Chapter 8.8, Problem 25WE

To determine

To show : the given polynomial has a zero between $1.5$ and $1.6$ and approximate it to the nearest hundredth.

Explanation of Solution

Given information :

The polynomial:

  $P(x)=x^3-3x+1$

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 .

Proof :

As per the problem,

Consider the given polynomial:

  $P(x)=x^3-3x+1$

For $x=1.5$

  $\begin{array}{l}P(x)=x^3-3x+1\text{ [Write the polynomial]}\\ $ P(1.5)={(1.5)}^3-3(1.5)+1\text{ [Substitute }x=1.5]$P(1.5)=-0.125\end{array}$

For $x=1.6$

  $\begin{array}{l}P(x)=x^3-3x+1\text{ [Write the polynomial]}\\ $ P(1.5)={(1.5)}^3-3(1.5)+1\text{ [Substitute }x=1.5]$P(1.6)=0.296\end{array}$

As $P(1.5)<0$ and $P(1.6)>0$ , there lies at least one number between $1.5$ and $1.6$ for which $P(x)=0$

Make a table for $1.5\le x\le 1.6$

  

The table of values above, shows that the positive root is between $1.53$ and $1.54$ Since $P(1.53)$ is closer to $0$ than $P(1.54)$ , it can be assumed that root is closer to $1.53$ . Therefore, to the nearest hundredth, the positive root of P is $1.53$ .

So, the polynomial $P(x)=x^3-3x+1$ has a zero at $x=1.53$