Chapter 2.3, Problem 31E

Solution

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

For the polynomial $f\left(x\right)=44x^4-65x^3+x^2+17x+3$ , the matching graph is option $\left(c\right)$ and the function has zeros at $x=1\left(\text{multiplicity}\text{2}\right),-\frac{1}{4},\frac{3}{11}$ .

Given:$f\left(x\right)=44x^4-65x^3+x^2+17x+3$

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 $f\left(x\right)=44x^4-65x^3+x^2+17x+3$ 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 $f\left(x\right)=44x^4-65x^3+x^2+17x+3$

By Rational root test the possible zeros are

  $\begin{array}{l}\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{11},\pm \frac{1}{22},\pm \frac{1}{44}\\ \pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm \frac{3}{11},\pm \frac{3}{22},\pm \frac{3}{44}\end{array}$

Use the synthetic division method to solve the function.

  $\begin{array}{l}1\overline{)\begin{array}{l}44,-65,1,17,3\\ 0,44,-21,-20,-3\end{array}}\\ 44,-21,-20,-3,0\end{array}$

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

  $44x^4-65x^3+x^2+17x+3=\left(x-1\right)\left(44x^3-21x^2-20x-3\right)$

By Rational root test the possible zeros for $\left(44x^3-21x^2-20x-3\right)$ are

  $\begin{array}{l}\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm \frac{1}{11},\pm \frac{1}{22},\pm \frac{1}{44}\\ \pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm \frac{3}{11},\pm \frac{3}{22},\pm \frac{3}{44}\end{array}$

Use the synthetic division method to solve the function.

  $\begin{array}{l}1\overline{)\begin{array}{l}44,-21,-20,-3\\ 0,44,-23,3\end{array}}\\ 44,23,3,0\end{array}$

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

  $\begin{array}{c}44x^4-65x^3+x^2+17x+3=0\\ \left(x-1\right)\left(44x^3-21x^2-20x-3\right)=0\\ {\left(x-1\right)}^2\left(44x^2+23x+3\right)=0\\ {\left(x-1\right)}^2\left(4x+1\right)\left(11x-3\right)=0\\ x-1=0\left(\text{multiplicity}2\right)\text{or}\\ 4x+1=0\text{or}\\ 11x-3=0\end{array}$

Thus, the values of $x$ are $x=1\left(\text{multiplicity}2\right),-\frac{1}{4},\frac{3}{11}$

The graph $a$ and $b$ does not have limit $\underset{x\to -\infty }{\lim }f\left(x\right)=\infty$ and $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$ . The graph $d$ does not have zeros at $x=1\left(\text{multiplicity}\text{2}\right),-\frac{1}{4},\frac{3}{11}$ . Therefore, the graph is $c$ .

  

The function has zeros at $x=1\left(\text{multiplicity}\text{2}\right),-\frac{1}{4},\frac{3}{11}$ .