Chapter 4.5, Problem 39E

To determine

To find: all real zeros of the given function using the graph.

Answer to Problem 39E

The real zeros are $1,\frac{-1-\sqrt{17}}{2}$ , and $\frac{-1+\sqrt{17}}{2}$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=4x^3-20x+16$

A graph is given as

  

Calculation:

Consider the function.

  $f\left(x\right)=4x^3-20x+16$

The above function is of degree 3, so it has 3 solutions.

Here, the leading coefficient is $4$ and constant term is $16$ .

The possible rational zeros are

  $\pm 1,\pm 2,\pm 4,\pm 8,\pm 16,\pm \frac{1}{2}\text{and}\pm \frac{1}{4}$ .

From the graph it can be concluded that one reasonable zero is $x=1$ .

For $x=1$

  $\begin{array}{ccccc}1& 4& 0& -20& 16\\ & & 4& 4& -16\\ & 4& 4& -16& 0\end{array}$

Therefore, $x-1$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}f\left(x\right)=4x^3-20x+16\\ =4\left(x-1\right)\left(x^2+x-4\right)\end{array}$

Now, solve $x^2+x-4=0$ using quadratic formula.

  $\begin{array}{c}{x}^2+x-4=0\\ x=\frac{-1\pm \sqrt{1^2-4\cdot \left(-4\right)\cdot 1}}{2\cdot 1}\\ x=\frac{-1\pm \sqrt{17}}{2}\end{array}$

Therefore, all real zeros are $1,\frac{-1-\sqrt{17}}{2}$ , and $\frac{-1+\sqrt{17}}{2}$ .