Chapter 4.6, Problem 39E

To determine

To find: the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Answer to Problem 39E

The possible numbers of zeros for $g\left(x\right)$ are summarized in below table:

Positive real zeros	Negative real zeros	Imaginary zeros	Total zeros
3	3	0	6
3	1	2	6
1	3	2	6
1	1	4	6

Explanation of Solution

Given information:

Given function,

  $g\left(x\right)=x^6+x^5-3x^4+x^3+5x^2+9x-18$

Calculation:

Consider the function $g\left(x\right)=x^6+x^5-3x^4+x^3+5x^2+9x-18$ .

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

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

  $\begin{array}{c}g\left(-x\right)={\left(-x\right)}^6+{\left(-x\right)}^5-3{\left(-x\right)}^4+{\left(-x\right)}^3+5{\left(-x\right)}^2+9\left(-x\right)-18\\ =x^6-x^5-3x^4-x^3+5x^2-9x-18\end{array}$

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

The possible numbers of zeros for $g\left(x\right)$ are summarized in below table:

Positive real zeros	Negative real zeros	Imaginary zeros	Total zeros
3	3	0	6
3	1	2	6
1	3	2	6
1	1	4	6