Chapter 8.8, Problem 13WE

To determine

To calculate : the real root of the given equation $P(x)=0$ and approximate it to the nearest tenth.

Answer to Problem 13WE

The real root of the given equation is $-1.5$

Explanation of Solution

Given information :

The polynomial equation:

  $P(x)=x^3+x^2+1$

Formula used :

Intermediate-Value Theorem:

If P is a polynomial function with real coefficients, and m is any number between P (a) and P ( b ), then there is at least one number c between a and b for which P ( c ) = m .

Calculation :

As per the given problem,

Consider the given equation:

  $P(x)=x^3+x^2+1$

By Descartes’s rule $P(x)=0$ has no positive roots. Make a table using negative values of x.

The table of values at the left below shows that $P(-2)<0$ and $P(-1)>0$ . Therefore, for some r between $-2$ and $-1$ , $P(r)=0$ .

  

The table of values at the right above, shows that r is between $-1.5$ and $-1.4$ . Since $P(-1.5)$ is closer to $0$ than $P(-1.4)$ , it can be assumed that r is closer to $-1.5$ . Therefore, to the nearest tenth, the real zero of P is $-1.5$