Chapter 2.7, Problem 90E

a.

To determine

To find the points on the line at which the polynomial is zero.

Answer to Problem 90E

  $\boxed{a}$And $\boxed{b}$ .

Explanation of Solution

Given information:

Consider the polynomial $\left(x-a\right)\left(x-b\right)$ and the real number line shown below:

  

Identify the points on the line at which the polynomial is zero.

Calculation:

Here, we will consider the polynomial $\left(x-a\right)\left(x-b\right)$ and the real number line:

  

From the above real number line, we can observed that the polynomial $\left(x-a\right)\left(x-b\right)$ is zero at points $\boxed{a}$and $\boxed{b}$ .

Hence, $\boxed{a}$and $\boxed{b}$ are the points on the line at which the polynomial is zero.

b.

To determine

To write the sign of each factor and the sign of the product.

Answer to Problem 90E

  $\boxed{Positive}$

  $\boxed{Negative}$

  $\boxed{Positive}$

Explanation of Solution

Given information:

Consider the polynomial $\left(x-a\right)\left(x-b\right)$ and the real number line shown below:

  

In each of the three subintervals of the line, write the sign of each factor and the sign of the product.

Calculation:

From the above real number line, we can observed that the three sub intervals are $\left(-\infty ,a\right),\left(a,b\right)$ and $\left(b,\infty \right)$ , and the two factors are $\left(x-a\right)$ and $\left(x-b\right)$ .

Now, for the sub interval $\left(-\infty ,a\right),x<a$

  $\left(x-a\right)$ will always be negative.

If, $x<b$

So, $\left(x-b\right)$ will also be negative.

Hence, as both $\left(x-a\right)$ and $\left(x-b\right)$ will be negative, their product $\left(x-a\right)\left(x-b\right)$ will be $\boxed{Positive}$.

Now, similarly for the sub interval $\left(a,b\right),x>a$

  $\left(x-a\right)$ will always be positive

If, $x<b$

So, $\left(x-b\right)$ will be negative.

Hence, as $\left(x-a\right)$ is positive and $\left(x-b\right)$ is negative, their product $\left(x-a\right)\left(x-b\right)$ will be $\boxed{Negative}$ .

Now, for the sub interval $\left(b,\infty \right),x>a$

  $\left(x-a\right)$ will always be positive

If, $x>b$

So, $\left(x-b\right)$ will also be positive.

Hence, as both $\left(x-a\right)$ and $\left(x-b\right)$ will be positive, their product $\left(x-a\right)\left(x-b\right)$ will also be $\boxed{Positive}$ .

c.

To determine

To find at what values does the polynomial change signs?

Answer to Problem 90E

  $\boxed{\left(-\infty ,a\right)to\left(a,b\right)}$Or from $\boxed{\left(a,b\right)to\left(b,\infty \right)}$.

Explanation of Solution

Given information:

Consider the polynomial $\left(x-a\right)\left(x-b\right)$ and the real number line shown below:

  

At what values does the polynomial change signs?

Calculation:

The polynomial $\left(x-a\right)\left(x-b\right)$ will change sign if the value of $x$ changes from the interval $\boxed{\left(-\infty ,a\right)to\left(a,b\right)}$or from $\boxed{\left(a,b\right)to\left(b,\infty \right)}$ .