## Precalculus (6th Edition) Blitzer

We have a maximum of $3$ positive real roots and a maximum of $1$ negative real root.
We need to use Descartes' Rule. Looking at the coefficients of the initial function, we count 3 changes in signs. Thus, we have a maximum of $3$ positive real roots. Plug in $-x$ into the function: $(-x)= 3(-x)^4-2(-x)^3+5(-x)^2+(-x)-9$ or, $f(-x)= 3x^4+2x^3+5x^2-x-9$ We have one change in sign. Thus, we have a maximum of $1$ negative real root. So, we have a maximum of $3$ positive real roots and a maximum of $1$ negative real root.