Chapter 6.7, Problem 2CYP

To determine

To Find: The possible number of the positive real zeros and imaginary zeros of $h\left(x\right)=2x^5+x^4+3x^3-4x^2-x+9$ .

Answer to Problem 2CYP

Thefunction has either 2 positive real roots and 2 real negative root or the function have 2 real positive root or 3 negative root and 2 imaginary roots.

Explanation of Solution

Given:

Thegiven function is $h\left(x\right)=2x^5+x^4+3x^3-4x^2-x+9$ .

Calculation:

Consider the given function is,

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

The above function has degree four so there are four roots of the function. The first step is to determine the number of positive roots of the function for this consider the change in sign in the functions are $+,+,-,-,+$ . The sign changes two times this shows that there are either 2 positive real roots or 1 real positive root.

The number of possible negative roots of the function is obtained as,

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

In the above function the sign changes only three time this shows that there is only three negative real root.

Thus, the function has either 2 positive real roots and 2 real negative root or the function have 2 real positive root or 3 negative root and 2 imaginary roots.