Chapter EP, Problem 6.7.8EP

To determine

To calculate: The number of positive real zeroes, negative real zeroes and imaginary zeroes of the function $f\left(x\right)=3x^4-5x^3+2x^2-7x+5$ .

Answer to Problem 6.7.8EP

The number of positive real zeroes, negative real zeroes and imaginary zeroes of the function $f\left(x\right)=3x^4-5x^3+2x^2-7x+5$ are tabulated below,

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 4& 0& 0& 4+0+0=4\\ 2& 0& 2& 2+0+2=4\\ 0& 0& 4& 0+0+4=4\end{array}$

Explanation of Solution

Given information:

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

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

Observe that degree of polynomial is 4, so the functions has 4 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 $f\left(x\right)=3x^4-5x^3+2x^2-7x+5$

The coefficients are $+3,-5,+2,-7,+5$ .

There are 4 sign changes, so there are 4, 2 or 0 positive real zeros.

Now,

  $\begin{array}{c}f\left(-x\right)=3{\left(-x\right)}^4-5{\left(-x\right)}^3+2{\left(-x\right)}^2-7\left(-x\right)+5\\ f\left(-x\right)=3x^4+5x^3+2x^2+7x+5\end{array}$

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

The coefficients are $+3,+5,+2,+7,+5$ .

There is no sign change, so there are no negative real zeros.

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

  $\begin{array}{cccc}\begin{array}{c}\text{Number of Positive}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of Negative}\\ \text{Real zeros}\end{array}& \begin{array}{c}\text{Number of}\\ \text{Imaginary zeros}\end{array}& \begin{array}{c}\text{Total Number}\\ \text{of zeros}\end{array}\\ 4& 0& 0& 4+0+0=4\\ 2& 0& 2& 2+0+2=4\\ 0& 0& 4& 0+0+4=4\end{array}$