Question

How could you use Descartes' Rule and the Fundamental Theorem of Algebra to predict the number of complex roots to a polynomial, as well as find the number of possible positive and negative real roots to a polynomial?

Answer 1

$\bf f(x)=x^5\stackrel{\downarrow }{-}x^4\stackrel{\downarrow }{+}x^3\stackrel{\downarrow }{-}x^2\stackrel{\downarrow }{+}5\qquad \impliedby \textit{4 sign changes}$

$\bf f(-x)=(-x)^5-(-x)^4+(-x)^3-(-x)^2+5~~ \begin{cases} (-x)^5=(-x)(-x)(-x)\\ \qquad \qquad (-x)(-x)\\ \qquad \qquad -x^5\\ (-x)^4=(-x)(-x)(-x)(-x)\\ \qquad \qquad x^4\\ (-x)^3=(-x)(-x)(-x)\\ \qquad \qquad -x^3\\ (-x)^2=(-x)(-x)\\ \qquad \qquad x^2 \end{cases} \\\\\\ f(-x)=-x^5-x^4-x^3-x^2\stackrel{\downarrow }{+}5\qquad \impliedby \textit{1 sign change}$

so the number of positive real roots is either 4, or (4-2) 2, or (2-2) 0. And the negative real roots are only 1. Any slack gets picked up by the versatile complex twins.

4 real positive ones and 1 negative real one

or

2 real positive ones and 1 negative real one and 2 complex ones

or

0 real positive ones and 1 negative real one and 4 complex ones.