Find $m$ such that a given polynomial has all its roots real 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 $m \in (-\infty, 0) \cup (2, \infty)$
 A: 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,
$$m\in (-\infty , -1]\cup [3,\infty )$$
A: $x_1^2+x_2^2+x_3^2\ge0$ 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.
A: 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
