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.

Question

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?

Answer 1

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.

Answer 2

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.