Chapter 1.4, Problem 25E

To determine

To calculate: The product of the rational expression,

  $\frac{x^2-2x-15}{x^2-9}\cdot \frac{x+3}{x-5}$

Answer to Problem 25E

The product of rational expression is $\frac{\left(x+3\right)}{\left(x-3\right)}$ .

Explanation of Solution

Given information:

The expression is given as:

  $\frac{x^2-2x-15}{x^2-9}\cdot \frac{x+3}{x-5}$

Formula used:

For the rational expression:

Fractions property for multiplying rational expression:

  $\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

Product formula: $\left(A+B\right)\left(A-B\right)=A^2-B^2$

Factoring trinomials: The factor of rational expression which contain three terms is of the from $x^2+bx+c$ ,

  $\left(x+r\right)\left(x+s\right)=x^2+\left(r+s\right)x+rs$

Choose the values of $r$ and $s$ which satisfied these equations $r+s=b$ and $rs=c$

Calculation:

Consider the,rational expression

  $\frac{x^2-2x-15}{x^2-9}\cdot \frac{x+3}{x-5}$

Use the fraction property,

  $\frac{A}{B}\cdot \frac{C}{D}=\frac{AC}{BD}$

  $\frac{\left(x^2-2x-15\right)\left(x+3\right)}{\left(x^2-9\right)\left(x-5\right)}$

Factor the first term in numeratorbyFactoring trinomials method,

  $\begin{array}{l}\frac{\left(x^2-5x+3x-15\right)\left(x+3\right)}{\left(x^2-9\right)\left(x-5\right)}\\ \frac{\left(x-5\right)\left(x+3\right)\left(x+3\right)}{\left(x^2-9\right)\left(x-5\right)}\end{array}$

Apply the product formula in denominator term,

  $\frac{\left(x-5\right)\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x-5\right)}$

Cancel common factors from the numerator and denominator,

  $\frac{\left(x+3\right)}{\left(x-3\right)}$

Thus, the product of rational expression is $\frac{\left(x+3\right)}{\left(x-3\right)}$