$, we find that the remainder is equal to $x\left(\dfrac{2n}{3}-\dfrac{2m^2}{9}\right)$. Now, as $x_1x_2=\dfrac{n}{3}<0$, we have $P(x_1)P(x_2)=x_1x_2\left(\dfrac{2n}{3}-\dfrac{2m^2}{9}\right)^2<0$ and we're done.
 
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