Chapter 2.6, Problem 42E

Solution

To match: all real zeros of the function.

  

Zeros: $1,1+\sqrt{3},1-\sqrt{3}$

Graph C.

Given information: 

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

Formula used:

The Rational Zero Theorem:

If $f\left(x\right)=a_n{x}^n+\cdots +a_{1}x+a_0$ consists of integer coefficients, then the shape of each rational zero of $f$ is as follows:

  $\frac{p}{q}=\frac{\text{ factor of constant term }{a}_0}{\text{ factor of leading coefficient }{a}_n}$

Calculation:

Consider the function

  $g\left(x\right)=x^3-3x^2+2$

The constant term is $2$ , and its factors are

  $\pm 1,\pm 2$

The leading coefficient is $1$ , and its factors are

  $\pm 1$

Therefore, all rational zeros are (write all combinations of factors of $2$ and factors of $1$ ):

  $\pm \frac{1}{1},\pm \frac{2}{1}$

Simplify all fractions:

  $\pm 1,\pm 2$

Try $1$ is a zero and uses synthetic substitution.

So, we were right! $f(1)=0$ , so, $1$ is a zero of $f\left(x\right)$ , and

  $f\left(x\right)=\left(x-1\right)\left(x^2-2x-2\right)$

(The numbers in the last row of the table for synthetic substitution are the coefficients of a quotient when dividing $f\left(x\right)$ by $y-1$ .

To find other zeros, solve the equation

  $x^2-2x-2=0$

The solution of the equation

  $a{x}^2+bx+c=0$ is

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here $a=1,b=-2$ and $c=-2$ , so

  $\begin{array}{c}x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4(1)(-2)}}{2(1)}\\ =\frac{2\pm \sqrt{4+8}}{2}\\ =\frac{2\pm \sqrt{12}}{2}\\ =\frac{2\pm 2\sqrt{3}}{2}\\ =1\pm \sqrt{3}\end{array}$

(Used

  $\begin{array}{c}\sqrt{12}=\sqrt{4\cdot 3}\\ =\sqrt{4}\cdot \sqrt{3}\\ =2\sqrt{3}\end{array}$

So, other zeros are

  $x=1+\sqrt{3}\text{ and }x=1-\sqrt{3}$

So, because of the zeros, is the graph of this function.