Chapter 2.3, Problem 29E

Solution

To Match: The given polynomial function with its graph and approximate all of the real zeros of the function.

For the polynomial $f\left(x\right)=20x^3+8x^2-83x+55$ , the matching graph is option $\left(a\right)$ and the function has zeros at $x=-2.5,x=1,x=\mathrm{1.1.}$

Given:$f\left(x\right)=20x^3+8x^2-83x+55$

For any polynomial function $f\left(x\right)=a_n{x}^n+\dots +a_{1}x+a_0$ , the limits $\underset{x\to \infty }{\lim }f\left(x\right)$ and $\underset{x\to -\infty }{\lim }f\left(x\right)$ are determined by the degree $n$ of the polynomial and its leading coefficients $a_n$ .

Here the function $f\left(x\right)=20x^3+8x^2-83x+55$ is a cubic function and the leading coefficients is $a_n>0$ .

Therefore, $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$ .

Solve $f\left(x\right)=20x^3+8x^2-83x+55$

By Rational root test the possible zeros are

  $\begin{array}{l}\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{5},\pm \frac{1}{10},\pm \frac{1}{20}\\ \pm 5,\pm \frac{5}{2},\pm \frac{5}{4},\pm \frac{5}{5},\pm \frac{5}{10},\pm \frac{5}{20}\\ \pm 11,\pm \frac{11}{2},\pm \frac{11}{4},\pm \frac{11}{5},\pm \frac{11}{10},\pm \frac{11}{20}\\ \pm 55,\pm \frac{55}{2},\pm \frac{55}{4},\pm \frac{55}{5},\pm \frac{55}{10},\pm \frac{55}{20}\end{array}$

Use the synthetic division method to solve the function.

  $\begin{array}{l}1\overline{)\begin{array}{l}20,8,-83,55\\ 0,20,28,-55\end{array}}\\ 20,28,-55,0\end{array}$

So, the function has zero at $x=1$ .

  $\begin{array}{c}20x^3+8x^2-83x+55=0\\ \left(x-1\right)\left(20x^2+28x-55\right)=0\\ \left(x-1\right)\left(2x+5\right)\left(10x-11\right)=0\\ x-1=0\text{ or }\\ 2x+5=0\text{ or}\\ 10x-11=0\end{array}$

Thus, the values of $x$ are $x=1,-2.5,1.1$

The graph $b$ does not have zeros at $1,1.1,-2.5$ . The graph $c$ and $d$ does not have $\underset{x\to -\infty }{\lim }f\left(x\right)=-\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$ . Therefore, the graph is $a$ .

  

The function has zeros at $x=-2.5,x=1,x=\mathrm{1.1.}$