Chapter EP, Problem 6.7.6EP

To determine

To calculate: The number of positive real zeroes, negative real zeroes and imaginary zeroes of the function $f\left(x\right)=-2x^6-5x^5+8x^2-3x+1$ .

Answer to Problem 6.7.6EP

The number of positive real zeroes, negative real zeroes and imaginary zeroes of the function $f\left(x\right)=-2x^6-5x^5+8x^2-3x+1$ are tabulated below,

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 3& 1& 2& 3+1+2=6\\ 1& 1& 4& 1+1+4=6\end{array}$

Explanation of Solution

Given information:

The function $f\left(x\right)=-2x^6-5x^5+8x^2-3x+1$ .

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

Calculation:

Consider the function $f\left(x\right)=-2x^6-5x^5+8x^2-3x+1$ .

Observe that degree of polynomial is 6, so the functions has 6 zeros which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of $P\left(x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(x\right)=-2x^6-5x^5+8x^2-3x+1$

The coefficients are $-2,-5,+8,-3,+1$ .

There are 3 sign changes, so there are 3 or 1 positive real zeros.

Now,

  $\begin{array}{c}f\left(-x\right)=-2{\left(-x\right)}^6-5{\left(-x\right)}^5+8{\left(-x\right)}^2-3\left(-x\right)+1\\ f\left(-x\right)=-2x^6+5x^5+8x^2+3x+1\end{array}$

Descartes’ rule of signs states that consider a polynomial $P\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of $P\left(-x\right)$ or is less than this by an even number.

So, count the number of times the sign changes between the coefficients of $f\left(-x\right)=-2x^6+5x^5+8x^2+3x+1$ .

The coefficients are $-2,+5,+8,+3,+1$ .

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

Next, construct a table with possible combinations of real and imaginary zeros.

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 3& 1& 2& 3+1+2=6\\ 1& 1& 4& 1+1+4=6\end{array}$