Playing around with a graphing calculator, it seems that the quartic polynomial $$P(x)=x^4+ax^3+bx-1$$ always has exactly 2 real (distinct) roots, where $a,b \neq 0$. However, this is something that I've not been able to formally prove.

Is it actually the case that there will always be exactly 2 real distinct roots, or is it possible for some $(a,b)$ such that it has exactly 4 real distinct roots?

Things I have so far:

  • There are points of inflexions at $x=0$ and $x=\frac{-a}{2}$.
  • $\begingroup$ Are you counting multiplicity? $x^4 + 2x^3 - 2x - 1 = (x+1)^3 (x-1)$ and has four roots. $\endgroup$
    – user14972
    Aug 9, 2016 at 13:27
  • $\begingroup$ I should have clarified. I am not counting multiplicities. In other words, I'd like to see a case where there are four distinct real roots, as I've been unable to find one. $\endgroup$
    – Trogdor
    Aug 9, 2016 at 13:29

2 Answers 2


Note that if $a=-b$ then $$x^4-b x^3+bx-1=(x-1)(x+1)(x^2-bx+1)$$ So if $b^2-4>0$, that is $|b|>2$ you will have $4$ distinct real roots (if $x=\pm1$ and $x^2-bx+1=0$ then $b=\pm2$).


Wolframalpha says that

$$x^4 - 6 x^3 + 5 x - 1 = 0$$

has four real roots.

This was found by starting with Descartes' rule of signs which says this sign pattern has either 1 or 3 positive zeroes, and then playing with the coefficients. The main thing is that the coefficients had to be larger, so as to dwarf the $-1$ term.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .