Chapter 5, Problem 56E

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 $1$ positive real zeroand $2$ negative real zeros.

Given information :

Thepolynomial function is $P\left(x\right)=5x^3+7x^2-2x-1$ .

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 polynomialis $P\left(x\right)=5x^3+7x^2-2x-1$ .

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.

  $\left\{\begin{array}{c}P\left(x\right)=5x^3+7x^2-2x-1\\ \text{Signs of coefficients}=++--\\ \text{Number of sign changes}=1\end{array}\right\}$

So, there is only one positive real zero.

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.

  $\left\{\begin{array}{c}P\left(x\right)=5x^3+7x^2-2x-1\\ P\left(x\right)=-5x^3+7x^2+2x-1\\ \text{Signs of coefficients}=-++-\\ \text{Number of sign changes}=2\end{array}\right\}$

So, there are two negative real zeros.

Therefore, the given polynomial function has $1$ positive real zeroand $2$ negative real zeros.