Find the nature of the roots of the equation $ 6x^4-25x^3+81x^2-9x-13=0$ using Descartes's rule of signs

Question

Let $f(x)= 6x^4-25x^3+81x^2-9x-13=0$
According to Descartes's rule of signs There are three changes of signs in $f(x)$.Therefore, $f(x)=0$ may have three positive roots or one positive root and two imaginary roots.
Again,There are three changes of signs in $f(-x)$.So $f(x)=0$ has one negative root or three negative roots.Now my question is how to conclude my answer with exact nature of the roots.I think some derivative work is needed to conclude the answer but i couldn't figure out.
Thanks in advance.

Answer 1

There are three changes of signs in $f(x)$. Therefore, $f(x)=0$ may have three positive roots or one positive root

Correct.

and two imaginary roots

This does not follow, yet. For all you know at this point, $f$ could have either (a) three positive and one negative root, or (b) one positive and three negative roots, or (c) one positive, one negative and two non-real complex roots.

Again, There are three changes of signs in $f(-x)$.

No, there is only one change of signs, so $f$ has exactly one negative root. This eliminates possibility (b) which leaves (a) and (c) to decide.

One way to see that not all roots can be real is notice that, by Vieta's relations, the sum of the squares of all roots is $\,(25/6)^2 - 2 \cdot(81/6) \lt 0\,$, while a sum of squares of real numbers cannot be negative. This eliminates possibility (a) and leaves (c) one positive, one negative and two non-real complex roots.

Answer 2

$f(x)=(2x-1)(3x+1)(x^2-4x+13)$

so you have roots at $x=\frac{1}{2}, x=\frac{-1}{3}, x=\frac{4\pm\sqrt{16-52}}{2}$

or after tidying

$x=\frac{1}{2}, x=\frac{-1}{3}, x=2\pm3i$

Answer 3

If you have three positive roots, you cannot also have three negative roots, or one negative root and two imaginary one. Same thing if you start with three negative roots. You can therefore conclude that you have one positive, one negative, and two complex roots.

Note As noticed by @JasonDeVito, you don't have three changes of sign in $f(-x)$. So we can only conclude that you have exactly one negative root.