Chapter 2.6, Problem 25E

Solution

To find: all real zeros of the function.

The real zeros of the function $3,\frac{3}{2},-1$ ..

Given information: 

The function $g\left(x\right)=2x^3-7x^2+9$ .

Calculation:

Consider the function

  $g\left(x\right)=2x^3-7x^2+9$

First, let's use the Rational Zeros theorem to find the possible rational zeros of the function.

  $\begin{array}{c}\frac{p}{q}=\frac{\text{ factors of }9}{\text{ factors of }2}\\ =\frac{\pm 1,\pm 3,\pm 9}{\pm 1,\pm 2}\\ =\pm 1,\pm 3,\pm 9,\pm \frac{1}{2},\pm \frac{3}{2},\pm \frac{9}{2}\end{array}$

Let's utilize synthetic division to determine which potential rational zeros are actually the function's rational zeros. Let's begin with $1$ .

However, it doesn't work. $3$ is a zero of the function.

  $\begin{array}{l}1\overline{)\begin{array}{llll}2\hfill & -7\hfill & 0\hfill & 9\hfill \\ \hfill & 2\hfill & -5\hfill & -5\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 2\hfill & -5\hfill & -5\hfill & 4\hfill \end{array}\\ 3\overline{)\begin{array}{llll}2\hfill & -7\hfill & 0\hfill & 9\hfill \\ \hfill & 6\hfill & -3\hfill & -9\hfill \end{array}}\\ \begin{array}{lllll}\hfill & 2\hfill & -1\hfill & -3\hfill & 0\hfill \end{array}\end{array}$

Now, try synthetic division anymore since the last row (without the last $0$ ) of the synthetic division by $3$ is a quadratic now: $2x^2-x+3$ .

Choose this to $0$ and solve for $x$ using

  $\begin{array}{c}2x^2-x-3=0\\ \left(2x-3\right)\left(x+1\right)=0\\ x=\frac{3}{2}\text{ or }x=-1\end{array}$

Therefore, the zeros are $3,\frac{3}{2},-1$ .