Chapter 4.8, Problem 19E

To determine

To find: all real zeros of the function.

Answer to Problem 19E

The zeros are $x=-\frac{1}{2},x=-4$ and $x=1$ ..

Explanation of Solution

Given information:

Given function,

  $h\left(x\right)=2x^3+7x^2-5x-4$

Calculation:

Consider the function $h\left(x\right)=2x^3+7x^2-5x-4$ :

Input function using key $2X\wedge 3+7X{x}^2-5X-4$ .

  

Now press 2ndGRAPH to get table:

  

Use the location principle.

From the table $h\left(-1\right)>0$ and $h\left(0\right)<0$ . So, by location principle function has a zero between 0 and $-1$ .

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

The only possible rational zero between 0 and $-1$ is $-\frac{1}{2}$ .

For $x=-\frac{1}{2}$ ,

  $\begin{array}{ccccc}-\frac{1}{2}& 2& 7& -5& -4\\ & & -1& -3& 4\\ & 2& 6& -8& 0\end{array}$

So, one factor is $x+\frac{1}{2}$ and other is $2x^2+6x-8$ .

Writing as a product of factors,

  $\begin{array}{c}h\left(x\right)=2x^3+7x^2-5x-4\\ =\left(x+\frac{1}{2}\right)\left(2x^2+6x-8\right)\\ =2\left(x+\frac{1}{2}\right)\left(x^2+3x-4\right)\\ =2\left(x+\frac{1}{2}\right)\left(x+4\right)\left(x-1\right)\end{array}$

Therefore, the zeros are $x=-\frac{1}{2},x=-4$ and $x=1$ .