Chapter 4, Problem 21RE

To determine

To calculate: The bounds to the real zeros of polynomial function $f(x)=x^3-x^2-4x+2$ .Also, obtain a complete graph of f .

Answer to Problem 21RE

The smaller bound is $5$ .

Explanation of Solution

Given information:

The polynomial function $x^3-x^2-4x+3$ .

Formula used:

Theorem: Bounds of zeros

Let f be a polynomial function whose leading coefficient is 1.

  $f(x)=x^n+a_{n-1}{x}^{n-1}+\mathrm{.....}+a_{1}x+a_0$

A bound M on the real zeros of f is the smaller of the two numbers

  $\boxed{Max\left\{1,\left|a_0\right|+\left|a_1\right|+\mathrm{.....}+\left|a_{n-1}\right|\right\},1+Max\left\{\left|a_0\right|,\left|a_1\right|,\mathrm{...},\left|a_{n-1}\right|\right\}}$

Where $Max\left\{\right\}$ means to choose the largest entry in {}.

Calculation:

Consider ,

  $f(x)=x^3-x^2-4x+3$

The leading coefficient is 1. The degree of the polynomial function is 3.

Now,

  $\begin{array}{l}{a}_3=1\\ a_2=-1\\ a_1=-4\\ a_0=3\end{array}$

Using the theorem we get,

  $Max\left\{1,\left|a_0\right|+\left|a_1\right|+\mathrm{.....}+\left|a_{n-1}\right|\right\}=Max\left\{1,\left|a_0\right|+\left|a_1\right|+\left|a_2\right|\right\}$

  $\begin{array}{l}=Max\left\{1,\left|-3\right|+\left|-4\right|+\left|-1\right|\right\}\\ =Max\left\{1,3+4+1\right\}\\ =Max\left\{1,8\right\}\\ =8\end{array}$

  $1+Max\left\{\left|a_0\right|,\left|a_1\right|,\mathrm{.....},\left|a_{n-1}\right|\right\}=1+Max\left\{\left|a_0\right|,\left|a_1\right|,\left|a_2\right|\right\}$

  $\begin{array}{l}=1+Max\left\{\left|-3\right|,\left|-4\right|,\left|-1\right|\right\}\\ =1+Max\left\{3,4,1\right\}\\ =1+4\\ =5\end{array}$

The smaller of the two numbers , 5 is the bound. Every zero of f lies between $-5$ and $5$ .

Using the above information the graph is drawn:-