Chapter 4.4, Problem 3CFU

To determine

To write: A polynomial function $f\left(x\right)$ whose coefficients have three sign changes and find the number of sign changes that $f\left(-x\right)$ has.

Answer to Problem 3CFU

This function either has:

3 or 1 positive real zeros,

0 negative real zeros,

2 or 0 imaginary zeros

Explanation of Solution

Given information: A polynomial function $f\left(x\right)$ whose coefficients have three sign changes.

Calculation:

Function that we made up

  $\begin{array}{l}f\left(x\right)=x^3-2x^2+x-2\\ f\left(x\right)=+x^3-2x^2+x-2\end{array}$

Pay attention to the signs in front of each term. Each time a sign changes from + to − or vice versa that counts as 1 change.

3 sign changes = 3 or 1 $\ast$positive real zeros

They have asked for 3 sign changes and they are:

  $\begin{array}{l}1)+x^3\text{ to }-2x^2,\\ 2)-2x^2\text{ to }+x,\text{ and }\\ \text{3) }+x\text{ to }-2\end{array}$

However many # of changes you find, there is also a possibility of two less than that number $\left(3-2=1\right)$ . But you cannot have a negative amount of possibilities.

Now do $f\left(-x\right)$

Note there are no sign changes for this function.

  $\begin{array}{l}f\left(-x\right)={\left(-x\right)}^3-2{\left(-x\right)}^2+\left(-x\right)-2\\ f\left(-x\right)=-x^3-x^2-x-2\end{array}$

0 sign change = There are 0 negative real zeros.

The degree of the function is 3, so we know there are 3 zeros possible.

These are the only possible combinations given what we just found

  $f\left(x\right)=x^3-2x^2+x-2=3\text{ zeros}$

* 3 positive real zeros, 0 negative real zero, 0 imaginary zeros.

* 1 positive real zero, 0 negative real zero, 2 imaginary zeros.

This function either has:

3 or 1 positive real zeros,

0 negative real zeros,

2 or 0 imaginary zeros