I have the following polynomial with real coefficients:
$$f(x)=x^3-(m+2)x^2+(m^2+1)x-1$$
I have to find all real $m$'s so that all of the roots of $f$ are real.
Trying to guess a root didn't get me anywhere.
I computed $x_1^2+x_2^2+x_3^2$ using Vieta's relations to be $-(m-2)^2+6$. This has to be positive if the roots are real, so $m\in[-\sqrt6+2, \sqrt6+2]$.
I tried using the derivative of $f$ and Rolle's theorem, but the calculations get complicated quite fast. I managed to prove that m has to be somewhere in the interval $(-\sqrt\frac32+1, \sqrt\frac32+1)$, though I can't guarantee that this is correct. I could continue this way and I'll probably reach a solution sooner or later, but I hope there's a much more elegant solution that I've missed.
Thanks for your help!
Best Answer
I shall assume that we want three different real roots Consider $$f(x)=x^3-(m+2)x^2+(m^2+1)x-1$$ The first condition is that $$f'(x)=3x^2-2(m+2)x+(m^2+1)$$ shows two real roots which are $$x_\pm=\frac{1}{3} \left(m+2\pm\sqrt{-2 m^2+4 m+1}\right)$$ This gives the first condition $$-2 m^2+4 m+1 > 0$$
Now, you need that $$f(x_-) \times f(x_+) <0$$ that is to say $$3 m^6-4 m^5+6 m^4-22 m^3-9 m^2+26 m+23 < 0$$ which cannot be solved. Numerical calculations give $$1.558 < m < 1.756 $$