Chapter 2.5, Problem 93E

To determine

To find: The possible numbers of positive and negative real zeros of the function by using the Descartes’s Rule of Signs.

Answer to Problem 93E

From the Descartes Rule of Signs, the polynomial $f\left(x\right)=-5x^3+x^2-x+5$ has only one positive real zero and has one negative real zero.

Explanation of Solution

Given information:

The given polynomial function is $f\left(x\right)=-5x^3+x^2-x+5$ .

Let us consider the following polynomial function.

  $f\left(x\right)=-5x^3+x^2-x+5$

The variation of sign in $f\left(x\right)$ has only one.

  $f\left(x\right)=\overline{)\begin{array}{l}-5x^3+x^2-x+5\\ -+-+\end{array}}$

Also for $f\left(-x\right)$ we have

  $\begin{array}{c}f\left(x\right)=-5{\left(-x\right)}^3+{\left(-x\right)}^2-\left(-x\right)+5\\ =\overline{)\begin{array}{c}+5x^3+x^2+x+5\\ ++++\end{array}}\end{array}$

No sign change for $f\left(-x\right)$ .

Therefore, form the Descartes’ Rule of Signs, the polynomial $f\left(x\right)=-5x^3+x^2-x+5$ has three or one positive real zero, and has no negative real zeros.