Chapter 5.5, Problem 31PPSE

Solution

To calculate : The possible number of positive real zeros and negative real zeros for the given polynomial function using Descartes' Rule of Signs.

The given polynomial function has $2\text{ or }0$ positive real zero and $1$ negative real zero.

Given information :

The polynomial function is $P\left(x\right)=9x^3-4x^2+10$ .

Concept used:

Descartes' Rule of Signs: It is applied to find the number of positive real zeros and negative real zeros of a polynomial functions by analysing the signs of the coefficients.

Calculation :

The given polynomial is $P\left(x\right)=9x^3-4x^2+10$ .

The number of positive real zeros is equal to the number of sign changes in consecutive coefficients of $P\left(x\right)$ or less than that by an even number.

  $\left\{\begin{array}{c}P\left(x\right)=9x^3-4x^2+10\\ \text{Signs of coefficients}=+-+\\ \text{Number of sign changes}=2\end{array}\right\}$

So, the function has two or no positive real zeros.

The number of negative real zeros is equal to the number of sign changes in consecutive coefficients of $P\left(-x\right)$ or less than that by an even number.

  $\left\{\begin{array}{c}P\left(x\right)=9x^3-4x^2+10\\ P\left(-x\right)=-9x^3-4x^2+10\\ \text{Signs of coefficients}=--+\\ \text{Number of sign changes}=1\end{array}\right\}$

So, there is only one negative real zero.

Therefore, the given polynomial function has $2\text{ or }0$ positive real zero and $1$ negative real zero.