Chapter 6.7, Problem 27PPS

To determine

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

Answer to Problem 27PPS

  $\begin{array}{cccc}\begin{array}{l}\text{Number of Positive}\\ \text{Real Zeros}\end{array}& \begin{array}{l}\text{Number of Negative}\\ \text{Real Zeros}\end{array}& \text{Number of Imaginary Zeros}& \text{Total Number of Zeros}\\ 2& 2& 0& 4\\ 2& 0& 2& 4\\ 0& 2& 2& 4\\ 0& 0& 4& 4\end{array}$

Explanation of Solution

Given information:

The given equation is $f\left(x\right)=x^4-5x^3+2x^2+5x+7$ .

The given equation has degree 4, therefore, it has 4 zeros. To find the positive real zeros, count the number of changes in sign for the coefficients of $f\left(x\right)$ .

  $f\left(x\right)=x^4-5x^3+2x^2+5x+7$ 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.

  $f\left(x\right)=x^4+5x^3+2x^2-5x+7$ 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 negative real zeros.

The table of possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{l}\text{Number of Positive}\\ \text{Real Zeros}\end{array}& \begin{array}{l}\text{Number of Negative}\\ \text{Real Zeros}\end{array}& \text{Number of Imaginary Zeros}& \text{Total Number of Zeros}\\ 2& 2& 0& 4\\ 2& 0& 2& 4\\ 0& 2& 2& 4\\ 0& 0& 4& 4\end{array}$