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.

  • 3
    $\begingroup$ I don't think there are three sign changes in $f(-x)$ - just one. $\endgroup$ Aug 29, 2018 at 19:55
  • $\begingroup$ yes.I notice it now.but any problem arise with those case then how could we solve it? @Jason DeVito $\endgroup$
    – emonHR
    Aug 29, 2018 at 20:08

3 Answers 3



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

or after tidying

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

  • $\begingroup$ I just only need the nature of roots not them.please use Descartes's rule of sign to show it. $\endgroup$
    – emonHR
    Aug 29, 2018 at 19:56
  • 2
    $\begingroup$ The problem you have with Descartes' rule of signs is that It is not a complete criterion, because it does not provide the exact number of positive or negative roots. So my answer gives you all you need. $\endgroup$
    – Mandelbrot
    Aug 29, 2018 at 19:59

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

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.


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.

  • $\begingroup$ Why couldn't there be $3$ positive and $1$ negative root? $\endgroup$ Aug 29, 2018 at 20:11
  • $\begingroup$ For $f(-x)$ you can have 3 negative roots or 1 negative root and two imaginary. So if you have only one negative root, two imaginary and 3 positive roots, the polynomial has to have at least 6 roots. $\endgroup$
    – Andrei
    Aug 29, 2018 at 20:20
  • $\begingroup$ I'm confused. If, we use the polynomial $f(x) = (x-4)(x-5)(x-6)(x+1) = x^4 - 14x^3 +59x^2 -46x - 120$, then $f(x)$ has the same sign pattern as OPs polynomial, but manifestly has $3$ positive and $1$ negative root. What am I missing? $\endgroup$ Aug 29, 2018 at 20:30
  • $\begingroup$ You are correct. I did not verify the statement that there are three sign changes in $f(-x)$. You have only one. My mistake. $\endgroup$
    – Andrei
    Aug 29, 2018 at 20:42
  • $\begingroup$ Oh - I wasn't even referencing that issue. But in any event, we agree that there is preicsely one negative root. And I think we also agree that the number of positive roots is three or one. How are we ruling out the case where there are three positive roots? I still feel like I'm missing something little... $\endgroup$ Aug 29, 2018 at 20:51

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