Chapter 4.8, Problem 21E

To determine

To find: all real zeros of the function.

Answer to Problem 21E

The zeros are $x=3,x=\frac{3}{4}$ and $x=-4$ .

Explanation of Solution

Given information:

Given function,

  $g\left(x\right)=4x^3+x^2-51x+36$

Calculation:

Consider the function $g\left(x\right)=4x^3+x^2-51x+36$ :

Input function using key $4X\wedge 3+X{x}^2-51X+36$ .

  

Now press 2ndGRAPH to get table:

  

Use the location principle.

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

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

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

For $x=3$ ,

  $\begin{array}{ccccc}3& 4& 1& -51& 36\\ & & 12& 39& -36\\ & 4& 13& -12& 0\end{array}$

So, one factor is $x-3$ and other is $4x^2+12x-12$ .

Writing as a product of factors,

  $\begin{array}{c}g\left(x\right)=4x^3+x^2-51x+36\\ =\left(x-3\right)\left(4x^2+13x-12\right)\\ =\left(x-3\right)\left(x+4\right)\left(4x-3\right)\end{array}$

Therefore, the zeros are $x=3,x=\frac{3}{4}$ and $x=-4$ .