I'm using Descartes' rule of signs to find the number of roots of $$f(x)=x^4+2x^2+x+8$$
Since there are no sign changes, there should be no positive roots. To find negative roots I apply the rule for $f(-x)$ which is $$f(-x)=x^4+2x^2-x+8$$
which has $1$ sign change. Then I use the fact that if the number of sign changes is zero or one, the number of positive roots equals the number of sign changes to conclude that this polynomial has $1$ root. However, plotting this function shows that it has no roots!
What mistake am I making here?
Best Answer
The error lies in the fact that there are two sign changes: one from $2$ to $-1$ and another one from $-1$ to $8$.
As a side note, you don't have to graph it to see that it has no roots. This follows from the fact that$$(\forall x\in\Bbb R):f(x)=x^4+x^2+\left(x+\frac12\right)^2+\frac{31}4>\frac{31}4.$$