Chapter 2.3, Problem 32E

Solution

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

For the polynomial $f\left(x\right)=4x^4-8x^3-19x^2+23x-6$ , the matching graph is option $\left(d\right)$ and the function has zeros at $x=-2,3,\frac{1}{2}\left(\text{multiplicity}2\right)$ .

Given:$f\left(x\right)=4x^4-8x^3-19x^2+23x-6$

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 $f\left(x\right)=4x^4-8x^3-19x^2+23x-6$ 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 $f\left(x\right)=4x^4-8x^3-19x^2+23x-6$

By Rational root test the possible zeros are

  $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 2,\pm 3,\pm \frac{3}{2},\pm \frac{3}{4},\pm 6$

Use the synthetic division method to solve the function.

  $\begin{array}{l}-2\overline{)\begin{array}{l}4,-8,-19,-23,-6\\ 0,-8,32,-26,6\end{array}}\\ 4,-16,13,-3,0\end{array}$

So, the function has zero at $x=-2$ .

  $4x^4-8x^3-19x^2+23x-6=\left(x+2\right)\left(4x^3-16x^2+13x-3\right)$

By Rational root test the possible zeros for $\left(4x^3-16x^2+13x-3\right)$ are

  $\pm 1,\pm \frac{1}{2},\pm \frac{1}{4},\pm 3,\pm \frac{3}{2},\pm \frac{3}{4}$

Use the synthetic division method to solve the function.

  $\begin{array}{l}3\overline{)\begin{array}{l}4,-16,13,-3\\ 0,12,-12,3\end{array}}\\ 4,-4,1,0\end{array}$

So, the function has zero at $x=3$ .

  $\begin{array}{c}4x^4-8x^3-19x^2+23x-6=0\\ \left(x+2\right)\left(4x^3-16x^2+13x-3\right)=0\\ \left(x+2\right)\left(x-3\right)\left(4x^2-4x+1\right)=0\\ \left(x+2\right)\left(x-3\right){\left(2x-1\right)}^2=0\\ x+2=0\text{or}\\ x-3=0\text{or}\\ 2x-1=0\end{array}$

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

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 $c$ does not have zeros at $x=-2,3,\frac{1}{2}\left(\text{multiplicity}2\right)$ . Therefore, the graph is $d$ .

  

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