Chapter 8.8, Problem 26WE

To determine

To show : the given polynomial has a zero between $-2.6$ and $-2.5$ and approximate it to the nearest hundredth.

Explanation of Solution

Given information :

The polynomial:

  $P(x)=x^3+x^2-3x+2$

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 .

Proof :

As per the problem,

Consider the given polynomial:

  $P(x)=x^3+x^2-3x+2$

For $x=-2.6$

  $\begin{array}{l}P(x)=x^3+x^2-3x+2\text{ [Write the polynomial]}\\ $ P(-2.6)={(-2.6)}^3+{(-2.6)}^2-3(-2.6)+2\text{ [Substitute }x=-2.6]$P(-2.6)=-1.016\end{array}$

For $x=-2.5$

  $\begin{array}{l}P(x)=x^3+x^2-3x+2\text{ [Write the polynomial]}\\ $ P(-2.5)={(-2.5)}^3+{(-2.5)}^2-3(-2.5)+2\text{ [Substitute }x=-2.5]$P(-2.5)=0.125\end{array}$

As $P(-2.6)<0$ and $P(-2.5)>0$ , there lies at least one number between $-2.6$ and $-2.5$ for which $P(x)=0$

Make a table for $-2.6\le x\le -2.5$

  

The table of values above, shows that the negative root is between $-2.52$ and $-2.51$ Since $P(-2.51)$ is closer to $0$ than $P(-2.52)$ , it can be assumed that root is closer to $-2.51$ . Therefore, to the nearest hundredth, the positive root of P is $-2.51$ .

So, the polynomial $P(x)=x^3+x^2-3x+2$ has a zero at $x=-2.51$