Chapter 11.2, Problem 62PPE

To determine

Write two rational expressions. Find their product.

Answer to Problem 62PPE

  $\frac{4x^2-36x+81}{2x^2}·\frac{4x+16}{2x-9}=4$

Explanation of Solution

Given:

Write two rational expressions. Find their product.

Concept Used:

Rational expression is in the form $\frac{\text{P}}{\text{Q}}$ , where $\text{Q}\ne 0$ , denominator cannot be zero in any rational expression.

A rational expression is considered simplified if the numerator and denominator have no factors in common.

Step 1: Factor the numerator and the denominator.

Step 2: List restricted values.

Step 3: Cancel common factors.

Step 4: Simplify and note any restricted values not implied by the expression.

Let the two rational expressions are: $\frac{4x^2-36x+81}{2x^2}\text{and}\frac{4x+16}{2x-9}$

Calculation:

Let the two rational expressions are: $\frac{4x^2-36x+81}{2x^2}\text{and}\frac{4x+16}{2x-9}$

Now find their product

  $\begin{array}{l}\frac{4x^2-36x+81}{2x^2}\text{and}\frac{4x+16}{2x-9}\\ \\ \frac{4x^2-36x+81}{2x^2}·\frac{4x+16}{2x-9}\\ \\ \frac{(2x-9)(2x-9)}{(x+4)(2x-9)}·\frac{4(x+4)}{2x-9}\\ \\ \frac{(2x-9)}{(2x-9)}·\frac{(2x-9)}{(2x-9)}·\frac{(x+4)}{(x+4)}·\frac{4}{1}=4\end{array}$

Thus, $\frac{4x^2-36x+81}{2x^2}·\frac{4x+16}{2x-9}=4$