Chapter 4.6, Problem 41E

To determine

To find: the zeros which is not a possible classification for $f\left(x\right)=x^5-4x^3+6x^2+2x-6$

Answer to Problem 41E

The correct option is C.

Explanation of Solution

Given information:

Given options are,

1. Three positive real zeros, two negative real zeros, and zero imaginary zeros.
2. Three positive real zeros, zeros negative real zeros, and two imaginary zeros.
3. One positive real zero, four negative real zeros, and zero imaginary zeros.
4. One positive real zeros, two negative real zeros, and two imaginary zeros.

Calculation:

Consider the function $f\left(x\right)=x^5-4x^3+6x^2+2x-6$ .

The coefficient in $f\left(x\right)$ have 3 sign change, so $f\left(x\right)$ has three or one positive real zero.

Now find $f\left(-x\right)$ by substituting $x=-x$ .

  $\begin{array}{c}f\left(-x\right)={\left(-x\right)}^5-4{\left(-x\right)}^3+6{\left(-x\right)}^2+2\left(-x\right)-6\\ =-x^5+4x^3+6x^2-2x-6\end{array}$

The coefficient in $f\left(-x\right)$ have 1 sign change, so $f\left(x\right)$ has two or zero negative real zero not four.

The classification given in option C is not possible.

Therefore, the correct option is C.