Find the number of real roots of the polynomial $$f(x)=x^5+x^3-2x+1$$ If I use Descarte's Rule then I get $$f(x)=x^5+x^3-2x+1$$ there can't be more than two positive real roots. Again $$f(-x)=-x^5-x^3+2x+1$$ there can't be more than one negative real root.


It's obvious there is one negative real root, because the graph goes through $(0,1)$, and the end behavior of the graph is down on the left hand side.

Of the remaining four roots, at least two and at most four are complex (because complex roots come in pairs.)

The derivative shows that the only positive real local minimum occurs at $x=\sqrt{0.4}$, but checking this in the equation shows that it is still above the $x$ axis. So the graph does not cross the x-axis on the positive real numbers, and the remaining four roots are complex.

  • $\begingroup$ Can you explain why there is four complex roots not two ? $\endgroup$ – Argha Aug 8 '12 at 17:38
  • $\begingroup$ Sure, and you also alerted me to an oversight I made. Luckily it doesn't change the outcome. The strategy I used was to look for the local minima on the positive numbers, and use that with the fact the graph goes up forever on the right. From the derivative you can conclude there is only one local minimum over the positive numbers, and it occurs above the x-axis. Clearly it's impossible that the graph dips below the x-axis, because it would eventually have to turn around and go up, creating a minimum with a negative y coordinate. $\endgroup$ – rschwieb Aug 8 '12 at 17:47

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