Question

Using the Descartes Rule of Signs, describe the real zeroes of the function.

Answer 1

as you already know, Descartes rule of signs check the times the sign changes or f(x) and for f(-x)

$\bf \stackrel{\textit{positive roots}}{f(x)=}\underset{change}{2x^5-}x^4\underset{change}{-2x^3+}4x^2\underset{change}{+x-2} \\\\\\ \stackrel{\textit{negative roots}}{f(~-x~)=}-2x^5\underset{change}{-x^4+}2x^3\underset{change}{+4x^2-}x-2$

by the fundamental theorem of algebra, the polynomial has a degree of 5, so it has to have at most 5 zeros/solutions/roots.

f(x) has 3 sign changes, notice, that means, it has either 3, or (3-2), 1 positive zeros.

for f(-x), recall that x³, will be (-x)³ = (-x)(-x)(-x) = -x³, so in short, if the exponent is ODD, the sign changes for that term, if it's EVEN, it doesn't change.

so for f(-x), we have 2 sign changes, meaning, it has either 2 or (2-2), 0 negative roots.

the slack is picked up by the complex roots.

so

3 positive, 2 negative, 0 complex

or

1 positive, 2 negative, 2 complex  *recall complex always come in pairs*

or

3 positive, 0 negative, 2 complex

or

1 positive, 0 negative, 4 complex.