Algebra Precalculus – Finding Roots of the Polynomial Equation Using Descartes’ 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

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

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

or after tidying

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