Chapter 4.5, Problem 40E

To determine

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

Answer to Problem 40E

The real zeros are $-\frac{3}{2},-\frac{5}{2}$ , and $4$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=4x^3-49x-60$

A graph is given as

  

Calculation:

Consider the function.

  $f\left(x\right)=4x^3-49x-60$

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

Here, the leading coefficient is $4$ and constant term is $-60$ .

The possible rational zeros are

  $\pm 1,\pm 2,\pm 4,\pm 6,\pm 10,\pm 12,\pm 15,\pm 20,\pm 30,\pm 60,\pm \frac{1}{2}\text{,}\mathrm{...}\text{,and}\pm \frac{15}{2}$ .

From the graph it can be concluded that one reasonable zero is $x=4$ .

For $x=4$

  $\begin{array}{ccccc}4& 4& 0& -49& -60\\ & & 16& 64& 60\\ & 4& 16& 15& 0\end{array}$

Therefore, $x-4$ is a factor.

Factor completely using the result of above synthetic division.

  $\begin{array}{c}f\left(x\right)=4x^3-49x-60\\ =\left(x-4\right)\left(4x^2+16x-15\right)\end{array}$

Now, solve $4x^2+16x-15=0$ using quadratic formula.

  $\begin{array}{c}4x^2+16x-15=0\\ x=\frac{-16\pm \sqrt{{16}^2-4\cdot \left(-15\right)\cdot 4}}{2\cdot 4}\\ x=-\frac{3}{2},-\frac{5}{2}\end{array}$

Therefore, all real zeros are $-\frac{3}{2},-\frac{5}{2}$ , and $4$ .