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Let $f = X^3 + mX^2 + mX + 1$ be a polynomial with real coefficients and $m \in \mathbb{R}$. Find $m$ such that all of $f$'s roots are real.

I could only think about having the following condition: $$x_1^2 + x_2^2 + x_3^2 \geq 0$$

This way, I've got $m \in (-\infty, 0) \cup (2, \infty)$

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    $\begingroup$ hint: we know $x=-1$ is a root. $\endgroup$ – lulu Jun 21 '17 at 9:44
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Hint: Try $X=-1$, and thus factorize the cubic into quadratic polynomial.


Answer

You'll get $$X^3+mX^2+mX+1=(X+1)(X^2+(m-1)X+1)$$ using long division. Which means that $$\Delta=(m-1)^2-4\ge 0$$ $$\implies m^2-2m-3\ge0$$ $$\implies m\ge 3 \;or\; m\le -1$$ And using the notation of set theory, $$m\in (-\infty , -1]\cup [3,\infty )$$

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$x_1^2+x_2^2+x_3^2\ge0$ does not imply that the roots are all real.

As implied by the BAI's answer, when $m=2$, the roots of the cubic equation are $-1$ and the roots of

$$X^2+X+1=0$$

So we may take $x_1=-1$, $\displaystyle x_2=\frac{-1+\sqrt{3}i}{2}$ and $\displaystyle x_2=\frac{-1-\sqrt{3}i}{2}$.

$$x_1^2+x_2^2+x_3^2=1+\frac{-1-\sqrt{3}i}{2}+\frac{-1+\sqrt{3}i}{2}=0$$

But only $-1$ is a real root.

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we have $\mathop {\lim }\limits_{x \to + \infty } f(x) = + \infty$ and $\mathop {\lim }\limits_{x \to - \infty } f(x) = + \infty$. The following of intermedicate value theorem, $\exists x \in\mathbf{R} $ so that f(x)=0. Therefore, roof of this equation ussually exist, for all m

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