Chapter 4.8, Problem 18E

To determine

To find: all real zeros of the function.

Answer to Problem 18E

The zeros are $x=-2,x=2$ and $x=3$ .

Explanation of Solution

Given information:

Given function,

  $f\left(x\right)=x^3-3x^2-4x+12$

Calculation:

Consider the function $f\left(x\right)=x^3-3x^2-4x+12$ :

Input function using key $X\wedge 3-3X{x}^2-4X+12$ .

  

Now press 2ndGRAPH to get table:

  

Use the location principle.

From the table $f\left(-1\right)>0$ and $f\left(-3\right)<0$ . So, by location principle function has a zero between $-1$ and $-3$ .

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

The only possible rational zero between $-1$ and $-3$ is $-2$ .

For $x=-2$ ,

  $\begin{array}{ccccc}-2& 1& -3& -4& 12\\ & & -2& 10& -12\\ & 1& -5& 6& 0\end{array}$

So, one factor is $x+2$ and other is $x^2-5x+6$ .

Writing as a product of factors,

  $\begin{array}{c}f\left(x\right)=x^3-3x^2-4x+12\\ =\left(x+2\right)\left(x^2-5x+6\right)\\ =\left(x+2\right)\left(x-3\right)\left(x-2\right)\end{array}$

Therefore, the zeros are $x=-2,x=2$ and $x=3$ .