Chapter 1.1, Problem 62E

Solution

a)

To graph: The equation $y=\left(x+2\right)\left(x^2-4x+4\right)$ .

Given Information:

The equation $y=\left(x+2\right)\left(x^2-4x+4\right)$ .

Graph:

The graph of $y=\left(x+2\right)\left(x^2-4x+4\right)$ using graphing calculator is shown in figure 1.

  

Figure 1

b)

To confirm: Algebraically that $y=\left(x+2\right)\left(x^2-4x+4\right)$ has zeros only at $x=-2$ and $x=2$ .

Given Information:

The equation $y=\left(x+2\right)\left(x^2-4x+4\right)$ .

Concept Used:

Zeroes of any equation is obtained by equating the equation to 0 and simplifying for the values of x.

Calculation:

Equate $y=\left(x+2\right)\left(x^2-4x+4\right)$ to 0 and simplify as follows.

  $\begin{array}{c}y=0\\ \left(x+2\right)\left(x^2-4x+4\right)=0\\ \left(x+2\right){\left(x-2\right)}^2=0\end{array}$

Thus, by the zero product rule, either $x+2=0$ or ${\left(x-2\right)}^2=0$ which implies $x=-2$ and $x=2$ .

Therefore, $y=\left(x+2\right)\left(x^2-4x+4\right)$ has zeros only at $x=-2$ and $x=2$ .

c)

To graph: The equation $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ .

Given Information:

The equation $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ .

Graph:

The graph of $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ using graphing calculator is shown in figure 2.

  

Figure 2

d)

To confirm: Algebraically that $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ has zeros only at $x=-2$ .

Given Information:

The equation $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ .

Concept Used:

Zeroes of any equation is obtained by equating the equation to 0 and simplifying for the values of x.

Calculation:

Equate $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ to 0 to obtain $\left(x+2\right)\left(x^2-4x+4.01\right)=0$ .

Thus, by the zero product rule, either $x+2=0$ or $x^2-4x+4.01=0$

Now, find the discriminant of $x^2-4x+4.01=0$ using $D=b^2-4ac$ .

  $\begin{array}{c}D={\left(-4\right)}^2-4\left(1\right)\left(4.01\right)\\ =16-16.04\\ =-0.4<0\end{array}$

Since the discriminant is negative it implies no real solutions of $x^2-4x+4.01=0$ exists.

Thus, $x+2=0$ which implies $x=-2$ is the only zero of $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ .

Therefore, $y=\left(x+2\right)\left(x^2-4x+4.01\right)$ has zeros only at $x=-2$ .

e)

To graph: The equation $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ .

Given Information:

The equation $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ .

Graph:

The graph of $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ using graphing calculator is shown in figure 3.

  

Figure 3

f)

To confirm: Algebraically that $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ has three zeroes.

Given Information:

The equation $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ .

Concept Used:

Zeroes of any equation is obtained by equating the equation to 0 and simplifying for the values of x.

Calculation:

Equate $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ to 0 to obtain $\left(x+2\right)\left(x^2-4x+3.99\right)=0$ .

Thus, by the zero product rule, either $x+2=0$ or $x^2-4x+3.99=0$

So, $x+2=0$ which implies $x=-2$ .

Now, find the discriminant of $x^2-4x+3.99=0$ using $D=b^2-4ac$ .

  $\begin{array}{c}D={\left(-4\right)}^2-4\left(1\right)\left(3.99\right)\\ =16-15.96\\ =0.04>0\end{array}$

Since the discriminant is positive it implies there are two roots/zeroes of $x^2-4x+3.99=0$ .

Therefore, $y=\left(x+2\right)\left(x^2-4x+3.99\right)$ has three real zeros.