Show that the polynomial $f(x) = 3x^4 - 2 x^3 + x^2 + ax - 1$ with $a \in \mathbb{R}$ does not have all of its roots real.

Consider the polynomial:

$$f(x) = 3x^4 - 2x^3 + x^2 + ax - 1$$

with $$a \in \mathbb{R}$$ and the roots $$x_1, x_2, x_3, x_4 \in \mathbb{C}$$.

I have to show that the polynomial $$f$$ cannot have all of its roots real. I tried looking for something obvious by using Vieta's formulas:

$$V_1 = x_1 + x_2 + x_3 + x_4 = \dfrac{2}{3}$$

$$V_2 = x_1x_2 + x_1x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = \dfrac{1}{3}$$

$$V_3 = x_1x_2x_3 + x_1 x_2 x_4 + x_2 x_3 x_4 = -\dfrac{a}{3}$$

$$V_4 = x_1 x_2 x_3 x_4 = -\dfrac{1}{3}$$

But looking at all of this, it doesn't look like there would be a problem if all of the roots would be real. Since $$a \in \mathbb{R}$$, even $$V_3$$ would look plausible.

So, how can I approach this problem? How can I show that all of the roots cannot be real?

• Can you use calculus or not? The calculus answer is that the second derivative has no roots so the first derivative has at most one root so the function has at most two roots. – Ian Feb 16 '20 at 17:21

Non calculus approach-

$$V_1 = x_1 + x_2 + x_3 + x_4 = \dfrac{2}{3}$$

Implies $$(x_1 + x_2 + x_3 + x_4)^2= \dfrac{4}{9}$$

Implies $$(x_1)^2 + (x_2)^2 + (x_3)^2 + (x_4)^2 + 2(x_1x_2 + x_1x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4) = \dfrac{4}{9}$$

However- $$x_1x_2 + x_1x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = \dfrac{1}{3}$$

Implies $$(x_1)^2 + (x_2)^2 + (x_3)^2 + (x_4)^2 = \dfrac{-2}{9}$$

Implies that not all the roots can be purely real.

Because $$f''(x)=36x^2-12x+2>0.$$

If all roots are real, so by the Rolle's theorem $$f''$$ has two real roots, which is a contradiction.

• Can down-voter explain us, why did you do it? – Michael Rozenberg Feb 16 '20 at 17:53