Chapter 6.7, Problem 32PPS

To determine

State the possible number of real zeros, negative real zeros, and imaginary zeros of each function.

Answer to Problem 32PPS

The function could have the following:

1. 2 positive real zeros, 1 negative real zero and 2 imaginary zeros.
2. 0 positive real zeros, 1 negative real zero and 4 imaginary zeros.

Explanation of Solution

Given information:

The given equation is $f\left(x\right)=-x^5+14x^3+18x-36$ .

The given equation has degree 5, therefore, it has 5 zeros. To find the positive real zeros, count the number of changes in sign for the coefficients of $f\left(x\right)$ .

  $f\left(x\right)=-x^5+14x^3+18x-36$ Function $f\left(x\right)$

  $f\left(x\right)=-++-$ Show the sign of each term.

There are 2 sign changes, so there are 2 or 0 positive real zeros.

  $f\left(x\right)=x^5-14x^3-18x-36$ Function $f\left(-x\right)$

  $f\left(x\right)=+---$ Show the sign of each term.

There is 1 sign change, so there is 1 negative real zero.

The function could have the following:

1. 2 positive real zeros, 1 negative real zero and 2 imaginary zeros.
2. 0 positive real zeros, 1 negative real zero and 4 imaginary zeros.