Chapter 4.8, Problem 22E

To determine

To find: all real zeros of the function.

Answer to Problem 22E

The zeros are $x=-3,x=-\frac{1}{2}$ and $x=5$

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=2x^3-3x^2-32x-15$

Calculation:

Consider the function $f\left(x\right)=2x^3-3x^2-32x-15$ :

Input function using key $2X\wedge 3-3X{x}^2-32X-15$ .

  

Now press 2ndGRAPH to get table:

  

Use the location principle.

From the table $f\left(-4\right)<0$ and $f\left(-2\right)>0$ . So, by location principle function has a zero between $-4$ and $-2$ .

Because function is polynomial of degree 3, so it has three zeros.

The only possible rational zero between $-4$ and $-2$ is $-3$ .

For $x=-3$ ,

  $\begin{array}{ccccc}-3& 2& -3& -32& -15\\ & & -6& 27& 15\\ & 2& -9& -5& 0\end{array}$

So, one factor is $x+3$ and other is $2x^2-9x-5$ .

Writing as a product of factors,

  $\begin{array}{c}f\left(x\right)=2x^3-3x^2-32x-15\\ =\left(x+3\right)\left(2x^2-9x-5\right)\\ =\left(x+3\right)\left(x-5\right)\left(2x+1\right)\end{array}$

Therefore, the zeros are $x=-3,x=-\frac{1}{2}$ and $x=5$ .