Chapter 8.4, Problem 45PPSE

Solution

a.

To calculate:The simplified expression of $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ .

The area of the rectangle is $\frac{6\left(a+1\right)}{a-3}$ square units.

Given information:

The expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ .

Formula Used:

To simplify the ration expressions, identify the restrictions of the function.

To multiply the rational expressions, look for the numerator and denominator to see if there are any common factors that can be cancelled out.

To divide the rational expressions, multiply the dividend by the reciprocal of the divisor.

Calculation:

Consider the expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ ,

Factor the numerator.

  $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}=\frac{\left(2x^n-1\right)\left(2x^n+1\right)}{2x^n-1}$

Simplify the expression.

  $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}=2x^n+1$

Thus, the simplified expression of $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ is $2x^n+1$ .

b.

To identify: The value of the expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ that is always off from the simplified expression obtained in part (a).

The value of the expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ that is always off from the simplified expression obtained in part (a) is $2x^n+1$ .

Given information:

The expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ .

Explanation:

Consider the expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ ,

From part (a), the simplified expression is $2x^n+1$ .

In the given expression, $x^n$ is always integer as $x$ is always an integer. $2x^n$ will be an even integer because any number with multiple of 2 is always even.

Now, $2x^n+1$ will be the odd integer as an even integer added with 1 is always odd.

Thus, the value of the expression $\frac{{\left(2x^n\right)}^2-1}{2x^n-1}$ that is always off from the simplified expression obtained in part (a) is $2x^n+1$ .